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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% The Project Gutenberg EBook of An Introduction to Nonassociative Algebras, by 
+% R. D. Schafer                                                           %
+%                                                                         %
+% This eBook is for the use of anyone anywhere at no cost and with        %
+% almost no restrictions whatsoever.  You may copy it, give it away or    %
+% re-use it under the terms of the Project Gutenberg License included     %
+% with this eBook or online at www.gutenberg.org                          %
+%                                                                         %
+%                                                                         %
+% Title: An Introduction to Nonassociative Algebras                       %
+%                                                                         %
+% Author: R. D. Schafer                                                   %
+%                                                                         %
+% Release Date: April 24, 2008 [EBook #25156]                             %
+%                                                                         %
+% Language: English                                                       %
+%                                                                         %
+% Character set encoding: ASCII                                           %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK NONASSOCIATIVE ALGEBRAS ***   %
+%                                                                         %
+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+
+\def\ebook{25156}
+\newtoks\PGheader
+{\catcode`\#11\relax\catcode`\L\active\obeylines\obeyspaces%
+\global\PGheader={%
+The Project Gutenberg EBook of An Introduction to Nonassociative Algebras, by 
+R. D. Schafer
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: An Introduction to Nonassociative Algebras
+
+Author: R. D. Schafer
+
+Release Date: April 24, 2008 [EBook #25156]
+
+Language: English
+
+Character set encoding: ASCII
+
+*** START OF THIS PROJECT GUTENBERG EBOOK NONASSOCIATIVE ALGEBRAS ***
+}}
+\AtBeginDocument{\CreditsLine{%
+Produced by David Starner, David Wilson, Suzanne Lybarger
+and the Online Distributed Proofreading Team at
+http://www.pgdp.net
+}}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+%%                                                                       %%
+%% Packages and substitutions:                                           %%
+%%                                                                       %%
+%% memoir:   Advanced book class. Required.                              %%
+%% memhfixc: Part of memoir; needed to work with hyperref. Required.     %%
+%% amsmath:  AMS mathematics enhancements. Required.                     %%
+%% amssymb:  extra AMS mathematics symbols. Required.                    %%
+%% amsthm:   configurable theorem environments. Required.                %%
+%% hyperref: Hypertext embellishments for pdf output. Required.          %%
+%%           Driver option needs to be set explicitly.                   %%
+%% indentfirst: Standard package to indent first line following chapter/ %%
+%%              section headings. Recommended.                           %%
+%%                                                                       %%
+%%                                                                       %%
+%% Producer's Comments: A fairly straightforward text, with modern       %%
+%%                      notation and very few unusual layouts, other     %%
+%%                      than indented equation tags and fixed tabstops   %%
+%%                                                                       %%
+%% Things to Check:                                                      %%
+%%                                                                       %%
+%% hyperref driver option matches workflow: OK                           %%
+%% color driver option matches workflow (color package is called         %%
+%%    by hyperref, so may rely on color.cfg): OK                         %%
+%% Spellcheck: OK                                                        %%
+%% Smoothreading pool: No                                                %%
+%% LaCheck: OK                                                           %%
+%% Lprep/gutcheck: OK                                                    %%
+%% PDF pages:  81                                                        %%
+%% PDF page size: 499 x 709pt (b5)                                       %%
+%% PDF bookmarks: created but closed by default                          %%
+%% PDF document info: filled in                                          %%
+%% 1 overfull hbox due to lengthy inline formula (unavoidable)           %%
+%%                                                                       %%
+%% pdflatex x3                                                           %%
+%%                                                                       %%
+%% Compile History:                                                      %%
+%%                                                                       %%
+%% Apr 07: dcwilson.                                                     %%
+%%         Compiled with pdfLaTeX THREE times.                           %%
+%%         MiKTeX 2.7, Windows XP Pro                                    %%
+%%                                                                       %%
+%%                                                                       %%
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\listfiles
+
+\makeatletter
+
+\documentclass[b5paper,12pt,twoside,openany,onecolumn]{memoir}[2005/09/25]
+
+\setlrmarginsandblock{2.3cm}{2.6cm}{*}
+\setulmarginsandblock{3.1cm}{2.2cm}{*}
+\setlength{\headsep}{1cm}
+\setlength{\footskip}{0.6cm}
+\fixthelayout
+\typeoutlayout
+%
+% font stuff
+% this is the section most likely to require modification
+%
+% Courier, for the PG licence stuff
+\DeclareRobustCommand\ttfamily % Courier, for the PG licence stuff
+        {\not@math@alphabet\ttfamily\mathtt
+         \fontfamily{pcr}\fontencoding{T1}\selectfont}
+
+\setlength{\overfullrule}{\z@} % hide black boxes for PPV version!
+
+% mathematics
+\usepackage[leqno]{amsmath}[2000/07/18]
+\usepackage[psamsfonts]{amssymb}[2002/01/22]
+\usepackage{amsthm}[2004/08/06]
+
+% The original uses uppercase; we use bold instead
+\newtheoremstyle{schafer}% name
+  {}%      Space above, empty = `usual value'
+  {}%      Space below
+  {\upshape}% Body font
+  {\parindent}%         Indent amount (empty = no indent, \parindent = para indent)
+  {\bfseries}% Thm head font
+  {.}%        Punctuation after thm head
+  {.5em}%     Space after thm head: " " = normal interword space;
+        %       \newline = linebreak
+  {#1\if!#3!\else\ \DPanchor{#1:#3}\fi\thmnote{#3}}% Thm head spec and destination for hyperlinking
+\theoremstyle{schafer}
+\newtheorem*{theorem}{Theorem}% we hardcode the number in the note
+\newtheorem*{proposition}{Proposition}% we hardcode the number in the note
+\newtheorem*{lemma}{Lemma}% we hardcode the number in the note
+\newtheorem*{corollary}{Corollary}% we hardcode the number in the note
+\newtheorem*{exercise}{Exercise}% we hardcode the number in the note
+\makeatletter
+\renewenvironment{proof}[1][\proofname]{\par
+  \normalfont \topsep6\p@\@plus6\p@\relax
+  \trivlist
+  \item[\hskip\labelsep\indent
+        \itshape % obviously this differs form the original typescript, but makes proofs easier to find
+    #1\@addpunct{:}]\ignorespaces
+}{%
+  \endtrivlist
+}
+\DeclareMathOperator{\sgn}{sgn}
+\DeclareMathOperator{\diag}{diag}
+\DeclareMathOperator{\trace}{trace}
+\newcommand{\type}[1]{\mathrm{#1}}
+% to kill the fraktur, use \newcommand{\algebra}[1]{#1} instead
+\newcommand{\algebra}[1]{\mathfrak{#1}}
+% note \H and \L are already defined in such a way that redefining them is almost impossible
+% we need the normal \S, and \SS is like \H and \L
+\newcommand{\A}{\algebra{A}}
+\newcommand{\B}{\algebra{B}}
+\newcommand{\C}{\algebra{C}}
+\newcommand{\D}{\algebra{D}}
+\newcommand{\E}{\algebra{E}}
+\newcommand{\G}{\algebra{G}}
+\newcommand{\HH}{\algebra{H}}
+\newcommand{\I}{\algebra{I}}
+\newcommand{\J}{\algebra{J}}
+\newcommand{\LL}{\algebra{L}}
+\newcommand{\M}{\algebra{M}}
+\newcommand{\N}{\algebra{N}}
+\newcommand{\R}{\algebra{R}}
+\newcommand{\Ss}{\algebra{S}}
+\newcommand{\T}{\algebra{T}}
+\newcommand{\Q}{\algebra{Q}}
+\newcommand{\Z}{\algebra{Z}}
+% we only want inline equations to break at an operator where
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+\def\allowbreaks{\binoppenalty700\relax}
+%
+\medmuskip = 4mu plus 2mu minus 2mu % stop these shrinking to nothing
+\def\dotsc{\allowbreak\ldots}
+\let\dotm\cdot
+
+% for (possibly aligned) equations with tags and/or commentary
+% should be replaced with amsmath equivalent if I can find one!
+\newskip\@Centering \@Centering=0pt plus 1000pt minus 1000pt
+%   eqn number right-aligned at 2.25 normal parindent less 1 quad
+% & (skip, default 2em) LHS
+% & RHS
+% & commentary text starting at [distance from right margin, default 10em] or right-aligned if wider
+% \\
+% \begin{myalign}[optional commentary tabstop, measured from right margin](optional equation indent)
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+  \intertext@\@ifnextchar[{\my@lign}{\my@lign[10em]}}{\]\aftergroup\ignorespaces}
+\def\my@lign[#1]{\@ifnextchar({\my@l@gn[#1]}{\my@l@gn[#1](2em)}}
+\def\my@l@gn[#1](#2)#3\end{\displ@y \tabskip=\displaywidth
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+    \hbox to#1{\box0\hfil}\fi\tabskip=\z@\crcr
+  #3\crcr}\end}
+\def\Intertext#1{\ifvmode\else\\\fi\noalign{\nobreak\noindent\strut#1}}
+% for simple tags in myalign; includes hyper-destination
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+% to hyperlink to an equation: \tagref [chapter.](eqn)
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+\def\t@gref[#1](#2){(\hyperlink{eqn:#1#2}{#2})}
+
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+    \@xp\@ifempty\@xp{\transcribersNotes}%
+      {\renewcommand{\transcribersNotes}{note}}
+      {\renewcommand{\transcribersNotes}{notes}}}}
+\newcommand{\CreditsLine}[1]{\newcommand{\thePr@ductionTeam}{#1}}
+
+\def\makehalftitlepage{% the boilerplate header
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+  \pagestyle{empty}
+  \pagenumbering{Alph} % to ensure unique hyperref page anchors
+  \begin{PGboilerplate}[\tiny] % 8pt for B5
+    \PGlicencelink
+    \the\PGheader
+  \end{PGboilerplate}
+  \cleartorecto
+  \endgroup}
+
+\def\makecopyrightpage{% production credits and transcriber's notes
+  \begingroup\pagestyle{empty}
+  \null\vfil
+  \begin{center}
+    \thePr@ductionTeam
+  \end{center}
+  \vfil\vfil
+  \vbox{\Small\hsize=.75\textwidth\parindent=\z@\parskip=.5em
+  \textit{Transcriber's \transcribersNotes}\par\medskip\raggedright
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+  \newpage\endgroup}
+
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+  \setlength{\midchapskip}{\z@}
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+  \renewcommand{\thechapter}{\@Roman\c@chapter.}
+  \renewcommand{\printchaptertitle}[1]{\chaptitlefont ##1\ignorespaces}
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+{}
+\else
+\def\PG#1 #2.png#3
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+\let\PGx\PG
+\fi
+
+% PDF stuff: links, document info, etc
+% Originally coded with a PostScript workflow in mind;
+% if default driver given in hyperref.cfg is not suitable,
+% add appropriate explicit option to hyperref call
+\usepackage[final,colorlinks]{hyperref}[2003/11/30]
+% we check if the driver is the useless (for pdf) "hypertex",
+% and if so we force dvips instead and issue a warning
+\def\@tempa{hypertex}
+\ifx\@tempa\Hy@defaultdriver
+  \GenericWarning{*** }{***\MessageBreak
+   Inappropriate driver for hyperref specified: assuming dvips.\MessageBreak
+   You should amend the source code if using another driver.\MessageBreak
+   \expandafter\@gobble\@gobble}
+  \Hy@SetCatcodes\input{hdvips.def}\Hy@RestoreCatcodes
+\fi
+\usepackage{memhfixc}[2004/05/13]
+\providecommand\ebook{1wxyz}% temporary, will be overridden by WWer
+\hypersetup{pdftitle=The Project Gutenberg eBook \#\ebook: An Introduction to Nonassociative Algebras,
+  pdfsubject=NSF Institute in Algebra,
+  pdfauthor=Richard D. Schafer,
+  pdfkeywords={David Starner, Suzanne Lybarger, David Wilson,
+               Project Gutenberg Online Distributed Proofreading Team},
+  pdfstartview=Fit,
+  pdfstartpage=1,
+  pdfpagemode=UseNone,
+  pdfdisplaydoctitle,
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+  menucolor=\ifdraftdoc blue\else black\fi,
+  citecolor=\ifdraftdoc blue\else black\fi,
+  urlcolor=\ifdraftdoc magenta\else black\fi}
+
+% slight modification of hyperref's command,
+% for adding explicit bookmarks and destinations
+\newcommand\DPpdfbookmark[3][0]{\rlap{\hyper@anchorstart{#3}\hyper@anchorend
+     \Hy@writebookmark{}{#2}{#3}{#1}{toc}}}
+\newcommand\DPanchor[1]{\rlap{\hyper@anchorstart{#1}\hyper@anchorend}}
+\newcommand\DPlabel[1]{\DPanchor{#1}\label{#1}}
+% hyperref seems to use the most recent section anchor instead of page
+% anchors in a \pageref, so we have to fix that; using AtBeginDocument
+% so hyperref doesn't clobber it
+\AtBeginDocument{\def\T@pageref#1{\@safe@activestrue\expandafter\@setpgref
+  \csname r@#1\endcsname\@secondoffive{#1}\@safe@activesfalse}
+  \def\@setpgref#1#2#3{\ifx #1\relax\protect\G@refundefinedtrue\nfss@text
+    {\reset@font\bfseries ??}\@latex@warning{Reference `#3' on page \thepage
+     \space undefined}\else\expandafter\Hy@setpgref@link
+     #1\@empty\@empty\@nil{#2}\fi}
+  \def\Hy@setpgref@link#1#2#3#4#5#6\@nil#7{\begingroup\toks0={\hyper@@link
+   {#5}{page.#2}}\toks1=\expandafter{#7{#1}{#2}{#3}{#4}{#5}}\edef\x
+   {\endgroup\the\toks0{\the\toks1}}\x}}
+
+% for itemized lists without hanging indent
+\def\Itemize{\leftmargini\z@\list{}{\labelwidth\parindent
+    \itemindent2\parindent\advance\itemindent\labelsep\parsep\z@\itemsep6\p@\@plus3\p@
+    \def\makelabel##1{\rlap{\LeftBracket##1\RightBracket}\hss}%
+    \let\LeftBracket(\let\RightBracket)}}%
+\let\endItemize\endlist
+\def\IItemize{\leftmargini\z@\list{}{\labelwidth1.5\parindent\advance\labelwidth-\labelsep
+    \itemindent2.5\parindent
+    \def\makelabel##1{\rlap{\LeftBracket##1\RightBracket}\hss}%
+    \let\LeftBracket\empty\let\RightBracket\empty}}%
+\let\endIItemize\endlist
+
+% for the bibliography
+\def\bysame{\leavevmode\hbox to3em{\hrulefill}\thinspace}
+\nobibintoc
+\renewcommand{\prebibhook}{\DPpdfbookmark[0]{Bibliography}{Biblio*1}\markboth{\bibname}{\bibname}}
+\renewcommand{\bibsection}{\bigskip\small\prebibhook}
+\setbiblabel{#1.\hfill}
+\setlength\bibindent{1.8em}
+\def\@openbib@code{\advance\leftmargin\bibindent
+      \itemindent -\bibindent
+      \listparindent \itemindent}
+% to change from commas to semicolons in multiple citations
+\def\@citex[#1]#2{\leavevmode
+  \let\@citea\@empty
+  \@cite{\@for\@citeb:=#2\do
+    {\@citea\def\@citea{;\penalty\@m\ }%
+     \edef\@citeb{\expandafter\@firstofone\@citeb\@empty}%
+     \if@filesw\immediate\write\@auxout{\string\citation{\@citeb}}\fi
+     \@ifundefined{b@\@citeb}{\hbox{\reset@font\bfseries ?}%
+       \G@refundefinedtrue
+       \@latex@warning
+         {Citation `\@citeb' on page \thepage \space undefined}}%
+       {\hbox{\csname b@\@citeb\endcsname}}}}{#1}}
+
+% bits and pieces
+\emergencystretch=12pt
+\let\Small\footnotesize
+\let\SMALL\scriptsize
+\let\ThoughtBreak\bigskip
+
+\makeatother
+
+\begin{document}
+\PG--File: 001.png---\******\*******\************\******\------------------
+
+\iffalse
+
+\begin{center}
+AN INTRODUCTION TO
+
+NONASSOCIATIVE ALGEBRAS
+
+\vspace{1cm}
+
+R. D. Schafer
+
+Massachusetts Institute of Technology
+
+\vspace{4cm}
+
+An Advanced Subject-Matter Institute in Algebra  \\
+%[** Typo?---------^. The next page doesn't have the hyphen.]
+Sponsored by  \\
+The National Science Foundation
+
+\vspace{2cm}
+
+Prepared for Multilithing by  \\
+Ann Caskey
+
+
+\vspace{2cm}
+
+The Department of Mathematics  \\
+Oklahoma State University  \\
+Stillwater, Oklahoma  \\
+1961
+\end{center}
+\fi
+
+\makehalftitlepage
+\frontmatter
+
+\title{AN INTRODUCTION TO\\
+NONASSOCIATIVE ALGEBRAS}
+
+\author{R.\,D. Schafer}
+\affiliation{Massachusetts Institute of Technology}
+\subtitle{An Advanced Subject-Matter Institute in Algebra\\
+Sponsored by\\
+The National Science Foundation}
+
+\date{Stillwater, Oklahoma, 1961}
+
+\maketitle
+
+\newpage
+
+\transcribersnote{This e-text was created from scans of the multilithed
+book published by the Department of Mathematics at Oklahoma State University
+in 1961. The book was prepared for multilithing by Ann Caskey.}
+
+\transcribersnote{The original was typed rather than typeset, which somewhat
+limited the symbols available; to assist the reader we have here adopted the
+convention of denoting algebras etc by fraktur symbols, as followed by the
+author in his substantially expanded version of the work published under the
+same title by Academic Press in 1966.}
+
+\transcribersnote{\SMALL Minor corrections to punctuation and spelling and minor modifications to layout
+are documented in the \LaTeX\ source.}
+
+\makecopyrightpage
+\cleartorecto
+
+
+
+\PG--File: 002.png---\******\*******\************\******\------------------
+
+
+
+
+\null\vfil\DPpdfbookmark[0]{Preface}{Preface*1}
+These are notes for my lectures in July, 1961, at the Advanced Subject
+Matter Institute in Algebra which was held at Oklahoma State University in
+the summer of 1961.
+
+Students at the Institute were provided with reprints of my paper,
+\emph{Structure and representation of nonassociative algebras} (Bulletin of the
+American Mathematical Society, vol.~61 (1955), pp.~469--484), together with
+copies of a selective bibliography of more recent papers on non\-associative
+algebras. These notes supplement \S\S3--5 of the 1955 Bulletin \hyperlink{cite.Ref64}{article},
+bringing the statements there up to date and providing detailed proofs of
+a selected group of theorems. The proofs illustrate a number of important
+techniques used in the study of nonassociative algebras.
+
+\bigskip
+\begin{flushright}
+\textsc{R.\,D.\ Schafer}
+\end{flushright}
+
+\bigskip
+\begin{flushleft}
+\small
+Stillwater, Oklahoma  \\
+July 26, 1961
+\end{flushleft}
+
+\mainmatter
+
+\PGx--File: 003.png---\******\******\********\********\---------------------
+\chapter{Introduction} % I.
+
+By common consent a ring $\R$ is understood to be an additive abelian
+group in which a multiplication is defined, satisfying
+\begin{myalign}
+\tag{1} &&(xy)z = x(yz)           &for all $x,y,z$ in $\R$
+\Intertext{and}
+\tag{2} &&(x+y)z = xz+yz,\qquad z(x+y) = zx+zy \\ % break not in original
+  &&&for all $x,y,z$ in $\R$,
+\end{myalign}
+while an algebra $\A$ over a field $F$ is a ring which is a vector space over
+$F$ with
+\begin{myalign}
+\tag{3} &&\alpha(xy) = (\alpha x)y = x(\alpha y)  &for all $\alpha$ in $F$, $x,y$ in $\A$,\\
+\end{myalign}
+so that the multiplication in $\A$ is bilinear. Throughout these notes,
+however, the associative law \tagref(1) will fail to hold in many of the algebraic
+systems encountered. For this reason we shall use the terms ``ring'' and
+``algebra'' for more general systems than customary.
+
+We define a \emph{ring} $\R$ to be an additive abelian group with a second
+law of composition, multiplication, which satisfies the distributive
+laws \tagref(2). We define an \emph{algebra} $\A$ over a field $F$ to be a vector space
+over $F$ with a bilinear multiplication (that is, a multiplication satisfying
+\tagref(2) and \tagref(3)). We shall use the name \emph{associative ring} (or \emph{associative
+algebra}) for a ring (or algebra) in which the associative law \tagref(1) holds.
+
+In the general literature an algebra (in our sense) is commonly
+referred to as a \emph{nonassociative algebra} in order to emphasize that \tagref(1)
+is not being assumed. Use of this term does not carry the connotation
+that \tagref(1) fails to hold, but only that \tagref(1) is not assumed to hold. If \tagref(1)
+is actually not satisfied in an algebra (or ring), we say that the algebra
+(or ring) is \emph{not associative}, rather than nonassociative.
+
+As we shall see in \chaplink{2}, a number of basic concepts which are familiar
+from the study of associative algebras do not involve associativity in any
+\PG--File: 004.png---\************\******\********\******\-----------------
+way, and so may fruitfully be employed in the study of nonassociative
+algebras. For example, we say that two algebras $\A$ and $\A'$ over $F$ are
+\emph{isomorphic} in case there is a vector space isomorphism $x \leftrightarrow x'$ between
+them with
+\begin{myalign}
+\tag{4} && (xy)' = x'y'&for all $x, y$ in $\A$.
+\end{myalign}
+
+Although we shall prove some theorems concerning rings and infinite-dimensional
+algebras, we shall for the most part be concerned with finite-dimensional
+algebras. If $\A$ is an algebra of dimension $n$ over $F$, let
+$u_1, \dotsc, u_n$ be a basis for $\A$ over $F$. Then the bilinear multiplication in $\A$
+is completely determined by the $n^3$ \emph{multiplication constants} $\gamma_{ijk}$ which
+appear in the products
+\begin{myalign}
+\tag{5} &&
+  u_i u_j = \sum_{k=1}^n \gamma_{ijk} u_k,   &$\gamma_{ijk}$ in $F$.
+\end{myalign}
+We shall call the $n^2$ equations \tagref(5) a \emph{multiplication table}, and shall
+sometimes have occasion to arrange them in the familiar form of such a
+table:
+\[
+\begin{array}{c| ccccc}
+         & u_1    & \dots  & u_j     & \dots  & u_n \rule{0pt}{3ex}  \\
+\hline
+  u_1    &        &        & \vdots  \\
+  \vdots &        &        & \vdots  \\
+  u_i
+&\multicolumn{5}{c}{\dots\ \ \sum \gamma_{ijk} u_k\ \ \dots}  \\
+  \vdots &        &        & \vdots  \\
+  u_n    &        &        & \vdots
+\end{array}
+\]
+
+\PG--File: 005.png---\*******\******\********\******\----------------------
+The multiplication table for a one-dimensional algebra $\A$ over $F$ is
+given by $u_1^2 =\gamma u_1 (\gamma = \gamma_{111})$. There are two cases: $\gamma=0$ (from which
+it follows that every product $xy$ in $\A$ is $0$, so that $\A$ is called a \emph{zero
+algebra}), and $\gamma \ne 0$. In the latter case the element $e = \gamma^{-1}u_1$ serves
+as a basis for $\A$ over $F$, and in the new multiplication table we have
+$e^2 = e$. Then $\alpha \leftrightarrow \alpha e$ is an isomorphism between $F$ and this one-dimensional
+algebra $\A$. We have seen incidentally that any one-dimensional algebra is
+associative. There is considerably more variety, however, among the
+algebras which can be encountered even for such a low dimension as two.
+
+Other than associative algebras the best-known examples of algebras
+are the Lie algebras which arise in the study of Lie groups. A \emph{Lie algebra}
+$\LL$ over $F$ is an algebra over $F$ in which the multiplication is \emph{anticommutative},
+that is,
+\begin{myalign}
+\tag{6}
+&& x^2 = 0  & (implying $xy = -yx$),
+\intertext{and the \emph{Jacobi identity}}
+\tag{7}
+&& (xy)z + (yz)x + (zx)y = 0   &for all $x, y, z$ in $\LL$
+\intertext{is satisfied.  If $\A$ is any associative algebra over $F$, then the \emph{commutator}}
+\tag{8}
+&& [x,y] = xy - yx
+\Intertext{satisfies }
+\tag{6$'$}
+&& [x,x] = 0
+\Intertext{and }
+\tag{7$'$}
+&& \bigl[ [x,y],z \bigr]
++ \bigl[ [y,z],x \bigr]
++ \bigl[ [z,x],y \bigr] = 0.
+\end{myalign}
+Thus the algebra $\A^-$ obtained by defining a new multiplication \tagref(8) in the
+same vector space as $\A$ is a Lie algebra over $F$. Also any subspace of $\A$
+which is closed under commutation \tagref(8) gives a subalgebra of $\A^-$, hence a
+Lie algebra over $F$. For example, if $\A$ is the associative algebra of
+all $n \times n$ matrices, then the set $\LL$ of all skew-symmetric matrices in $\A$
+is a Lie algebra of dimension $\frac{1}{2}n(n-1)$. The Birkhoff-Witt theorem states
+\PG--File: 006.png---\********\******\********\********\-------------------
+that any Lie algebra $\LL$ is isomorphic to a subalgebra of an (infinite-dimensional)
+algebra $\A^-$ where $\A$ is associative. In the general literature
+the notation $[x,y]$ (without regard to \tagref(8)) is frequently used, instead of
+$xy$, to denote the product in an arbitrary Lie algebra.
+
+In these notes we shall not make any systematic study of Lie algebras.
+A number of such accounts exist (principally for characteristic $0$, where most
+of the known results lie). Instead we shall be concerned upon occasion with
+relationships between Lie algebras and other non\-associative algebras which
+arise through such mechanisms as the \emph{derivation algebra}. Let $\A$ be any
+algebra over $F$. By a \emph{derivation} of $\A$ is meant a linear operator $D$ on $\A$
+satisfying
+\begin{myalign}
+\tag{9}  &&(xy)D = (xD)y + x(yD) &for all $x,y$ in $\A$.
+\end{myalign}
+The set $\D(\A)$ of all derivations of $\A$ is a subspace of the associative
+algebra $\E$ of all linear operators on $\A$. Since the commutator $[D, D']$
+of two derivations $D$, $D'$ is a derivation of $\A$, $\D(\A)$ is a subalgebra of
+$\E^-$; that is, $\D(\A)$ is a Lie algebra, called the \emph{derivation algebra} of $\A$.
+
+Just as one can introduce the commutator \tagref(8) as a new product to
+obtain a Lie algebra $\A^-$ from an associative algebra $\A$, so one can
+introduce a symmetrized product
+\begin{myalign}
+\tag{10} && x * y = xy + yx
+\end{myalign}
+in an associative algebra $\A$ to obtain a new algebra over $F$ where the
+vector space operations coincide with those in $\A$ but where multiplication
+is defined by the commutative product $x * y$ in \tagref(10). If one is content
+to restrict attention to fields $F$ of characteristic not two (as we shall
+be in many places in these notes) there is a certain advantage in writing
+\begin{myalign}
+\tag{10$'$} &&   x\dotm y = \tfrac12 (xy + yx)
+\end{myalign}
+to obtain an algebra $\A^+$ from an associative algebra $\A$ by defining products
+by \tagref(10$'$) in the same vector space as $\A$. For $\A^+$ is isomorphic under the
+\PG--File: 007.png---\*******\************\********\********\--------------
+mapping $a \to \frac12 a$ to the algebra in which products are defined by \tagref(10).
+At the same time powers of any element $x$ in $\A^+$ coincide with those in $\A$:
+clearly $x \dotm x = x^2$, whence it is easy to see by induction on $n$ that
+$x \dotm x \dotm \dots \dotm x \text{ ($n$ factors)} = (x \dotm \dots \dotm x) \dotm (x \dotm \dots \dotm x) =
+x^i \dotm x^{n-i} = \frac12 (x^i x^{n-i} + x^{n-i} x^i) = x^n$.
+
+If $\A$ is associative, then the multiplication in $\A^+$ is not only
+commutative but also satisfies the identity
+\begin{myalign}
+\tag{11}  &&(x \dotm y)\dotm(x \dotm x) = x\dotm\left[y\dotm(x\dotm x)\right] &for all $x, y$ in $\A^+$.
+\intertext{A (commutative) \emph{Jordan algebra} $\J$ is an algebra over a field $F$ in which
+products are \emph{commutative}:}
+\tag{12}  &&xy = yx               &for all $x, y$ in $\J$,
+\intertext{and satisfy the \emph{Jordan identity}}
+\tag{11$'$} &&(xy)x^2 = x(yx^2)    &for all $x, y$ in $\J$.
+\end{myalign}
+Thus, if $\A$ is associative, then $\A^+$ is a Jordan algebra. So is any
+subalgebra of $\A^+$, that is, any subspace of $\A$ which is closed under the
+symmetrized product \tagref(10$'$) and in which \tagref(10$'$) is used as a new multiplication
+(for example, the set of all $n \times n$ symmetric matrices). An algebra $\J$ over
+$F$ is called a \emph{special Jordan algebra} in case $\J$ is isomorphic to a subalgebra
+of $\A^+$ for some associative $\A$. We shall see that not all Jordan algebras
+are special.
+
+Jordan algebras were introduced in the early 1930's by a physicist,
+P.~Jordan, in an attempt to generalize the formalism of quantum mechanics.
+Little appears to have resulted in this direction, but unanticipated
+relationships between these algebras and Lie groups and the foundations of
+geometry have been discovered.
+
+The study of Jordan algebras which are not special depends upon
+knowledge of a class of algebras which are more general, but in a certain
+sense only slightly more general, than associative algebras. These are
+\PG--File: 008.png---\*******\*******\********\********\-------------------
+the \emph{alternative} algebras $\A$ defined by the identities
+\begin{myalign}
+\tag{13}  &x^2y &= x(xy)          &for all $x,y$ in $\A$
+\Intertext{and}
+\tag{14}  &yx^2 &= (yx)x          &for all $x,y$ in $\A$,
+\end{myalign}
+known respectively as the \emph{left} and \emph{right alternative laws}.  Clearly any
+associative algebra is alternative. The class of $8$-dimensional \emph{Cayley
+algebras} (or \emph{Cayley-Dickson algebras}, the prototype having been discovered
+in 1845 by Cayley and later generalized by Dickson) is, as we shall see,
+an important class of alternative algebras which are not associative.
+
+To date these are the algebras (Lie, Jordan and alternative) about
+which most is known. Numerous generalizations have recently been made,
+usually by studying classes of algebras defined by weaker identities.
+We shall see in \chaplink{2} some things which can be proved about completely
+arbitrary algebras.
+\PG--File: 009.png---\*******\************\********\********\--------------
+
+
+
+
+\chapter{Arbitrary Nonassociative Algebras} % II.
+
+Let $\A$ be an algebra over a field $F$. (The reader may make the
+appropriate modifications for a ring $\R$.) The definitions of the terms
+\emph{subalgebra}, \emph{left ideal}, \emph{right ideal}, (two-sided) \emph{ideal} $\I$, \emph{homomorphism},
+\emph{kernel} of a homomorphism, \emph{residue class algebra} $\A/\I$ (\emph{difference algebra}
+$\A-\I$), \emph{anti-isomorphism}, which are familiar from a study of associative
+algebras, do not involve associativity of multiplication and are thus
+immediately applicable to algebras in general. So is the notation $\B\C$
+for the subspace of $\A$ spanned by all products $bc$ with $b$ in $\B$, $c$ in $\C$
+($\B$, $\C$ being arbitrary nonempty subsets of $\A$); here we must of course
+distinguish between $(\A\B)\C$ and $\A(\B\C)$, etc.
+
+We have the \emph{fundamental theorem of homomorphism for algebras}:
+If $\I$ is an ideal of $\A$, then $\A/\I$ is a homomorphic image of $\A$ under the
+natural homomorphism
+\begin{myalign}
+\tag{1}  &&a \to  \overline{a} = a + \I,   &$a$ in $\A$, $a + \I$ in $\A/\I$.
+\intertext{Conversely, if $\A'$ is a homomorphic image of $\A$ (under the homomorphism}
+\tag{2}  &&a \to a',   &$a$ in $\A$, $a'$ in $\A'$),
+\end{myalign}
+then $\A'$ is isomorphic to $\A/\I$ where $\I$ is the kernel of the homomorphism.
+
+If $\Ss'$ is a subalgebra (or ideal) of a homomorphic image $\A'$ of $\A$, then
+the \emph{complete inverse image} of $\Ss'$ under the homomorphism \tagref(2)---that is, the
+set $\Ss = \{ s \in \A \mid s' \in \Ss'\}$---is a subalgebra (or ideal) of $\A$ which contains
+the kernel $\I$ of \tagref(2). If a class of algebras is defined by identities (as,
+for example, Lie, Jordan or alternative algebras), then any subalgebra or
+any homomorphic image belongs to the same class.
+
+We have the customary isomorphism theorems:
+
+\begin{Itemize}
+\item[i] If $\I_1$ and $\I_2$ are ideals of $\A$ such that $\I_1$ contains $\I_2$, then
+\PG--File: 010.png---\************\************\********\********\---------
+$(\A/\I_2)/(\I_1/\I_2)$ and $\A/\I_1$ are isomorphic.
+
+\item[ii] If $\I$ is an ideal of $\A$ and $\Ss$ is a subalgebra of $\A$, then $\I\cap\Ss$
+is an ideal of $\Ss$, and $(\I+\Ss)/\I$ and $\Ss/(\I\cap\Ss)$ are isomorphic.
+\end{Itemize}
+
+Suppose that $\B$ and $\C$ are ideals of an algebra $\A$, and that as a vector
+space $\A$ is the direct sum of $\B$ and $\C$ ($\A = \B + \C$, $\B\cap\C = 0$). Then $\A$ is
+called the \emph{direct sum} $\A = \B\oplus\C$ of $\B$ and $\C$ as algebras. The vector space
+properties insure that in a direct sum $\A = \B\oplus\C$ the components $b$, $c$ of
+$a = b + c$ ($b$ in $\B$, $c$ in $\C$) are uniquely determined, and that addition and
+multiplication by scalars are performed componentwise. It is the assumption
+that $\B$ and $\C$ are ideals in $\A = \B\oplus\C$ that gives componentwise multiplication
+as well:
+\begin{myalign}(0em)
+\tag{3}  &&(b_1 + c_1)(b_2 + c_2) = b_1b_2 + c_1c_2, &$b_i$ in $\B$, $c_i$ in $\C$.
+\end{myalign}
+For $b_1c_2$ is in both $\B$ and $\C$, hence in $\B\cap\C = 0$. Similarly $c_1b_2 = 0$,
+so \tagref(3) holds, (Although $\oplus$ is commonly used to denote vector space direct
+sum, it has been reserved in these notes for direct sum of ideals; where
+appropriate the notation $\perp$ has been used for orthogonal direct sum relative
+to a symmetric bilinear form.)
+
+Given any two algebras $\B$, $\C$ over a field $F$, one can construct an
+algebra $\A$ over $F$ such that $\A$ is the direct sum $\A = \B'\oplus\C'$ of ideals $\B'$,
+$\C'$ which are isomorphic respectively to $\B$, $\C$. The construction of $\A$ is
+familiar: the elements of $\A$ are the ordered pairs $(b, c)$ with $b$ in $\B$,
+$c$ in $\C$; addition, multiplication by scalars, and multiplication are
+defined componentwise:
+\begin{myalign} % tag (4) omitted from typescript
+&(b_1, c_1) + (b_2, c_2) &= (b_1 + b_2, c_1 + c_2),\\
+\tag{4}&\alpha(b, c) &= (\alpha b, \alpha c),\\
+&(b_1, c_1)(b_2, c_2) &= (b_1c_1, b_2c_2).
+\end{myalign}
+Then $\A$ is an algebra over $F$, the sets $\B'$ of all pairs $(b, 0)$ with $b$ in $\B$
+and $\C'$ of all pairs $(0, c)$ with $c$ in $\C$ are ideals of $\A$ isomorphic respectively
+\PG--File: 011.png---\************\************\******\********\-----------
+to $\B$ and $\C$, and $\A = \B'\oplus\C'$. By the customary identification of $\B$ with
+$\B'$, $\C$ with $\C'$, we can then write $\A = \B\oplus\C$, the direct sum of $\B$ and $\C$
+as algebras.
+
+As in the case of vector spaces, the notion of direct sum extends to
+an arbitrary (indexed) set of summands. In these notes we shall have
+occasion to use only finite direct sums $\A = \B_1\oplus\B_2\oplus\dotsb\oplus\B_t$.
+Here $\A$ is the direct sum of the vector spaces $\B_i$, and multiplication in
+$\A$ is given by
+\begin{myalign}(-0.25em)
+\tag{5}&&
+  (b_1 + b_2 + \dots + b_t) (c_1 + c_2 + \dots + c_t)
+= b_1c_1 + b_2c_2 + \dots + b_tc_t
+\end{myalign}
+for $b_i$, $c_i$ in $\B_i$. The $\B_i$ are ideals of $\A$. Note that (in the case of a
+vector space direct sum) the latter statement is equivalent to the fact that
+the $\B_i$ are subalgebras of $\A$ such that
+\begin{myalign}
+\tag{6}&&
+  \B_i\B_j = 0 & for $i \ne j$.
+\end{myalign}
+
+An element $e$ (or $f$) in an algebra $\A$ over $F$ is called a \emph{left} (or \emph{right})
+\emph{identity} (sometimes \emph{unity element}) in case $ea = a$ (or $af = a$) for all $a$ in $\A$.
+If $\A$ contains both a left identity $e$ and a right identity $f$, then $e = f$
+($= ef$) is a (two-sided) \emph{identity} $1$. If $\A$ does not contain an identity
+element $1$, there is a standard construction for obtaining an algebra $\A_1$
+which does contain $1$, such that $\A_1$ contains (an isomorphic copy of) $\A$ as
+an ideal, and such that $\A_1/\A$ has dimension $1$ over $F$. We take $\A_1$ to be the
+set of all ordered pairs $(\alpha, a)$ with $\alpha$ in $F$, $a$ in $\A$; addition and
+multiplication by scalars are defined componentwise; multiplication is
+defined by
+\begin{myalign}[0em]
+\tag{7}
+  &&(\alpha, a)(\beta, b)
+= (\alpha\beta,\ \beta a + \alpha b + ab), &$\alpha, \beta$ in $F$, $a, b$ in $\A$.\\
+\end{myalign}
+Then $\A_1$ is an algebra over $F$ with identity element $1 = (1, 0)$. The set
+$\A'$ of all pairs $(0, a)$ in $\A_1$ with $a$ in $\A$ is an ideal of $\A_1$ which is
+isomorphic to $\A$. As a vector space $\A_1$ is the direct sum of $\A'$ and the
+$1$-dimensional space $F1 = \{\alpha 1 \mid \alpha \text{ in } F\}$. Identifying $\A'$ with its isomorphic
+\PG--File: 012.png---\*******\*****\********\********\---------------------
+image $\A$, we can write every element of $\A_1$ uniquely in the form $\alpha 1 + a$
+with $\alpha$ in $F$, $a$ in $\A$, in which case the multiplication \tagref(7) becomes
+\begin{myalign}
+\tag{7$'$}&& (\alpha 1 + a)(\beta 1 + b) = (\alpha\beta)1 + (\beta a + \alpha b + ab).
+\end{myalign}
+We say that we have \emph{adjoined a unity element} to $\A$ to obtain $\A_1$. (If $\A$ is
+associative, this familiar construction yields an associative algebra $\A_1$
+with $1$. A similar statement is readily verifiable for (commutative)
+Jordan algebras and for alternative algebras. It is of course not true
+for Lie algebras, since $1^2 = 1 \ne 0$.)
+
+Let $\B$ and $\A$ be algebras over a field $F$. The \emph{Kronecker product}
+$\B\otimes_F\A$ (written $\B\otimes\A$ if there is no ambiguity) is the tensor product
+$\B\otimes_F\A$ of the vector spaces $\B$, $\A$ (so that all elements are sums
+$\sum b \otimes a$, $b$ in $\B$, $a$ in $\A$, multiplication being defined by distributivity and
+\begin{myalign}[0em]
+\tag{8}  &&(b_1 \otimes a_1)(b_2 \otimes a_2) = (b_1 b_2) \otimes (a_1 a_2), &$b_i$ in $\B$, $a_i$ in $\A$.\\
+\end{myalign}
+If $\B$ contains $1$, then the set of all $1 \otimes a$ in $\B\otimes\A$ is a subalgebra of
+$\B\otimes\A$ which is isomorphic to $\A$, and which we can identify with $\A$ (similarly,
+if $\A$ contains $1$, then $\B\otimes\A$ contains $\B$ as a subalgebra). If $\B$ and $\A$ are
+finite-dimensional over $F$, then $\dim (\B\otimes\A) = (\dim\B)(\dim\A)$.
+
+We shall on numerous occasions be concerned with the case where $\B$
+is taken to be a field (an arbitrary extension $K$ of $F$). Then $K$ does contain
+$1$, so $\A_K = K \otimes_F\A$ contains $\A$ (in the sense of isomorphism) as a subalgebra
+over $F$. Moreover, $\A_K$ is readily seen to be an algebra over $K$, which is
+called the \emph{scalar extension} of $\A$ to an algebra over $K$. The properties of
+a tensor product insure that any basis for $\A$ over $F$ is a basis for $\A_K$
+over $K$. In case $\A$ is finite-dimensional over $F$, this gives an easy
+representation for the elements of $\A_K$. Let $u_1, \dotsc, u_n$ be any basis for
+$\A$ over $F$.  Then the elements of $\A_K$ are the linear combinations
+\begin{myalign}
+\tag{9} &&\sum \alpha_i u_i\quad (=\sum \alpha_i \otimes u_i), &$\alpha_i$ in $K$,
+\end{myalign}
+where the coefficients $\alpha_i$ in \tagref(9) are uniquely determined. Addition and
+\PG--File: 013.png---\*******\*****\********\********\---------------------
+multiplication by scalars are performed componentwise. For multiplication
+in $\A_K$ we use bilinearity and the multiplication table
+\begin{myalign}
+\tag{10} &&u_i u_j = \sum \gamma_{ijk}\; u_k, &$\gamma_{ijk}$ in $F$.
+\end{myalign}
+The elements of $\A$ are obtained by restricting the $\alpha_i$ in \tagref(9) to elements of
+$F$.
+
+For finite-dimensional $\A$, the scalar extension $\A_K$ ($K$ an arbitrary
+extension of $F$) may be defined in a non-invariant way (without recourse
+to tensor products) by use of a basis as above. Let $u_1, \dotsc, u_n$ be any
+basis for $\A$ over $F$; multiplication in $\A$ is given by the multiplication
+table \tagref(10). Let $\A_K$ be an $n$-dimensional algebra over $K$ with the same
+multiplication table (this is valid since the $\gamma_{ijk}$, being in $F$, are in
+$K$). What remains to be verified is that a different choice of basis for
+$\A$ over $F$ would yield an algebra isomorphic (over $K$) to this one. (A non-invariant
+definition of the Kronecker product of two finite-dimensional
+algebras $\A$, $\B$ may similarly be given.)
+
+For the classes of algebras mentioned in the \hyperlink{chapter.1}{Introduction} (Jordan
+algebras of characteristic $\ne 2$, and Lie and alternative algebras of
+arbitrary characteristic), one may verify that algebras remain in the
+same class under scalar extension---a property which is not shared by
+classes of algebras defined by more general identities (as, for example,
+in \chaplink{5}).
+
+Just as the \emph{commutator} $[x,y] = xy-yx$ measures commutativity (and
+lack of it) in an algebra $\A$, the \emph{associator}
+\begin{myalign}
+\tag{11} &&(x,y,z) = (xy)z - x(yz)
+\end{myalign}
+of any three elements may be introduced as a measure of associativity
+(and lack of it) in $\A$. Thus the definitions of alternative and Jordan
+algebras may be written as
+\begin{myalign}
+&&(x,x,y) = (y,x,x) = 0     &for all $x,y$ in $\A$
+\PGx--File: 014.png---\*******\*******\********\********\-------------------
+\Intertext{and}
+&&[x,y] = (x,y,x^2) = 0  &for all $x,y$ in $\A$.
+\end{myalign}
+Note that the associator $(x,y,z)$ is linear in each argument. One identity
+which is sometimes useful and which holds in any algebra $\A$ is
+\begin{myalign}
+\tag{12}&& a(x,y,z) + (a,x,y)z = (ax,y,z) - (a,xy,z) + (a,x,yz)\\
+&&& for all $a,x,y,z$ in $\A$. %[.** punctuation assumed: off edge of scan]
+\end{myalign}
+
+The \emph{nucleus} $\G$ of an algebra $\A$ is the set of elements $g$ in $\A$ which
+associate with every pair of elements $x,y$ in $\A$ in the sense that
+\begin{myalign}[0em]
+\tag{13}&&(g,x,y) = (x,g,y) = (x,y,g) = 0  &for all $x,y$ in $\A$.\\
+\end{myalign}
+It is easy to verify that $\G$ is an associative subalgebra of $\A$. $\G$ is a
+subspace by the linearity of the associator in each argument, and
+$\allowbreaks(g_1g_2,x,y) = g_1(g_2,x,y) + (g_1,g_2,x)y + (g_1,g_2x,y) - (g_1,g_2,xy) = 0$
+by \tagref(13), etc.
+
+The \emph{center} $\C$ of $\A$ is the set of all $c$ in $\A$ which commute and associate
+with all elements; that is, the set of all $c$ in the nucleus $\G$ with the
+additional property that
+\begin{myalign}
+\tag{14} && xc = cx  &for all $x$ in $\A$.
+\end{myalign}
+This clearly generalizes the familiar notion of the center of an associative
+algebra. Note that $\C$ is a commutative associative subalgebra of $\A$.
+
+Let $a$ be any element of an algebra $\A$ over $F$. The \emph{right multiplication}
+$R_a$ of $\A$ which is determined by $a$ is defined by
+\begin{myalign}
+\tag{15}   &&R_a : x \to xa &for all $x$ in $\A$.
+\end{myalign}
+Clearly $R_a$ is a linear operator on $\A$. Also the set $R(\A)$ of all right
+multiplications of $\A$ is a subspace of the associative algebra $\E$ of all
+linear operators on $\A$, since $a \to R_a$ is a linear mapping of $\A$ into $\E$.
+(In the familiar case of an associative algebra, $R(\A)$ is a subalgebra of
+$\E$, but this is not true in general.) Similarly the \emph{left multiplication}
+$L_a$ defined by
+\begin{myalign}
+\tag{16}  &&L_a : x \to ax    &for all $x$ in $\A$
+\end{myalign}
+\PG--File: 015.png---\*******\********\****\********\----------------------
+is a linear operator on $\A$, the mapping $a \to L_a$ is linear, and the set
+$L(\A)$ of all left multiplications of $\A$ is a subspace of $\E$.
+
+We denote by $\M(\A)$, or simply $\M$, the enveloping algebra of $R(\A) \cup L(\A)$;
+that is, the (associative) subalgebra of $\E$ generated by right and left
+multiplications of $\A$. $\M(\A)$ is the intersection of all subalgebras of $\E$
+which contain both $R(\A)$ and $L(\A)$. The elements of $\M(\A)$ are of the form
+$\sum S_1 \dotsm S_n$ where $S_i$ is either a right or left multiplication of $\A$. We
+call the associative algebra $\M = \M(\A)$ the \emph{multiplication algebra} of $\A$.
+
+It is sometimes useful to have a notation for the enveloping algebra
+of the right and left multiplications (of $\A$) which correspond to the
+elements of any subset $\B$ of $\A$; we shall write $\B^*$ for this subalgebra of $\M(\A)$.
+That is, $\B^*$ is the set of all $\sum S_1 \dotsm S_n$, where $S_i$ is either $R_{b_i}$, the
+right multiplication of $\A$ determined by $b_i$ in $\B$, or $L_{b_i}$. Clearly $\A^* = \M(\A)$,
+but note the difference between $\B^*$ and $\M(\B)$ in case $\B$ is a proper subalgebra
+of $\A$---they are associative algebras of operators on different spaces ($\A$
+and $\B$ respectively).
+
+An algebra $\A$ over $F$ is called \emph{simple} in case $0$ and $\A$ itself are the
+only ideals of $\A$, and $\A$ is not a zero algebra (equivalently, in the presence
+of the first assumption, $\A$ is not the zero algebra of dimension $1$). Since
+an ideal of $\A$ is an invariant subspace under $\M = \M(\A)$, and conversely, it
+follows that $\A$ is simple if and only if $\M \ne 0$ is an irreducible set of
+linear operators on $\A$. Since $\A^2$ ($= \A\A$) is an ideal of $\A$, we have $\A^2 = \A$
+in case $\A$ is simple.
+
+An algebra $\A$ over $F$ is a \emph{division algebra} in case $\A \ne 0$ and the
+equations
+\begin{myalign}
+\tag{17}&& ax = b,\quad ya = b &($a \ne 0$, $b$ in $\A$)
+\end{myalign}
+have unique solutions $x$, $y$ in $\A$; this is equivalent to saying that, for any
+$a \ne 0$ in $\A$, $L_a$ and $R_a$ have inverses $L_a^{-1}$ and $R_a^{-1}$. Any division algebra is
+\PG--File: 016.png---\*******\*****\********\********\---------------------
+simple. For, if $\I \ne 0$ is merely a left ideal of $\A$, there is an element
+$a \ne 0$ in $\I$ and $\A \subseteq \A a \subseteq \I$ by \tagref(17), or $\I = \A$; also clearly $\A^2 \ne 0$. (Any
+associative division algebra $\A$ has an identity $1$, since \tagref(17) implies
+that the non-zero elements form a multiplicative group. In general, a
+division algebra need not contain an identity $1$.) If $\A$ has finite
+dimension $n \ge 1$ over $F$, then $\A$ is a division algebra if and only if $\A$ is
+\emph{without zero divisors} ($x \ne 0$ and $y \ne 0$ in $\A$ imply $xy \ne 0$), inasmuch as the
+finite-dimensionality insures that $L_a$ (and similarly $R_a$), being (1--1) for
+$a \ne 0$, has an inverse.
+
+In order to make the observation that any simple ring is actually
+an algebra, so the study of simple rings reduces to that of (possibly
+infinite-dimensional) simple algebras, we take for granted that the appropriate
+definitions for rings are apparent and we digress to consider any simple
+ring $\R$. The (associative) multiplication ring $\M = \M(\R) \ne 0$ is irreducible as
+a ring of endomorphisms of $\R$. Thus by Schur's Lemma the centralizer $\C'$
+of $\M$ in the ring $\E$ of all endomorphisms of $\R$ is an associative division
+ring. Since $\M$ is generated by left and right multiplications of $\R$, $\C'$
+consists of those endomorphisms $T$ in $\E$ satisfying $R_y T = TR_y$, $L_xT = TL_x$, or
+\begin{myalign}
+\tag{18} &&(xy)T = (xT)y = x(yT)  &for all $x, y$ in $\R$.
+\end{myalign}
+Hence $S$, $T$ in $\C'$ imply $(xy)ST = \left((xS)y\right)T = (xS)(yT) = \left(x(yS)\right)T = (xT)(yS)$.
+Interchanging $S$ and $T$, we have $(xy)ST = (xy)TS$, so that $zST = zTS$ for
+all $z$ in $\R^2 = \R$. That is, $ST = TS$ for all $S$, $T$ in $\C'$; $\C'$ is a field which
+we call the \emph{multiplication centralizer} of $\R$. Now the simple ring $\R$ may be
+regarded in a natural way as an algebra over the field $\C'$. Denote $T$ in $\C'$ by
+$\alpha$, and write $\alpha x = xT$ for any $x$ in $\R$. Then $\R$ is a (left) vector space over
+$\C'$. Also \tagref(18) gives the defining relations $\alpha(xy) = (\alpha x)y = x(\alpha y)$ for an
+algebra over $\C'$. As an algebra over $\C'$ (or any subfield $F$ of $\C'$), $\R$ is
+simple since any ideal of $\R$ as an algebra is \textit{a priori} an ideal of $\R$ as a ring.
+\PG--File: 017.png---\********\*****\******\********\----------------------
+
+Moreover, $\M$ is a dense ring of linear transformations on $\R$ over $\C'$
+(Jacobson, Lectures in Abstract Algebra, vol.~II, p.~274), so we have proved
+
+\begin{theorem}[1]
+Let $\R$ be a simple ring, and $\M$ be its multiplication ring.
+Then the multiplication centralizer $\C'$ of $\M$ is a field, and $\R$ may be regarded
+as a simple algebra over any subfield $F$ of $\C'$. $\M$ is a dense ring of linear
+transformations on $\R$ over $\C'$.
+\end{theorem}
+
+Returning now to any simple algebra $\A$ over $F$, we recall that the
+multiplication algebra $\M(\A)$ is irreducible as a set of linear operators on
+the vector space $\A$ over $F$. But (Jacobson, ibid) this means that $\M(\A)$ is
+irreducible as a set of endomorphisms of the additive group of $\A$, so that
+$\A$ is a simple ring. That is, the notions of simple algebra and simple
+ring coincide, and \hyperlink{Theorem:1}{Theorem~1} may be paraphrased for algebras as
+
+\begin{theorem}[1$'$]
+Let $\A$ be a simple algebra over $F$, and $\M$ be its
+multiplication algebra. Then the multiplication centralizer $\C'$ of $\M$ is a
+field (containing $F$), and $\A$ may be regarded as a simple algebra over $\C'$.
+$\M$ is a dense ring of linear transformations on $\A$ over $\C'$.
+\end{theorem}
+
+Suppose that $\A$ has finite dimension $n$ over $F$. Then $\E$ has dimension
+$n^2$ over $F$, and its subalgebra $\C'$ has finite dimension over $F$. That is, the
+field $\C'$ is a finite extension of $F$ of degree $r = (\C':F)$ over $F$. Then
+$n = mr$, and $\A$ has dimension $m$ over $\C'$. Since $\M$ is a dense ring of linear
+transformations on (the finite-dimensional vector space) $\A$ over $\C'$, $\M$ is the
+set of \emph{all} linear operators on $\A$ over $\C'$. Hence $\C'$ is contained in $\M$ in the
+finite-dimensional case. That is, $\C'$ is the center of $\M$ and is called the
+\emph{multiplication center} of $\A$.
+
+\begin{corollary}
+Let $\A$ be a simple algebra of finite dimension over $F$, and
+$\M$ be its multiplication algebra. Then the center $\C'$ of $\M$ is a field, a
+\PG--File: 018.png---\*********\******\****\********\----------------------
+finite extension of $F$. $\A$ may be regarded as a simple algebra over $\C'$.
+$\M$ is the algebra of all linear operators on $\A$ over $\C'$.
+\end{corollary}
+
+An algebra $\A$ over $F$ is called \emph{central simple} in case $\A_K$ is simple for
+every extension $K$ of $F$. Every central simple algebra is simple (take $K=F$).
+
+We omit the proof of the fact that any simple algebra $\A$ (of arbitrary
+dimension), regarded as an algebra over its multiplication centralizer $\C'$ (so
+that $\C'=F$) is central simple. The idea of the proof is to show that, for
+any extension $K$ of $F$, the multiplication algebra $\M(\A_K)$ is a dense ring of
+linear transformations on $\A_K$ over $K$, and hence is an irreducible set of
+linear operators.
+
+\begin{theorem}[2]
+The center $\C$ of any simple algebra $\A$ over $F$ is either $0$
+or a field. In the latter case $\A$ contains $1$, the multiplication centralizer
+$\C'= \C^* =\{ R_c\mid c \in \C \}$, and $\A$ is a central simple algebra over $\C$.
+\end{theorem}
+
+\begin{proof}
+Note that $c$ is in the center of any algebra $\A$ if and only if
+$R_c=L_c$ and $[L_c, R_y]=R_cR_y-R_{cy}=R_yR_c-R_{yc}=0$ for all $y$ in $\A$ or,
+more compactly,
+\begin{myalign}
+\tag{19} &&R_c=L_c,\quad R_cR_y=R_yR_c=R_{cy}&for all $y$ in $\A$.
+\intertext{Hence \tagref(18) implies that}
+\tag{20}&&cT \text{ is in } \C &for all $c$ in $\C$, $T$ in $\C'$.
+\intertext{For \tagref(18) may be written as}
+\tag{18$'$} &&R_yT=TR_y=R_{yT} &for all $y$ in $\A$
+\intertext{or, equivalently, as}
+\tag{18$''$} &&L_xT=L_{xT}=TL_x &for all $x$ in $\A$.
+\intertext{Then \tagref(18$'$) and \tagref(18$''$) imply $R_{cT}=TR_c=TL_c=L_{cT}$, together with $R_{cT}R_y=
+R_cTR_y=R_cR_{yT}=R_{c(yT)}=R_{(cT)y}$ and $R_yR_{cT}=R_yR_cT=R_cR_yT=R_cTR_y$ ($=R_{(cT)y}$),
+That is, \tagref(20) holds. Note also that \tagref(19) implies}
+\tag{21} &&L_xR_c=R_cL_x & for all $c$ in $\C$, $x$ in $\A$.
+\end{myalign}
+
+\PG--File: 019.png---\*********\*****\****\********\-----------------------
+Since $R_{c_1}R_{c_2}=R_{c_1c_2}$ ($c_i$ in $C$) by \tagref(19), the subalgebra $\C^*$ of $\M(\A)$ is
+just $\C^* = \{R_c \mid c \in \C\}$, and the mapping $c\to R_c$ is a homomorphism of $\C$ onto
+$\C^*$. Also \tagref(19) and \tagref(21) imply that $R_c$ commutes with every element of $\M$ so
+that $\C^* \subseteq \C'$. Moreover, $\C^*$ is an ideal of the (commutative) field $\C'$ since
+\tagref(18$'$) and \tagref(20) imply that $TR_c = R_{cT}$ is in $\C^*$ for all $T$ in $\C'$, $c$ in $\C$. Hence
+either $\C^*=0$ or $\C^*=C'$.
+
+Now $\C^*=0$ implies $R_c=0$ for all $c$ in $\C$; hence $\C=0$. For, if there
+is $c \ne 0$ in $\C$, then $\I=Fc \ne 0$ is an ideal of $\A$ since $\I\A=\A\I=0$. Then
+$\I=\A$, $\A^2=0$, a contradiction.
+
+In the remaining case $\C^*=\C'$, the identity operator $1_\A$ on $\A$ is in $\C'=\C^*$.
+Hence there is an element $e$ in $\C$ such that $R_e=L_e=1_\A$, or $ae=ea=a$ for
+all $a$ in $\A$; $\A$ has a unity element $1=e$. Then $c \to R_c$ is an isomorphism
+between $\C$ and the field $\C'$. $\A$ is an algebra over the field $\C$, and as such
+is central simple.
+\end{proof}
+
+For any algebra $\A$ over $F$, one obtains a \emph{derived series} of subalgebras
+$\A^{(1)} \supseteq \A^{(2)} \supseteq \A^{(3)} \supseteq \dotsb $ by defining
+ $\A^{(1)}=\A$, $\A^{(i+1)}=(\A^{(i)})^2$. $\A$ is
+called \emph{solvable} in case $\A^{(r)}=0$ for some integer $r$.
+
+\begin{proposition}[1]
+If an algebra $\A$ contains a solvable ideal $\I$, and if
+$\overline \A=\A/\I$ is solvable, then $\A$ is solvable.
+\end{proposition}
+
+\begin{proof}
+Since \tagref(1) is a homomorphism, it follows that $\overline{\A^2}={\overline \A}\vphantom{\A}^2$ and that
+$\overline{\A^{(i)}}={\overline \A}\vphantom{\A}^{(i)}$. Then ${\overline \A}\vphantom{\A}^{(r)}=0$ implies
+ $\overline {\A^{(r)}}=0$, or $\A^{(r)} \subseteq \I $. But $\I^{(s)}=0$
+for some $s$, so $\A^{(r+s)}=( \A^{(r)})^{(s)} \subseteq \I^{(s)}=0$. Hence $\A$ is solvable.
+\end{proof}
+
+\begin{proposition}[2]
+If $\B$ and $\C$ are solvable ideals of an algebra $\A$, then $\B+\C$
+is a solvable ideal of $\A$. Hence, if $\A$ is finite-dimensional, $\A$ has a unique
+maximal solvable ideal $\N$. Moreover, the only solvable ideal of $\A/\N$ is $0$.
+\PG--File: 020.png---\*************\*****\********\******\-----------------
+\end{proposition}
+
+\begin{proof} $\B + \C$ is an ideal because $\B$ and $\C$ are ideals. By the second
+isomorphism theorem $(\B+\C)/\C\cong \B/(\B\cap \C)$. But $\B/(\B\cap \C)$ is a homomorphic
+image of the solvable algebra $\B$, and is therefore clearly solvable. Then
+$\B + \C$ is solvable by \hyperlink{Proposition:1}{Proposition~1}. It follows that, if $\A$ is finite-dimensional,
+the solvable ideal of maximum dimension is unique (and contains every solvable
+ideal of $\A$). Let $\N$ be this maximal solvable ideal, and $\overline\G$ be any solvable
+ideal of $\overline \A = \A/\N$. The complete inverse image $\G$ of $\overline\G$ under the natural
+homomorphism of $\A$ onto $\overline\A$ is an ideal of $\A$ such that $\G/\N = \overline\G$. Then $\G$ is
+solvable by \hyperlink{Proposition:1}{Proposition~1}, so $\G\subseteq\N$. Hence $\G/\N = \overline\G = 0$.
+\end{proof}
+
+An algebra $\A$ is called \emph{nilpotent} in case there exists an integer $t$
+such that any product $z_1z_2\dotsm z_t$ of $t$ elements in $\A$, no matter how associated,
+is $0$. This clearly generalizes the concept of nilpotence as defined for
+associative algebras. Also any nilpotent algebra is solvable.
+
+\begin{theorem}[3] An ideal $\B$ of an algebra $\A$ is nilpotent if and only if the
+(associative) subalgebra $\B^*$ of $\M(\A)$ is nilpotent.
+\end{theorem}
+
+\begin{proof} Suppose that every product of $t$ elements of $\B$, no matter how
+associated, is $0$. Then the same is true for any product of more than $t$
+elements of $\B$. Let $T = T_1\dotsm T_t$ be any product of $t$ elements of $\B^*$. Then
+$T$ is a sum of terms each of which is a product of at least $t$ linear operators
+$S_i$, each $S_i$ being either $L_{b_i}$ or $R_{b_i}$ ($b_i$ in $\B$). Since $\B$ is an ideal of $\A$,
+$xS_1$ is in $\B$ for every $x$ in $\A$. Hence $xT$ is a sum of terms, each of which is
+a product of at least $t$ elements in $\B$. Hence $xT = 0$ for all $x$ in $\A$, or $T = 0$,
+$\B^*$ is nilpotent. For the converse we need only that $\B$ is a subalgebra of $\A$.
+We show by induction on $n$ that any product of at least $2^n$ elements in $\B$, no
+matter how associated, is of the form $bS_1 \dotsm S_n$ with $b$ in $\B$, $S_i$ in $\B^*$.
+For $n = 1$, we take any product of at least $2$ elements in $\B$. There is a
+final multiplication which is performed. Since $\B$ is a subalgebra, each
+\PG--File: 021.png---\*******\*****\********\******\-----------------------
+of the two factors is in $\B$: $bb_1 = bR_{b_1} = bS_1$.  Similarly in any product of
+at least $2^{n+1}$ elements of $\B$, no matter how associated, there is a final
+multiplication which is performed. At least one of the two factors is a
+product of at least $2^n$ elements of $\B$, while the other factor $b'$ is in $\B$.
+Hence by the assumption of the induction we have either $b'(bS_1\dotsm S_n) =
+bS_1\dotsm S_nL_{b'} = bS_1\dotsm S_{n+1}$ or $(bS_1\dotsm S_n)b' = bS_1\dotsm S_nR_{b'} = bS_1\dotsm S_{n+1}$, as
+desired. Hence, if any product $S_1\dotsm S_t$ of $t$ elements in $\B^*$ is $0$, any
+product of $2^t$ elements of $\B$, no matter how associated, is $0$. That is, $\B$
+is nilpotent.
+\PG--File: 022.png---\*******\*****\********\******\-----------------------
+\end{proof}
+
+
+
+
+\chapter{Alternative Algebras} % III.
+
+
+As indicated in the \hyperlink{chapter.1}{Introduction}, an \emph{alternative algebra} $\A$ over $F$ is
+an algebra in which
+\begin{myalign}
+\tag{1} &  x^2 y &= x(xy)  &for all $x, y$ in $\A$
+\Intertext{and}
+\tag{2} &  yx^2 &= (yx)x   &for all $x, y$ in $\A$.
+\end{myalign}
+In terms of associators, \tagref(1) and \tagref(2) are equivalent to
+\begin{myalign}
+\tag{1$'$} &  (x,x,y) &= 0   &for all $x, y$ in $\A$
+\Intertext{and}
+\tag{2$'$} &  (y,x,x) &= 0   &for all $x, y$ in $\A$.
+\end{myalign}
+In terms of left and right multiplications, \tagref(1) and \tagref(2) are equivalent to
+\begin{myalign}
+\tag{1$''$} &  L_{x^2} &= {L_x}^2    &for all $x$ in $\A$
+\Intertext{and}
+\tag{2$''$} &  R_{x^2} &= {R_x}^2    &for all $x$ in $\A$.
+\end{myalign}
+
+The associator $(x_1,x_2,x_3)$ ``alternates'' in the sense that, for any
+permutation $\sigma$ of $1, 2, 3$, we have $(x_{1\sigma},x_{2\sigma},x_{3\sigma}) = (\sgn \sigma)(x_1,x_2,x_3)$. To
+establish this, it is sufficient to prove
+\begin{myalign}
+\tag{3} &  (x,y,z) &= -(y,x,z)  &for all $x, y, z$ in $\A$
+\Intertext{and}
+\tag{4} &  (x,y,z) &= (z,x,y)   &for all $x, y, z$ in $\A$.
+\end{myalign}
+Now \tagref(1$'$) implies that $\allowbreaks(x+y,x+y,z) = (x,x,z) + (x,y,z) + (y,x,z) +
+(y,y,z) = (x,y,z) + (y,x,z) = 0$, implying \tagref(3). Similarly \tagref(2$'$) implies
+$(x,y,z) = -(x,z,y)$ which gives $(x,z,y) = (y,x,z)$. Interchanging $y$ and $z$,
+we have \tagref(4). The fact that the associator alternates is equivalent to
+\PG--File: 023.png---\*******\*****\********\*******\----------------------
+\begin{myalign}
+\tag{5}&&{\begin{aligned}
+  R_xR_y - R_{xy} &= L_{xy} - L_yL_x = L_yR_x - R_xL_y = \\
+     L_xL_y - L_{yx} &= R_yL_x - L_xR_y = R_{yx} - R_yR_x
+\end{aligned}}
+\end{myalign}
+for all $x, y$ in $\A$. It follows from \tagref(1$''$), \tagref(2$''$) and \tagref(5) that any scalar
+extension $\A_K$ of an alternative algebra $\A$ is alternative.
+
+Now \tagref(3) and \tagref(2$'$) imply
+\begin{myalign}
+\tag{6} &&  (x,y,x) = 0       &for all $x, y$ in $\A$;
+\Intertext{that is,}
+\tag{6$'$} && (xy)x = x(yx)   &for all $x, y$ in $\A$,
+\Intertext{or}
+\tag{6$''$} && L_xR_x = R_xL_x   &for all $x$ in $\A$.
+\end{myalign}
+Identity \tagref(6$'$) is called the \emph{flexible} law. All of the algebras mentioned in
+the \hyperlink{chapter.1}{Introduction} (Lie, Jordan and alternative) are flexible. The linearized
+form of the flexible law is
+\begin{myalign}
+\tag{6$'''$} && (x,y,z) + (z,y,x) = 0 &for all $x, y, z$ in $\A$.
+\end{myalign}
+
+We shall have occasion to use the Moufang identities
+\begin{myalign}
+\tag{7} &&     (xax)y = x\left[a(xy)\right],\\
+\tag{8} &&     y(xax) = \left[(yx)a\right]x,\\
+\tag{9} &&     (xy)(ax) = x(ya)x
+\end{myalign}
+for all $x, y, a$ in an alternative algebra $\A$ (where we may write $xax$ unambiguously
+by \tagref(6$'$)). Now $\allowbreaks(xax)y - x\left[a(xy)\right] = (xa,x,y) + (x,a,xy) = (-x,xa,y) - (x,xy,a) =
+-\left[x(xa)\right]y + x\left[(xa)y\right] - \left[x(xy)\right]a + x\left[(xy)a\right] = -(x^2a)y - (x^2y)a + x\left[(xa)y +
+(xy)a\right] = -(x^2,a,y) - (x^2,y,a) - x^2(ay) - x^2(ya) + x\left[(xa)y + (xy)a\right] = x\left[-x(ay)\right. -
+x(ya) + (xa)y + \left.(xy)a\right] = x\left[(x,a,y) + (x,y,a)\right] = 0$, establishing \tagref(7). Identity
+\tagref(8) is the reciprocal relationship (obtained by passing to the anti-isomorphic
+algebra, which is alternative since the defining identities are reciprocal).
+Finally \tagref(7) implies $\allowbreaks(xy)(ax)-x(ya)x = (x,y,ax) + x\left[y(ax)-(ya)x\right] =
+-(x,ax,y)-x(y,a,x) = -(xax)y + x\left[(ax)y\right.-\left.(y,a,x)\right] = -x\left[a(xy)\right.-(ax)y +
+\PGx--File: 024.png---\*******\*****\********\*******\----------------------
+\left.(y,a,x)\right] = -x\left[-(a,x,y) + (y,a,x)\right] = 0$, or \tagref(9) holds.
+
+\begin{theorem}[of Artin]
+The subalgebra generated by any two elements $x,y$
+of an alternative algebra $\A$ is associative.
+\end{theorem}
+
+\begin{proof}
+Define powers of a single element $x$ recursively by $x^1 = x$,
+$x^{i+1} = xx^i$. Show first that the subalgebra $F[x]$ generated by a single
+element $x$ is associative by proving
+\begin{myalign}
+\tag{10} &  x^ix^j &= x^{i+j}    &for all $x$ in $\A$ ($i, j = 1,2,3,\dots$).\\
+\intertext{We prove this by induction on $i$, but shall require the case $j = 1$:}
+\tag{11} &x^ix &= xx^i     &for all $x$ in $\A$ ($i = 1,2,\dots$).\\
+\end{myalign}
+Proving \tagref(11) by induction, we have $x^{i+1}x = (xx^i)x = x(x^ix) = x(xx^i) = xx^{i+1}$
+by flexibility and the assumption of the induction. We have \tagref(10) for $i = 1,2$
+by definition and \tagref(1). Assuming \tagref(10) for $i \geq 2$, we have $x^{i+1}x^j = (xx^i)x^j =
+\left[x(xx^{i-1})\right]x^j = \left[x(x^{i-1}x)\right]x^j = x\left[x^{i-1}(xx^j)\right] = x(x^{i-1}x^{j+1}) = xx^{i+j} = x^{i+j+1}$ by
+\tagref(11), \tagref(7) and the assumption of the induction. Hence $F[x]$ is associative.
+
+Next we prove that
+\begin{myalign}
+\tag{12}&   x^i(x^jy) &= x^{i+j}y   &for all $x,y$ in $\A$ ($i,j = 1,2,3,\dots$).\\
+\intertext{First we prove the case $j = 1$:}
+\tag{13}&   x^i(xy) &= x^{i+1}y    &for all $x,y$ in $\A$ ($i = 1,2,3,\dots$).\\
+\end{myalign}
+
+The case $i = 1$ of \tagref(13) is given by \tagref(1); the case $i = 2$ is $x^2(xy) = x\left[x(xy)\right] =
+(xxx)y = x^3y$ by \tagref(1) and \tagref(7). Then for $i \ge 2$, write the assumption \tagref(13) of the
+induction with $xy$ for $y$ and $i$ for $i + 1$: $x^{i-1}\left[x(xy)\right] = x^i(xy)$. Then
+$x^{i+1}(xy) = (xx^{i-1}x)(xy) = x\left[x^{i-1}\left\{x(xy)\right\}\right] = x\left[x^i(xy)\right] = (xx^ix)y = x^{i+2}y$ by
+\tagref(7). We have proved the case $j = 1$ of \tagref(12). Then with $xy$ written for $y$ in
+\tagref(12), the assumption of the induction is $x^{i+j}(xy) = x^i\left[x^j(xy)\right]$. It follows that
+$x^i(x^{j+1}y) = x^i\left[x^j(xy)\right] = x^{i+j}(xy) = x^{i+j+1}y$ by \tagref(13). Now \tagref(12) holds identically
+in y. Hence
+\begin{myalign}
+\tag{14}&&
+  x^i(x^jy^k) = (x^ix^j)y^k.
+\PGx--File: 025.png---\*************\********\********\*******\-------------
+\intertext{Reciprocally}
+\tag{15}&& (y^kx^j)x^i = y^k(x^jx^i).
+\intertext{Since the distributive law holds in $\A$, it is sufficient now to show that}
+\tag{16}&& (x^iy^k)x^j = x^i(y^kx^j)
+\end{myalign}
+in order to show that the subalgebra generated by $x,y$ is associative. But
+\tagref(14) implies $(x^i,y^k,x^j)=-(x^i,x^j,y^k)=0$.
+\end{proof}
+
+An algebra $\A$ over $F$ is called \emph{power-associative} in case the subalgebra
+$F[x]$ of $\A$ generated by any element $x$ in $\A$ is associative. Any alternative
+algebra is power-associative; the \hyperlink{Theorem:of Artin}{Theorem} of Artin also implies
+\begin{myalign}
+\tag{17} && {R_x}^j=R_{x^j},\qquad {L_x}^j=L_{x^j} &for all $x$ in $\A$.
+\end{myalign}
+
+An element $x$ in a power-associative algebra $\A$ is called \emph{nilpotent} in
+case there is an integer $r$ such that $x^r = 0$. An algebra (ideal) consisting
+only of nilpotent elements is called a \emph{nilalgebra} (\emph{nilideal}).
+
+\begin{theorem}[4]
+Any alternative nilalgebra $\A$ of finite dimension over $F$
+is nilpotent.
+\end{theorem}
+
+\begin{proof}
+Any subalgebra $\B$ of $\A$ is generated by a finite number of elements
+(for example, the elements in a basis for $\B$ over $F$). We prove by induction
+on the number of generators of $\B$ that $\B^*$ is nilpotent for all subalgebras $\B$;
+hence, in particular, for $\B = \A$. If $\B$ is generated by one element $x$, then
+by \tagref(6$''$) and \tagref(17) any $T$ in $\B^*$ is a linear combination of operators of the
+form
+\begin{myalign}
+\tag{18} &&{R_x}^{j_1},~ {L_x}^{j_2},~ {R_x}^{j_3}{L_x}^{j_4} &for $j_i \ge 1$.
+\end{myalign}
+Then, if $x^j = 0$, we have $T^{2j-1} = 0$,  $\B^*$ is nilpotent. Hence, by the
+assumption of the induction, we may take a maximal proper subalgebra $\B$ of
+$\A$ and know that $\B^*$ is nilpotent. But then there exists an element $x$ not in
+$\B$ such that
+\PG--File: 026.png---\*************\********\****\********\----------------
+\begin{myalign}
+\tag{19}&& x\B^* \subseteq\B.
+\end{myalign}
+For $\B^{*r} = 0$ implies that $\A\B^{*r}=0 \subseteq\B$, and there exists a smallest integer
+$m \ge 1$ such that $\A\B^{*m} \subseteq\B$. If $m = 1$, take $x$ in $\A$ but not in $\B$; if $m > 1$, take
+$x$ in $\A\B^{*m-1}$ but not in $\B$. Then \tagref(19) is satisfied. Since $\B$ is maximal, the
+subalgebra generated by $\B$ and $x$ is $\A$ itself. It follows from \tagref(19) that
+$\A = \B + F[x]$ so that $\M = \A^* = (\B + Fx)^*$. Put $y = b$ in \tagref(5) for any $b$ in $\B$.
+Then \tagref(19) implies that
+\begin{myalign}
+\tag{20}&&
+{\begin{aligned}
+R_xR_b &= R_{b_1}-R_bR_x,\quad R_xL_b=L_bR_x+R_bR_x-R_{b_2},\\
+L_xR_b &= R_bL_x+L_bL_x-L_{b_3},\quad L_xL_b=L_{b_1}-L_bL_x
+\end{aligned}}
+\end{myalign}
+for $b_i$ in $\B$. Equations \tagref(20) show that, in each product of right and left
+multiplications in $\B^*$ and $(Fx)^*$, the multiplication $R_x$ or $L_x$ may be
+systematically passed from the left to the right of $R_b$ or $L_b$ in a fashion
+which, although it may change signs and introduce new terms, preserves the
+number of factors from $\B^*$ and does not increase the number of factors from
+$(Fx)^*$. Hence any $T$ in $\A^*= (\B + Fx)^*$ may be written as a linear combination
+of terms of the form \tagref(18) and others of the form
+\begin{myalign}
+&&B_1,\quad B_2{R_x}^{m_1},\quad B_3{L_x}^{m_2},\quad B_4{R_x}^{m_3}{L_x}^{m_4}
+\end{myalign}
+for $B_i$ in $\B^*$, $m_i \ge 1$. Then if $\B^{*r} = 0$ and $x^j = 0$, we have $T^{r(2j-1)}=0$;
+for every term in the expansion of $T^{r(2j-1)}$ contains either an uninterrupted
+sequence of at least $2j-1$ factors from $(Fx)^*$ or at least $r$ factors $B_i$. In the
+latter case the $R_x$ or $L_x$ may be systematically passed from the left to the
+right of $B_i$ (as above) preserving the number of factors from $\B^*$, resulting
+in a sum of terms each containing a product $B_1 B_2 \dotsm B_r = 0$. Hence every
+element $T$ of the finite-dimensional associative algebra $\A^*$ is nilpotent.
+Hence $\A^*$ is nilpotent (Albert, Structure of Algebras, p.~23). Hence $\A$ is
+nilpotent by \hyperlink{Theorem:3}{Theorem~3}.
+\PG--File: 027.png---\********\******\****\********\-----------------------
+\end{proof}
+
+Any nilpotent algebra is solvable, and any solvable (power-assoc\-iative)
+algebra is a nilalgebra. By \hyperlink{Theorem:4}{Theorem~4} the concepts of nilpotent algebra,
+solvable algebra, and nilalgebra coincide for finite-dimensional alternative
+algebras. Hence there is a unique maximal nilpotent ideal $\N$ ($=$ solvable
+ideal $=$ nilideal) in any finite-dimensional alternative algebra $\A$; we call
+$\N$ the \emph{radical} of $\A$. We have seen that the radical of $\A/\N$ is $0$.
+
+We say that $\A$ is \emph{semisimple} in case the radical of $\A$ is $0$, and omit the
+proof that any finite-dimensional semisimple alternative algebra $\A$ is the
+direct sum $\A = \Ss_1 \oplus \dotsb \oplus \Ss_t$ of simple algebras $\Ss_i$. The proof is dependent
+upon the properties of the \emph{Peirce decomposition} relative to an idempotent $e$.
+
+An element $e$ of an (arbitrary) algebra $\A$ is called an \emph{idempotent} in
+case $e^2 = e \ne 0$.
+
+\begin{proposition}[3]
+Any finite-dimensional power-associative algebra, which
+is not a nilalgebra, contains an idempotent $e$ ($\ne 0$).
+\end{proposition}
+
+\begin{proof}
+$\A$ contains an element $x$ which is not nilpotent. The subalgebra
+$F[x]$ of $\A$ generated by $x$ is a finite-dimensional associative algebra which
+is not a nilalgebra. Then $F[x]$ contains an idempotent $e$ ($\ne 0$) (Albert, ibid),
+and therefore $\A$ does.
+\end{proof}
+
+By \tagref(1$''$) and \tagref(2$''$) $L_e$ and $R_e$ are idempotent operators on $\A$ which commute by
+\tagref(6$''$) (``commuting projections''). It follows that $\A$ is the vector space direct
+sum
+\begin{myalign}[6em]
+\tag{21}   &&\A = \A_{11} + \A_{10} + \A_{01} + \A_{00}
+\intertext{where $\A_{ij}$ ($i, j = 0, 1$) is the subspace of $\A$ defined by}
+\tag{22}  &&\A_{ij} = \{x_{ij} \mid ex_{ij} = ix_{ij},\; x_{ij} e = jx_{ij}\} &$i,j = 0, 1$.
+\end{myalign}
+Just as in the case of associative algebras, the decomposition of any element
+$x$ in $\A$ according to the \emph{Peirce decomposition} \tagref(21) is
+\begin{myalign}(0em)
+\tag{23}&&  x = exe + (ex - exe) + (xe - exe) + (x - ex - xe + exe).
+\end{myalign}
+\PG--File: 028.png---\********\********\****\********\---------------------
+We derive a few of the properties of the Peirce decomposition as follows:
+\begin{myalign}
+& (x_{ij}y_{ji})e &= (x_{ij},y_{ji},e ) + x_{ij}(y_{ji}e )\\
+&                 &= - (x_{ij},e,y_{ji} ) + x_{ij}(y_{ji}e )\\
+&                 &= - jx_{ij}y_{ji} + jx_{ij}y_{ji} + ix_{ij}y_{ji}\\
+&                 &= ix_{ij}y_{ji}
+\end{myalign}
+and similarly $e (x_{ij}y_{ji}) = ix_{ij}y_{ji}$, so
+\begin{myalign}
+\tag{24}     &&\A_{ij}\A_{ji} \subseteq \A_{ii},   &     $i,j=0,1$.
+\end{myalign}
+That is, $\A_{11}$ and $\A_{00}$ are subalgebras of $\A$, while $\A_{10}\A_{01} \subseteq\A_{11}$,
+ $\A_{01}\A_{10} \subseteq \A_{00}$.
+Also $x_{11}y_{00} = (ex_{11}e)y_{00} = e\left[x_{11}(ey_{00})\right] = 0$ by \tagref(7), and similarly $y_{00}x_{11} = 0$.
+Hence $\A_{11}$ and $\A_{00}$ are orthogonal subalgebras of $\A$.  Similarly $\A_{ii}\A_{ij} \subseteq \A_{ij}$,
+$\A_{ij}\A_{jj} \subseteq \A_{ij}$, etc.
+
+We wish to define the class of Cayley algebras mentioned in the
+\hyperlink{chapter.1}{Introduction}. We construct these algebras in the following manner. The
+procedure works slightly more smoothly if we assume that $F$ has characteristic
+$\ne 2$, so we make this restriction here although it is not necessary.
+
+An algebra $\A$ with $1$ over $F$ is called a \emph{quadratic algebra} in case
+$\A \ne F1$ and for each $x$ in $\A$ we have
+\begin{myalign}
+\tag{25}   &&x^2 - t(x)x + n(x)1 = 0, &$t(x)$, $n(x)$ in $F$.
+\end{myalign}
+If $x$ is not in $F1$, the scalars $t(x)$, $n(x)$ in \tagref(25) are uniquely determined;
+set $t(\alpha 1) = 2\alpha$, $n(\alpha 1) = \alpha^2$ to make the \emph{trace} $t(x)$ linear and the \emph{norm} $n(x)$
+a quadratic form.
+
+An \emph{involution} (\emph{involutorial anti-isomorphism}) of an algebra $\A$ is a linear
+operator $x \to \overline x$ on $\A$ satisfying
+\begin{myalign}[8em]
+\tag{26}  &&\overline{xy} = \overline{y}\: \overline{x},  \quad\overline{\overline{x}}= x  &for all $x, y$ in $\A$.
+\intertext{Here we are concerned with an involution satisfying}
+\tag{27}  &&x + \overline{x} \in F1,  \quad x\overline{x} (=\overline{x}x) \in F1 &for all $x$ in $\A$.
+\intertext{Clearly \tagref(27) implies \tagref(25) with}
+\PGx--File: 029.png---\****\********\****\********\-------------------------
+\tag{27$'$} &&x + \overline x = t(x)1, \quad x\overline x(=\overline{x}x) = n(x)1&for all $x$ in $\A$
+\end{myalign}
+(since $\overline{1} = 1$, we have $t(\alpha 1) = 2 \alpha$, $n(\alpha 1) = \alpha^2$ from \tagref(27)).
+
+Let $\B$ be an algebra with $1$ having dimension $n$ over $F$ and such that $\B$
+has an involution $x \to \overline{x}$ satisfying \tagref(27). We construct an algebra $\A$ of
+dimension $2n$ over $F$ with the same properties and having $\B$ as subalgebra
+(with $1 \in \B$) as follows: $\A$ consists of all ordered pairs $x = (b_1,b_2)$,
+$b_i$ in $\B$, addition and multiplication by scalars defined componentwise, and
+multiplication defined by
+\begin{myalign}
+\tag{28}&&(b_1,b_2)(b_3,b_4) = (b_1 b_3 + \mu b_4 \overline{b_2},\; \overline{b_1} b_4 + b_3 b_2)
+\end{myalign}
+for all $b_i$ in $\B$ and some $\mu \ne 0$ in $F$. Then $1 = (1,0)$ is a unity element
+for $\A$, $\B' = \{(b,0) \mid b \in \B\}$ is a subalgebra of $\A$ isomorphic to $\B$, $v = (0, 1)$
+is an element of $\A$ such that $v^2 = \mu1$ and $\A$ is the vector space direct sum
+$\A = \B' + v\B'$ of the $n$-dimensional vector spaces $\B'$, $v\B'$. Identifying $\B'$ with
+$\B$, the elements of $\A$ are of the form
+\begin{myalign}
+\tag{29} &&x = b_1 + v b_2 &($b_1$ in $\B$ uniquely determined by $x$),\\
+\end{myalign}
+and \tagref(28) becomes
+\begin{myalign}
+\tag{28$'$}&& (b_1 + v b_2)(b_3 + v b_4) = (b_1 b_3 + \mu b_4 \overline{b_2}) + v(\overline{b_1} b_4 + b_3 b_2 )
+\end{myalign}
+for all $b_i$ in $\B$ and some $\mu \ne 0$ in $F$. Defining
+\begin{myalign}
+\tag{30}&& \overline x = \overline{b_1} - v b_2,
+\end{myalign}
+we have $\overline {xy} = \overline y \:\overline x$ by \tagref(28$'$) since $b \to \overline b$ is an
+involution of $\B$; hence $x \to \overline x$ is an involution of $\A$. Also
+\begin{myalign}
+&&x + \overline x = t(x) 1, \qquad x \overline x(= \overline xx) = n(x)1
+\intertext{where, for $x$ in \tagref(29), we have}
+\tag{31} &&t(x) = t(b_1), \qquad n(x) = n(b_1) - \mu n(b_2).
+\end{myalign}
+
+Assume that the norm on $\B$ is a nondegenerate quadratic form; that is, the associated symmetric bilinear form
+\begin{myalign}
+\tag{32}&& (a,b) = \tfrac12 \left[n(a+b) - n(a) - n(b)\right] \quad  (= \tfrac12 t(a \overline b))
+\end{myalign}
+is nondegenerate (if $(a,b) = 0$ for all $b$ in $\B$, then $a = 0$). Then the norm
+\PG--File: 030.png---\*******\********\****\********\----------------------
+$n(x)$ on $\A$ defined by \tagref(31) is nondegenerate. For $y = b_3 + vb_4$ implies that
+$\allowbreaks (x,y) =
+\frac12\left[n(x+y)-n(x)-n(y)\right] = \frac12\left[n(b_1 + b_3)\right. - \mu n(b_2 + b_4) - n(b_1) + \mu n(b_2) -
+n(b_3) + \left.\mu n(b_4)\right] = (b_1,b_3) - \mu(b_2,b_4)$. Hence $(x,y)=0$ for all $y = b_3 +
+vb_4$ implies $(b_1,b_3) = \mu(b_2,b_4)$ for all $b_3, b_4$ in $\B$. Then $b_4=0$ implies
+$(b_1,b_3) = 0$ for all $b_3$ in $\B$, or $b_1 = 0$ since $n(b)$ is nondegenerate on $\B$;
+similarly $b_3 = 0$ implies $(b_2,b_4) = 0$ (since $\mu \ne 0$) for all $b_4$ in $\B$, or
+$b_2 = 0$. That is, $x = 0$; $n(x)$ is nondegenerate on $\A$.
+
+When is $\A$ alternative? Since $\A$ is its own reciprocal algebra, it is
+sufficient to verify the left alternative law \tagref(1$'$), which is equivalent
+to $(x,\overline{x},y) = 0$ since $(x,\overline{x},y) = (x,t(x)1-x, y) = -(x,x,y)$.
+Now $\allowbreaks % we use explicit sized barackets rather than \left and \right to facilitate linebreaks
+(x,\overline{x},y) =
+n(x)y - (b_1 + vb_2)\big[(\overline{b_1}b_3 - \mu b_4 \overline{b_2}) + v(b_1b_4 - b_3b_2)\big] = n(x)y - \big[b_1(\overline{b_1}b_3) -
+\mu b_1(b_4\overline{b_2}) + \mu(b_1b_4)\overline{b_2} - \mu(b_3b_2)\overline{b_2}\big] - v\big[\overline{b_1}(b_1b_4) - \overline{b_1}(b_3b_2) + (\overline{b_1}b_3)b_2 -
+\mu(b_4\overline{b_2})b_2\big] = n(x)y - \big[n(b_1)-\mu n(b_2)\big] (b_3 + vb_4) - \mu(b_1,b_4,\overline{b_2}) - v(\overline{b_1},b_3,b_2) =
+-\mu(b_1,b_4,\overline{b_2}) - v(\overline{b_1},b_3,b_2)$ by a trivial extension of the \hyperlink{Theorem:of Artin}{Theorem} of Artin.
+Hence $\A$ is alternative if and only if $\B$ is associative.
+
+The algebra $F1$ is not a quadratic algebra, but the identity operator on
+$F1$ is an involution satisfying \tagref(27); also $n(\alpha1)$ is nondegenerate on $F1$.
+Hence we can use an iterative process (beginning with $\B = F1$) to obtain by
+the above construction algebras of dimension $2^t$ over $F$; these depend
+completely upon the $t$ nonzero scalars $\mu_1,\mu_2,\dotsc,\mu_t$ used in the successive
+steps. The norm on each algebra is a nondegenerate quadratic form. The
+$2$-dimensional algebras $\Z = F1 + v_1(F1)$ are either quadratic fields over $F$
+($\mu_1$ a nonsquare in $F$) or isomorphic to $F \oplus F$ ($\mu_1$ a square in $F$). The
+$4$-dimensional algebras $\Q = \Z + v_2\Z$ are associative central simple algebras
+(called \emph{quaternion algebras}) over $F$; any $\Q$ which is not a division algebra
+is (by Wedderburn's theorem on simple associative algebras) isomorphic to the
+algebra of all $2\times2$ matrices with elements in $F$.
+\PG--File: 031.png---\*************\********\****\********\----------------
+
+We are concerned with the $8$-dimensional algebras $\C = \Q + v_3\Q$ which
+are called \emph{Cayley algebras} over $F$. Since any $\Q$ is associative, Cayley
+algebras are alternative. However, no Cayley algebra is associative.
+For $\Q$ is not commutative and there exist $q_1$, $q_2$ in $\Q$ such that $[q_1,q_2]\ne 0$;
+hence $(v_3,q_2,q_1) = (v_3q_2)q_1 - v_3(q_2q_1) = v_3[q_1,q_2] \ne 0$ by \tagref(28$'$).
+Thus this iterative process of constructing alternative algebras stops after
+three steps. The quadratic form $n(x)$ is nondegenerate; also it \emph{permits
+composition} in the sense that
+\begin{myalign}
+\tag{33} &&n(xy)=n(x)n(y) &for all $x, y$ in $\C$.
+\intertext{For $n(xy)1=(xy)(\overline{xy})=xy\overline{y}\:\overline{x}=n(y)x\overline{x}=n(x)n(y)1$. Also}
+\tag{34} &&t\left((xy)z\right) =t\left(x(yz)\right) &for all $x, y, z$ in $\A$.
+\end{myalign}
+For $\allowbreaks(x,y,z) = -(z,y,x) = (\overline{z},\overline{y},\overline{x})$ implies $(xy)z + \overline{z}(\overline{y}\:\overline{x}) = x(yz)+\break % heavyhandedness is the only solution here
+ (\overline{z}\:\overline{y})\overline{x}$,
+so that \tagref(34) holds.
+
+\begin{theorem}[5]
+Two Cayley algebras $\C$ and $\C'$ are isomorphic if and only if
+their corresponding norm forms $n(x)$ and $n'(x')$ are equivalent (that is, there
+is a linear mapping $x\to xH$ of $\C$ into $\C'$ such that
+\begin{myalign}
+\tag{35} &&n'(xH) = n(x) &for all $x$ in $\C$;
+\end{myalign}
+$H$ is necessarily (1--1) since $n(x)$ is nondegenerate).
+\end{theorem}
+
+\begin{proof}
+Suppose $\C$ and $\C'$ are isomorphic, the isomorphism being $H$.
+Then \tagref(25) implies $(xH)^2 - t(x)(xH) + n(x)1' = 0$ where $1'=1H$ is the unity
+element of $\C'$. But also $(xH)^2 - t'(xH)(xH) + n'(xH)1' = 0$. Hence
+$\left[t'(xH)-t(x)\right](xH) + \left[n(x) - n'(xH)\right]1' = 0$. If $x\notin F1$, then $xH\notin F1'$ and
+$n(x)=n'(xH)$. On the other hand $n(\alpha1)=\alpha^2=n'(\alpha1')$, and we have \tagref(35) for
+all $x$ in $\C$.
+
+For the converse we need to establish the fact that, if $\B$ is a proper
+subalgebra of a Cayley algebra $\C$, if $\B$ contains the unity element $1$ of $\C$, and
+if (relative to the nondegenerate symmetric bilinear form $(x,y)$ defined on $\C$ by
+\tagref(32)) $\B$ is a non-isotropic subspace of $\C$ (that is, $\B\cap\B^\perp=0$), then there is a
+\PG--File: 032.png---\************\********\****\********\-----------------
+subalgebra $\A = \B + v\B$ (constructed as above). For the involution $x \to \overline x$
+on $\C$ induces an involution on $\B$, since $\overline b=t(b)1-b$ is in $\B$ for all $b$ in
+$\B$. Also $\B$ non-isotropic implies $\C=\B \perp\B^{\perp}$ with $\B^{\perp}$ non-isotropic (Jacobson,
+Lectures in Abstract Algebra, vol.~II, p.~151; Artin, Geometric Algebra, p.~117).
+Hence there is a non-isotropic vector $v$ in $\B^{\perp}$, $n(v)=-\mu \ne 0$.
+Since $t(v) = t(v \overline1) = 2(v,1) = 0$, we have
+\begin{myalign}
+\tag{36} &&v^2=\mu 1,   &$\mu \ne 0$ in $F$.
+\end{myalign}
+Now $v\B \subseteq\B^\perp$ since \tagref(34) implies $(va,b)=\frac{1}{2} t\left((va)\overline b\right) = \frac{1}{2} t\left(v(a \overline b)\right)=(v,b \overline a)=0$
+for all $a, b$ in $\B$. Hence $\B \perp v\B$. Also $v\B$ has the same dimension as $\B$ since
+$b \to vb$ is (1--1). Suppose $vb = 0$; then $v(vb) = v^2b = \mu b = 0$, implying $b = 0$.
+In order to show that $\A = \B \perp v\B$ is the algebra constructed above, it remains
+to show that
+\begin{myalign}
+\tag{37} &&a(vb) = v(\overline ab ),\\
+\tag{38} &&(va)b = b(va),\\
+\tag{39} &&(va)(vb) = \mu b \overline a
+\intertext{for all $a, b$ in $\B$. Now $t(v) = 0$ implies $\overline v=-v$; hence $v$ in $\B^{\perp}$ implies
+$0 = 2(v,b) = t(v\overline b) = v\overline b + b\overline v = v\overline b -bv$, or}
+\tag{40} &&bv = v \overline b            &for all $b$ in $\B$.
+\end{myalign}
+Hence $\allowbreaks(v,\overline a, b) + (\overline a,v,b)=0=(v \overline a)b-v(\overline a b)+(\overline a v)b=(v \overline a)b-v(\overline a b)
++(va)b-\overline a(vb)=\left[t(a)1- \overline a\right]vb-v(\overline a b)$, establishing % [** typo:  extablishing]
+ \tagref(37). Applying the
+involution to $\overline b(va)=v(ba)$, and using \tagref(40), we have \tagref(38). Finally $(va)(vb) =
+(va)(\overline b v)= v(a\overline b)v=v^2(b\overline a)= \mu b\overline a$ by the Moufang identity \tagref(9).
+ Hence $\A =\B \perp\B^{\perp}$
+is the subalgebra specified. Since $\B$ and $\B^{\perp}$ are non-isotropic, so is $\A$.
+[Remark: we have shown incidentally that if $\Q$ is any quaternion subalgebra
+containing $1$ in a Cayley algebra $\C$, then $\Q$ may be used in the construction
+of $\C$ as $\C = \Q + v\Q$.]
+
+Now let $\C$ and $\C'$ have equivalent norm forms $n(x)$ and $n'(x')$. Let $\B$
+(and $\B'$) be as above. If $\B$ and $\B'$ are isomorphic under $H_0$,
+then the
+\PG--File: 033.png---\*************\******\********\*******\---------------
+restrictions of $n(x)$ and $n'(x')$ to $\B$ and $\B'$ are equivalent. Then by
+Witt's theorem (Jacobson, ibid, p.~162; Artin, ibid, p.~121), since $n(x)$
+and $n'(x')$ are equivalent, the restrictions of $n(x)$ and $n'(x')$ to $\B^\perp$ and
+$\B'^\perp$ are equivalent. Choose $v$ in $\B^\perp$ with $n(v) \ne 0$; correspondingly we have
+$v'$ in $\B'^\perp$ such that $n'(v') = n(v)$. Then $a + vb \to aH_0 + v'(bH_0)$ is an
+isomorphism of $\B\perp v\B$ onto $\B' \perp v'\B'$ by the construction above. Hence if
+we begin with $\B = F1$, $\B' = F1'$, repetition of the process gives successively
+isomorphisms between $\Z$ and $\Z'$, $\Q$ and $\Q'$, $\C$ and $\C'$.
+\end{proof}
+
+A Cayley algebra $\C$ is a division algebra if and only if $n(x) \ne 0$ for
+every  $x \ne 0$ in $\C$. For $x \ne 0$, $n(x) = 0$ imply $x\overline x = n(x)1 = 0$, $\C$ has zero
+divisors. Conversely, if $n(x) \ne 0$, then $\overline x(xy) = (\overline xx)y = n(x)y$ for all $y$
+implies $\frac{1}{n(x)}L_xL_{\overline x} = 1_\C$, $L_x^{-1} = \frac{1}{n(x)} L_{\overline x}$
+ and similarly $R_x^{-1} = \frac{1}{n(x)} R_{\overline x}$;
+hence if $n(x) \ne 0$ for all $x \ne 0$, then $\C$ is a division algebra.
+
+[Remark: If $F$ is the field of all real numbers, the norm form $n(x) =
+\sum {\alpha_i}^2$  for $x = \sum \alpha_i u_i$ clearly has the property above. Also there are alternative
+algebras $F1$, $\Z$, $\Q$, $\C$ with this norm form (take $\mu_i = -1$ at each step). Hence
+there are real alternative division algebras of dimensions $1$, $2$, $4$, $8$. It
+has recently been proved (see reference \cite{Ref12} of the appended bibliography
+of recent papers) that finite-dimensional real division algebras can have
+only these dimensions. It is not true, however, that the only finite-dimensional
+real division algebras are the four listed above; they are the
+only alternative ones. For other examples of finite-dimensional real division
+algebras (necessarily of these specified dimensions of course) see reference
+[23] in the bibliography of the 1955 Bulletin \hyperlink{cite.Ref64}{article}.]
+
+\begin{corollary} Any two Cayley algebras $\C$ and $\C'$ with divisors of zero
+are isomorphic.
+\PG--File: 034.png---\*************\********\****\********\----------------
+\end{corollary}
+
+\begin{proof} Show first that $\C$ has divisors of zero if and only if there
+is $w \notin F1$ such that $w^2 = 1$. For $1 - w \ne 0$, $1 + w \ne 0$ imply $(1 - w)(1 + w) =
+1 - w^2 = 0$ (note $t(w) = 0$ implies $\overline {1 \pm w} = 1 \mp w$ so that $n(1 \pm w ) =0$).
+Conversely, if $\C$ has divisors of zero, there exists $x \ne 0$ in $\C$ with $n(x) = 0$.
+Then $ x= \alpha 1 + u$, $u \in {(F1)}^ \perp =\{u \mid t(u)=0 \}$ implies $0 = n(x)1 = x\overline{x} =
+(\alpha 1 + u)(\alpha 1 - u) = \alpha^2 1 - u^2$.  If $\alpha \ne 0$, then $w = \alpha^{-1}u$ satisfies $w^2 =
+1$ ($w \notin F1$).  If $\alpha=0$, then $n(u) = 0$ so that $u$ is an isotropic vector in the
+non-isotropic space $(F1)^\perp$. Hence there exists $w$ in $(F1)^\perp$ with $n(w) = -1$
+(Jacobson, ibid, p.~154, ex.~3), or $w^2 = t(w)w - n(w)1 = 1$ ($w \notin F1$).
+
+Now let $e_1  = \frac{1}{2}(1- w)$, $e_2 = 1- e_1 = \frac{1}{2}(1+ w)$. Then ${e_1}^2 = e_1$, ${e_2}^2 = e_2$,
+$e_1e_2=e_2e_1 = 0$ ($e_1$ and $e_2$ are \emph{orthogonal idempotents}). Also $n(e_i)=0$ for $i=
+1,2$. Hence every vector in $e_i\C$  is isotropic since $n(e_ix) = n(e_i)n(x) = 0$.
+This means that $e_i\C$ is a totally isotropic subspace ($e_i\C \subseteq (e_i\C)^\perp$). Hence
+$\dim(e_i\C) \le \frac{1}{2} \dim\C = 4$ (Jacobson, p.~170; Artin, p.~122). But $x=1x =e_1x+
+e_2x$ for all $x$ in $\C$, so $\C=e_1\C+e_2\C$. Hence $\dim (e_i\C)=4$, and $n(x)$ has
+maximal Witt index $= 4 = \frac{1}{2}\dim\C$. Similarly $n'(x')$ has maximal Witt index $= 4$.
+Hence $n(x)$ and $n'(x')$ are equivalent (Artin, ibid). By \hyperlink{Theorem:5}{Theorem} 5, $\C$ and $\C'$
+are isomorphic.
+\end{proof}
+
+Over any field $F$ there is a Cayley algebra without divisors of zero
+(take $\mu=1$ so $v^2 = 1$). This unique Cayley algebra over $F$ is called the
+\emph{split Cayley algebra} over $F$.
+
+$F1$ is both the nucleus and center of any Cayley algebra. Also any
+Cayley algebra is simple (hence central simple over $F$). (This is obvious
+for all but the split Cayley algebra.) For, if $\I$ is any nonzero ideal of
+$\C$, there is $x \ne 0$ in $\I$. But $x$ is contained is some quaternion subalgebra
+$\Q$ of $\C$. Then $\Q\times \Q$ is an ideal of the simple algebra $\Q$. Hence $1\in\Q =
+\Q\times\Q \subseteq\I$, and $\I = \C$.
+
+We omit the proof of the fact that the only alternative central simple
+\PG--File: 035.png---\*******\*****\****\********\-------------------------
+algebras of finite dimension which are not associative are Cayley algebras.
+(Actually the following stronger result is known: any simple alternative
+ring, which is not a nilring and which is not associative, is a Cayley
+algebra over its center; in the finite-dimensional case the restriction
+eliminating nilalgebras is not required since \hyperlink{Theorem:4}{Theorem} 4 implies that $\A^2 \ne\A$
+for a finite-dimensional alternative nilalgebra). Hence the simple components
+$\Ss_i$ in a finite-dimensional semisimple alternative algebra are either
+associative or Cayley algebras over their centers.
+
+The derivation algebra $\D(\C)$ of any Cayley algebra of characteristic
+$\ne 3$ is a central simple Lie algebra of dimension $14$, called an \emph{exceptional
+Lie algebra of type G} (corresponding to the $14$-parameter complex exceptional
+simple Lie group $G_2$). The related subject of automorphisms of Cayley
+algebras is studied in \cite{Ref33}.
+\PG--File: 036.png---\************\***\****\********\----------------------
+
+\chapter{Jordan Algebras}% IV.
+
+In the \hyperlink{chapter.1}{Introduction} we defined a (commutative) Jordan algebra $\J$ over
+$F$ to be a commutative algebra in which the \emph{Jordan identity}
+\begin{myalign}
+\tag{1}       &&(xy)x^2 = x(yx^2)       &for all $x, y$ in $\J$
+\end{myalign}
+is satisfied. Linearization of \tagref(1) requires that we assume $F$ has
+characteristic $\ne 2$; we make this assumption throughout IV\@. It follows from
+\tagref(1) and the identities \tagref(2), \tagref(3) below that any scalar extension $\J_K$ of a
+Jordan algebra $\J$ is a Jordan algebra.
+
+Replacing $x$ in
+\begin{myalign}
+\tag{1$'$}    &&(x, y, x^2) = 0         &for all $x, y$ in $\J$
+\intertext{by $x + \lambda z$ ($\lambda \in F$), the coefficient of $\lambda$ is $0$ since $F$ contains at least three
+distinct elements, and we have}
+\tag{2}       &&2(x, y, zx) + (z, y, x^2) = 0     &for all $x, y, z$ in $\J$.
+\end{myalign}
+Replacing $x$ in \tagref(2) by $x + \lambda w$ ($\lambda \in F$), we have similarly (after dividing by
+$2$) the multilinear identity
+\begin{myalign}[0em](0em)
+\tag{3}      &&(x, y, wz) + (w, y, zx) + (z, y, xw) = 0  &for all $w, x, y, z$ in $\J$.\\
+\intertext{Recalling that $L_a = R_a$ since $\J$ is commutative, we see that \tagref(3) is equivalent to}
+\tag{3$'$}&&[R_x, R_{wz}] + [R_w, R_{zx}] + [R_z, R_{xw}] = 0  &for all $w, x, z$ in $\J$\\
+\Intertext{and to}
+\tag{3$''$} &&R_zR_{xy} - R_zR_yR_x + R_yR_{zx} - R_{y(zx)} + R_xR_{zy} - R_xR_yR_z = 0\\
+& &&for all $x, y, z$ in $\J$.\\
+\intertext{Interchange $x$ and $y$ in \tagref(3$''$) and subtract to obtain}
+\tag{4}    &&\left[R_z, [R_x, R_y]\right] = R_{(x, z, y)} = R_{z[R_x, R_y]}   &for all $x, y, z$ in $\J$.\\
+\end{myalign}
+Now \tagref(4) says that, for all $x, y$ in $\J$, the operator $[R_x, R_y]$ is a derivation of $\J$,
+since the defining condition for a derivation $D$ of an arbitrary algebra $\A$ may
+be written as
+\begin{myalign}
+&&[R_z, D] = R_{zD}           &for all $z$ in $\A$.
+\end{myalign}
+
+\PG--File: 037.png---\************\********\********\*******\--------------
+Our first objective is to prove that any Jordan algebra $\J$ is power-associative.
+As in \chaplink{3} we define powers of $x$ by $x^1 = x$, $x^{i+1} = xx^i$,
+and prove
+\begin{myalign}
+\tag{5} &&  x^ix^j = x^{i+j}      &for all $x$ in $\J$.
+\end{myalign}
+For any $x$ in $\J$, write $\G_x = R_x \cup R_{x^2}$. Then the enveloping algebra $\G_x^*$ is
+commutative, since the generators $R_x$, $R_{x^2}$ commute by \tagref(1). For $i \ge 2$, we
+put $y = x$, $z = x^{i-1}$ in \tagref(3$''$) to obtain
+\begin{myalign}
+\tag{6}&&  R_{x^{i+1}} = R_{x^{i-1}}R_{x^2} - R_{x^{i-1}}R_x^2 - R_x^2R_{x^{i-1}} + 2R_xR_{x^i}.
+\end{myalign}
+By induction on $i$ we see from \tagref(6) that $R_{x^i}$ is in $\G^*$ for $i = 3, 4,\dots$. Hence
+\begin{myalign}
+\tag{7} &&  R_{x^i}R_{x^j} = R_{x^j}R_{x^i}    &for $i, j = 1, 2, 3,\dots$
+\end{myalign}
+Then, in a proof of \tagref(5) by induction on $i$, we can assume that $x^ix^{j+1} =
+x^{i+j+1}$; then $x^{i+1}x^j = (xx^i)x^j = xR_{x^i}R_{x^j} = xR_{x^j}R_{x^i} = x^{j+1}x^i = x^{i+j+1}$
+as desired.
+
+One can prove, by a method similar to the proof of \hyperlink{Theorem:4}{Theorem~4} in \chaplink{3}
+(only considerably more complicated since the identities involved are more
+complicated), that any finite-dimensional Jordan nilalgebra is nilpotent.
+We omit the proof, which involves also a proof of the fact that
+\begin{myalign}
+\tag{8}&& \text{$R_x$ is nilpotent for any nilpotent $x$}\\
+&&&in a finite-dimensional Jordan algebra.\\
+\end{myalign}
+As in \chaplink{3}, this means that there is a unique maximal nilpotent ($=$~solvable $=$~nil)
+ideal $\N$ which is called the \emph{radical} of $\J$. Defining $\J$ to be \emph{semisimple} in case
+$\N = 0$, we have seen that $\J/\N$ is semisimple. The proof that any semisimple
+Jordan algebra $\Ss$ is a direct sum $\Ss = \Ss_1 \oplus \dotsb \oplus \Ss_t$ of simple $\Ss_i$ is quite
+complicated for arbitrary $F$; we shall use a trace argument to give a proof
+for $F$ of characteristic $0$.
+
+Let $e$ be an idempotent in a Jordan algebra $\J$. Put $i = 2$ and $x = e$ in
+\tagref(6) to obtain
+\begin{myalign}
+\tag{9} &&  2R_e^3 - 3R_e^2 + R_e = 0;
+\end{myalign}
+\PG--File: 038.png---\*********\********\********\*******\-----------------
+that is, $f(R_e) = 0$ where $f(\lambda) = (\lambda-1)(2\lambda-1)\lambda$. Hence the minimal polynomial
+for $R_e$ divides $f(\lambda)$, and the only possibilities for characteristic roots of
+$R_e$ are $1$, $\frac12$, $0$ ($1$ must occur since $e$ is a characteristic vector belonging to
+the characteristic root $1$: $eR_e = e^2 = e \ne 0$). Also the minimal polynomial
+for $R_e$ has simple roots. Hence $\J$ is the vector space direct sum
+\begin{myalign}
+\tag{10}  &\J &=  \J_1 + \J_{1/2} + \J_0\\
+\Intertext{where}
+\tag{11} & \J_i&= \left\{ x_i \mid x_ie = ix_i \right\},    &   $i = 1, 1/2, 0$.
+\end{myalign}
+Taking a basis for $\J$ adapted to the \emph{Peirce decomposition} \tagref(10), we see that
+$R_e$ has for its matrix relative to this basis the diagonal matrix
+$\diag \{1,1, \dots, 1,1/2, 1/2, \dots, 1/2, 0, 0, \dots, 0\}$ where the number of $1$'s
+is $\dim\J_1 > 0$ and the number of $1/2$'s is $\dim\J_{1/2}$. Hence
+\begin{myalign}
+\tag{12} &&  \trace R_e = \dim\J_1 + \tfrac12 \dim\J_{1/2}.
+\end{myalign}
+If $F$ has characteristic $0$, then $\trace R_e \ne 0$.
+
+A symmetric bilinear form $(x,y)$ defined on an arbitrary algebra $\A$ is
+called a \emph{trace form} (\emph{associative} or \emph{invariant} symmetric bilinear form) on $\A$ in
+case
+\begin{myalign}
+\tag{13} &&  (xy,z) = (x,yz)   &for all $x,y,z$ in $\A$.
+\end{myalign}
+If $\I$ is any ideal of an algebra $\A$ on which such a bilinear form is defined,
+then $\I^\perp$ is also an ideal of $\A$: for $x$ in $\I$, $y$ in $\I^\perp$, $a$ in $\A$ imply that $xa$ and
+$ax$ are in $\I$ so that $(x,ay) = (xa,y) = 0$ and $(x,ya) = (ya,x) = (y,ax) = 0$ by
+\tagref(13). In particular, the radical $\A^\perp =\{x \mid (x,y)=0 \text{ for all }y \in \A\}$ of
+the trace form is an ideal of $\A$.
+
+We also remark that it follows from \tagref(13) that $(xR_y,z) = (x,zL_y)$ and
+$(xL_y,z) = (z,yx) = (zy,x) = (x,zR_y)$ so that, for right (or left)
+multiplications $S_i$ determined by $b_i$,
+\begin{myalign}
+\tag{14} &&  (xS_1S_2 \dotsm S_h, y) = (x,yS_h'\dotsm S_2'S_1')
+\end{myalign}
+\PG--File: 039.png---\************\********\*****\********\----------------
+where $S'_i$ is the left (or right) multiplication determined by $b_i$; then, if $\B$
+is any subset of $\A$,
+\begin{myalign}
+\tag{15}
+&&(xT, y) = (x, yT')
+&for all $x, y$ in $\A$, $T$ in $\B^*$,\\
+\end{myalign}
+where $T'$ is in $\B^*$.
+
+\begin{theorem}[6]
+The radical $\N$ of any finite-dimensional Jordan algebra $\J$
+over $F$ of characteristic $0$ is the radical $\J^\perp$ of the trace form
+\begin{myalign}
+\tag{16}
+&&(x,y) = \trace R_{xy}
+&for all $x, y$ in $\J$.
+\end{myalign}
+\end{theorem}
+
+\begin{proof}
+Without any assumption on the characteristic of $F$ it follows
+from \tagref(4) that $(x,y)$ in \tagref(16) is a trace form: $(xy,z) - (x,yz) = \trace R_{(x,y,z)} = 0$
+since the trace of any commutator is $0$. Hence $\J^\perp$ is an ideal of $\J$. If $\J$
+were not a nilideal, then (by \hyperlink{Proposition:3}{Proposition~3}) $\J^\perp$ would contain an idempotent
+$e$ ($\ne 0$) and, assuming characteristic $0$, $(e, e) = \trace R_e \ne 0$ by \tagref(12), a
+contradiction. Hence $\J^\perp$ is a nilideal and $\J^\perp \subseteq\N$. Conversely, if $x$ is in $\N$,
+then $xy$ is in $\N$ for every $y$ in $\A$, and $R_{xy}$ is nilpotent by~(8). Hence $(x, y) =
+\trace R_{xy} = 0$ for all $y$ in $\A$; that is, $x$ is in $\J^\perp$. Hence $\N\subseteq\J^\perp$, $\N = \J^\perp$.
+\end{proof}
+
+\begin{theorem}[7]
+Let $\A$ be a finite-dimensional algebra over $F$ (of arbitrary
+characteristic) satisfying
+\begin{Itemize}
+\item[i\DPanchor{thm7:i}] there is a nondegenerate (associative) trace form $(x,y)$ defined on
+$\A$, and
+\item[ii\DPanchor{thm7:ii}] $\I^2 \ne 0$ for every ideal $\I \ne 0$ of $\A$.
+\end{Itemize}
+Then $\A$ is (uniquely) expressible as a direct sum $\A = \Ss_1 \oplus \dotsb \oplus \Ss_t$ of simple
+ideals $\Ss_i$.
+\end{theorem}
+
+\begin{proof}
+Let $\Ss$ ($\ne 0$) be a minimal ideal of $\A$. Since $(x,y)$ is a trace
+form, $\Ss^\perp$ is an ideal of $\A$. Hence the intersection $\Ss\cap \Ss^\perp$ is either $0$ or
+$\Ss$, since $\Ss$ is minimal. We show that $\Ss$ totally isotropic ($\Ss\subseteq \Ss^\perp$) leads to
+a contradiction.
+\PG--File: 040.png---\************\********\****\********\-----------------
+
+For, since $\Ss^2 \ne 0$ by (\hyperlink{thm7:ii}{ii}), we know that the ideal of $\A$ generated by
+$\Ss^2$ must be the minimal ideal $\Ss$. Thus $\Ss = \Ss^2 + \Ss^2\M$ where $\M$ is the multiplication
+algebra of $\A$. Any element $s$ in $\Ss$ may be written in the form $s = \sum(a_ib_i)T_i$
+for $a_i, b_i$ in $\Ss$, where $T_i = T'_i$ is the identity operator $1_\A$ or $T_i$ is in $\M$.
+For every $y$ in $\A$ we have by \tagref(15) that $(s, y) = \sum \left((a_ib_i)T_i, y \right) = \sum (a_ib_i, yT'_i) =
+\sum \left(a_i, b_i(yT'_i) \right) = 0$ since $b_i(yT'_i) \in \Ss \subseteq \Ss^\perp$. Then $s = 0$ since $(x, y)$ is
+nondegenerate; $\Ss = 0$, a contradiction. Hence $\Ss \cap \Ss^\perp = 0$; that is, $\Ss$ is
+non-isotropic. Hence $\A = \Ss \perp \Ss^\perp$ and $\Ss^\perp$ is non-isotropic. That is, $\A =
+\Ss \oplus \Ss^\perp$, the direct sum of ideals $\Ss$, $\Ss^\perp$, and the restriction of $(x, y)$ to $\Ss^\perp$
+is a nondegenerate (associative) trace form defined on $\Ss^\perp$. That is, (\hyperlink{thm7:i}{i})
+holds for $\Ss^\perp$ as well as $\A$. Moreover, any ideal of the direct summand $\Ss$ or
+$\Ss^\perp$ is an ideal of $\A$; hence $\Ss$ is simple and (\hyperlink{thm7:ii}{ii}) holds for $\Ss^\perp$. Induction on
+the dimension of $\A$ completes the proof.
+\end{proof}
+
+\begin{corollary} Any (finite-dimensional) semisimple Jordan algebra $\J$ over\DPanchor{cor:thm:7}
+$F$ of characteristic $0$ is (uniquely) expressible as a direct sum $\J = \Ss_1 \oplus \dotsb \oplus \Ss_t$
+of simple ideals $\Ss_i$.
+\end{corollary}
+
+\begin{proof} By \hyperlink{Theorem:6}{Theorem 6} the (associative) trace form \tagref(16) is nondegenerate;
+hence (\hyperlink{thm7:i}{i}) is satisfied. Also any ideal $\I$ such that $\I^2 = 0$ is nilpotent;
+hence $\I = 0$, establishing (\hyperlink{thm7:ii}{ii}).
+\end{proof}
+
+As mentioned above, the corollary is actually true for $F$ of characteristic
+$\ne 2$. What remains then, as far as the structure of semisimple Jordan algebras
+is concerned, is a determination of the central simple algebras. The first
+step in this is to show that every semisimple $\J$ (hence every simple $\J$) has
+a unity element $1$. Again the argument we use here is valid only for
+characteristic $0$, whereas the theorem is true in general.
+
+We begin by returning to the Peirce decomposition \tagref(10) of any Jordan
+algebra $\J$ relative to an idempotent $e$. The subspaces $\J_1$ and $\J_0$ are orthogonal
+\PG--File: 041.png---\************\********\****\********\-----------------
+subalgebras of $\J$ which are related to the subspace $\J_{1/2}$ as follows:
+\begin{myalign}
+\tag{17}  &&  \J_{1/2}\J_{1/2} \subseteq \J_1 + \J_0,\qquad \J_1\J_{1/2} \subseteq \J_{1/2},\qquad \J_0\J_{1/2} \subseteq \J_{1/2}.
+\end{myalign}
+For we may put $x = e$, $z = x_i \in \J_i$, $y = y_j \in \J_j$ in \tagref(2) to obtain $2i(e, y_j, x_i) +
+(x_i, y_j, e) = 0$, or $(1 - 2i)\left[(x_iy_j)e - j(x_iy_j)\right] = 0$, so that
+\begin{myalign}
+\tag{18}     &&\J_i\J_j \subseteq \J_j   &if $i \ne 1/2$.
+\end{myalign}
+Hence ${\J_1}^2 \subseteq \J_1$, ${\J_0}^2 \subseteq \J_0$, $\J_1\J_0 = \J_0\J_1 \subseteq
+\J_0 \cap \J_1 = 0$, so $\J_1$ and $\J_0$ are orthogonal
+subalgebras by \tagref(18), and also the last two inclusions in \tagref(17) hold. Put $x =
+x_{1/2}$, $z = y_{1/2}$, $y = w = e$ in \tagref(3) and write $x_{1/2}y_{1/2} = a = a_1 + a_{1/2} + a_0$ to
+obtain $\frac{1}{2}(x_{1/2}, e, y_{1/2}) + (e, e, a) + \frac{1}{2}(y_{1/2}, e, x_{1/2}) = (e, e, a) = 0$. Hence
+$ea - e(ea) = a_1 + \frac{1}{2}a_{1/2} - e(a_1 + \frac{1}{2}a_{1/2}) = a_1 + \frac{1}{2}a_{1/2} - a_1 - \frac{1}{4}a_{1/2} = \frac{1}{4}a_{1/2} = 0$.
+Hence $x_{1/2}y_{1/2} = a_1 + a_0 \in \J_1 + \J_0$, establishing \tagref(17).
+
+Now
+\begin{myalign}
+\tag{19}     &&\trace R_b = 0       &for all $b \in \J_{1/2}$.
+\end{myalign}
+For $b$ in $\J_{1/2}$ implies $\trace R_b = 2\trace R_{eb} = 2(e, b) = 2(e^2, b) = 2(e, eb) =
+(e, b)$ by \tagref(16) and \tagref(13), so $\trace R_b = (e, b) = 0$. Writing $x = x_1 + x_{1/2} + x_0$,
+$y = y_1 + y_{1/2} + y_0$ in accordance with \tagref(10), we have $xy = (x_1y_1 + x_{1/2}y_{1/2} +
+x_0y_0) + (x_1y_{1/2} + x_{1/2}y_1 + x_{1/2}y_0 + x_0y_{1/2})$ with the last term in parentheses
+in $\J_{1/2}$ by \tagref(17). Hence \tagref(19) implies that
+\begin{myalign}
+\tag{20}   &&   (x, y) = \trace R_{x_1y_1 + x_{1/2}y_{1/2} + x_0y_0}.
+\end{myalign}
+Now $x_{1/2}y_{1/2} = c = c_1 + c_0$ ($c_i$ in $\J_i$) implies $\trace R_{c_1} + \trace R_{c_0} =
+\trace R_c = (x_{1/2}, y_{1/2}) = 2(ex_{1/2}, y_{1/2}) = 2(e, x_{1/2}y_{1/2}) =
+2 \trace R_{e(c_1 + c_0)} \break % another heavyhanded solution
+= 2 \trace R_{c_1}$, so that $\trace R_{c_1} = \trace R_{c_0}$. Then \tagref(20)
+may be written as
+\begin{myalign}
+\tag{20$'$} &&      (x, y) = \trace R_{x_1y_1 + z_0}, &$z_0 = 2c_0 + x_0y_0$ in $\J_0$.
+\end{myalign}
+
+In any algebra $\A$ over $F$ an idempotent $e$ is called \emph{principal} in case
+there is no idempotent $u$ in $\A$ which is orthogonal to $e$ ($u^2 = u \ne 0$, $ue = eu = 0$);
+\PG--File: 042.png---\************\********\******\********\---------------
+that is, there is no idempotent $u$ in the subspace $\A_0 =
+\{x_0 \mid x_0 \in \A,\ ex_0 = x_0e = 0\}$. In a finite-dimensional Jordan algebra $\J$, this
+means that $e$ is a principal idempotent of $\J$ if and only if the subalgebra
+$\J_0$ (in the Peirce decomposition \tagref(10) relative to $e$) is a nilalgebra.
+
+Now any finite-dimensional Jordan algebra $\J$ which is not a nilalgebra
+contains a principal idempotent. For $\J$ contains an idempotent $e$ by
+\hyperlink{Proposition:3}{Proposition~3}. If $e$ is not principal, there is an idempotent $u$ in $\J_0$,
+$e' = e + u$ is idempotent, and $\J_{1, e'}$ (the $\J_1$ relative to $e'$) contains
+properly $\J_{1, e} = \J_1$. For $x_1$ in $\J_{1, e}$ implies $x_1e' = x_1(e + u) = x_1e + x_1u =
+x_1$, or $x_1$ is in $\J_{1, e'}$. That is, $\J_{1, e} \subseteq \J_{1, e'}$. But $u \in \J_{1, e'}$, $u \notin \J_{1, e}$.
+Then $\dim\J_{1, e} < \dim\J_{1, e'}$, and this process of increasing dimensions must
+terminate, yielding a principal idempotent.
+
+\begin{theorem}[8]
+Any semisimple (hence any simple) Jordan algebra $\J$ of finite
+dimension over $F$ of characteristic $0$ has a unity element~$1$.
+\end{theorem}
+
+\begin{proof}
+$\J$ has a principal idempotent $e$. Then $\J_0$ is a nilalgebra, so
+that $(x, y) = \trace R_{x_1y_1}$ by \tagref(20$'$) since $\trace R_{z_0} = 0$ by \tagref(8). Hence $x$ in
+$\J_{1/2} + \J_0$ implies $x_1 = 0$, $(x, y) = 0$ for all $y$ in $\J$, so $x$ is in $\J^\perp$.
+That is, $\J_{1/2} + \J_0 \subseteq \J^\perp = \N = 0$, or $\J = \J_1$, $e = 1$.
+\end{proof}
+
+If $\J$ contains $1$ and $e_1 \ne 1$, then $e_2 = 1 - e_1$, is an idempotent, and the
+Peirce decompositions relative to $e_1$ and $e_2$ coincide (with differing subscripts).
+We introduce a new notation: $\J_{11} = \J_{1, e_1}$ ($= \J_{0, e_2}$), $\J_{12} = \J_{1/2, e_1}$
+($= \J_{1/2, e_2}$), $\J_{22} = \J_{0, e_1}$ ($= \J_{1, e_2}$). More generally, if $1 = e_1 + e_2 + \dotsb +
+e_t$ for pairwise orthogonal idempotents $e_i$, we have the refined Peirce
+decomposition
+\begin{myalign}
+\tag{21}&&
+  \J = \sum_{i \le j} \J_{ij}
+\end{myalign}
+of $\J$ as the vector space direct sum of subspaces $\J_{ii} = \J_{1, e_i}$ ($1 \le i \le t$),
+\PG--File: 043.png---\************\*****\****\********\--------------------
+$\J_{ij} = \J_{1/2, e_i} \cap \J_{1/2, e_j}$ ($1 \le i < j \le t$); that is,
+\begin{myalign}
+\tag{22}&&{\begin{aligned}
+&\J_{ii} = \{x \mid x \in \J,\ xe_i = x\},\\
+&\J_{ij} = \J_{ji} = \{x \mid x \in \J,\ xe_i = \tfrac{1}{2}x = xe_j\},\quad i \ne j.
+\end{aligned}}
+\end{myalign}
+Multiplicative relationships among the $\J_{ij}$ are consequences of \tagref(17) and
+the statement preceding it.
+
+An idempotent $e$ in $\J$ is called \emph{primitive} in case $e$ is the only
+idempotent in $\J_1$ (that is, $e$ cannot be written as the sum $e = u + v$ of
+orthogonal idempotents), and \emph{absolutely primitive} in case it is primitive
+in any scalar extension $\J_K$ of $\J$. A central simple Jordan algebra $\J$ is
+called \emph{reduced} in case $1 = e_1 + \dotsb + e_t$ for pairwise orthogonal absolutely
+primitive idempotents $e_i$ in $\J$. In this case it can be shown that the
+subalgebras $\J_{ii}$ in the Peirce decomposition \tagref(22) are $1$-dimensional
+($\J_{ii} = Fe_i$) and that the subspaces $\J_{ij}$ ($i \ne j$) all have the same dimension.
+If $\J$ is a central simple algebra over $F$, there is a scalar extension $\J_K$
+which is reduced (for example, take $K$ to be the algebraic closure of $F$),
+and it can be shown that the number $t$ of pairwise orthogonal absolutely
+primitive idempotents $e_i$ in $\J_K$ such that $1 = e_1 + \dotsb + e_t$ is unique; $t$ is
+called the \emph{degree} of $\J$.
+
+We list without proof all (finite-dimensional) central simple Jordan
+algebras $\J$ of degree $t$ over $F$ of characteristic $\ne 2$. Recall from the
+\hyperlink{chapter.1}{Introduction} that $\J$ is a \emph{special Jordan algebra} in case $\J$ is isomorphic to
+a subalgebra of an algebra $\A^+$ where $\A$ is associative and multiplication in
+$\A^+$ is defined by
+\begin{myalign}
+\tag{23} &&  x \dotm y = \tfrac{1}{2}(xy + yx).
+\end{myalign}
+We say that each algebra is of \emph{type} $\type A$, $\type B$, $\type C$, $\type D$, or $\type E$ listed below.
+
+\begin{IItemize}
+\item[$\type A_\text{I}$.]  $\J \cong \A^+$  with $\A$ any central simple associative algebra (necessarily
+of dimension $t^2$ over $F$).
+\PG--File: 044.png---\************\***\****\********\----------------------
+
+\item[$\type A_\text{II}$.] Let $\A$ be any involutorial simple associative algebra over $F$, the
+involution being of the second kind (so that the center $\Z$ of $\A$ is a quadratic
+extension of $F$ and the involution induces a non-trivial automorphism on $\Z$
+(Albert, Structure of Algebras, p.~153)). Then $\J \cong \HH(\A)$, the $t^2$-dimensional
+subalgebra of self-adjoint elements in the $2t^2$-dimensional algebra $\A^+$. If
+$\J$ is of type $A_\text{I}$ or $A_\text{II}$, and if $K$ is the algebraic closure of $F$, then
+$\J_K \cong K_t^+$ where $K_t$ is the algebra of all $t \times t$ matrices with elements in $K$.
+
+\item[$\type B$, $\type C$.] Let $\A$ be any involutorial central simple associative algebra over
+$F$ (so the involution is of the first kind). Then $\J \cong \HH(\A)$, the subalgebra
+of self-adjoint elements in $\A^+$. There are two types ($\type B$ and $\type C$) which may be
+distinguished by passing to the algebraic closure $K$ of $F$, so that $\A_K$ is a
+total matrix algebra. In case $\type B$ the (extended) involution on $\A_K$ is
+transposition ($a \to a'$) so that $\A$ has dimension $t^2$ and $\J$ has dimension
+$\frac{1}{2}t(t+1)$ over $F$. In case $\type C$ the (extended) involution on $\A_K$ is $a \to g^{-1}a'g$
+where $g =\begin{pmatrix}
+0 &1_t\\
+-1_t &0
+\end{pmatrix}$ so that $\A$ has dimension $4t^2$ and $\J$ has dimension $2t^2 - t$
+over $F$.
+
+\item[$\type D$.] Let $(x, y)$ be any nondegenerate symmetric bilinear form on a vector
+space $\M$ of dimension $n \ge 2$. Then $\J$ is the vector space direct sum $\J = F1 + \M$,
+multiplication in the $(n+1)$-dimensional algebra $\J$ being defined by $xy = (x,y)1$
+for all $x, y$ in $\M$. Here $t = 2$ ($\dim J \ge 3$).
+
+\item[$\type E$.] The algebra $\C_3$ of all $3 \times 3$ matrices with elements in a Cayley algebra
+$\C$ over $F$ has the \emph{standard involution} $x \to\overline{x'}$ (conjugate transpose). The
+$27$-dimensional subspace $\HH(\C_3)$ of self-adjoint elements
+\begin{myalign}
+\tag{24} && {\begin{pmatrix}
+            \xi_1  &c       &\overline b\\
+            \overline c  &\xi_2  &a\\
+             b      &\overline a  &\xi_3
+           \end{pmatrix}}, &$\xi_i$ in $F$, $a, b, c$ in $\C$,
+\end{myalign}
+\PG--File: 045.png---\************\********\********\*******\--------------
+is a (central simple) Jordan algebra of degree $t = 3$ under the multiplication
+\tagref(23) where $xy$ is the multiplication in $\C_3$ (which is not associative). Then
+$\J$ is any algebra such that some scalar extension $\J_K \cong \HH(\C_3)_K$ ($= \HH((\C_K)_3)$). % missing closing ) added
+\end{IItemize}
+
+The possibility of additional algebras of degree $1$, mentioned in the
+1955 Bulletin \hyperlink{cite.Ref64}{article}, has been eliminated in reference \cite{Ref32} of the
+bibliography of more recent papers. The proof involves use of a two-variable
+identity which is easily seen to be true for special Jordan algebras. But
+any such identity is then true for arbitrary Jordan algebras since it has
+been proved that the free Jordan algebra with two generators is special
+\cite{Ref71, Ref38}, a result which is false for three generators \cite{Ref9}. The identity
+in question is
+\begin{myalign}
+\tag{25} &&   \{aba\}^2 = \left\{a\{ba^2b\}a\right\}  &for all $a, b$ in $\J$,
+\end{myalign}
+where $\{abc\}$ is defined in a Jordan algebra $\J$ by $\allowbreaks\{abc\} = (ab)c + (bc) a -
+(ac)b$, so that $\{aba\} = b(2{R_a}^2 - R_{a^2})$. Hence in $\A^+$ ($\A$ associative) we have $\{aba\} =
+2(b\dotm a)\dotm a - b\dotm a^2 = aba$, so that $\{aba\}^2 = aba^2ba = \left\{a\{ba^2b\}a\right\}$. Then \tagref(25) is
+satisfied in any special Jordan algebra (in particular, the free Jordan
+algebra with two generators) and thus in any Jordan algebra.
+
+Therefore all (finite-dimensional) separable Jordan algebras are known,
+and the Wedderburn decomposition theorem stated in the 1955 Bulletin \hyperlink{cite.Ref64}{article}
+is valid without restriction. Some of the computations employed in the
+original proof may be eliminated \cite{Ref79}.
+
+A central simple Jordan algebra of degree $2$ (that is, of type $\type D$) is a
+commutative quadratic algebra with $1$ ($a^2 - t(a)a + n(a)1 = 0$) having
+nondegenerate norm form $n(a)$, and conversely. For $a = \alpha 1 + x$, $x \in\M$, implies
+$a^2 - t(a)a + n(a)1 = 0$ where $t(a) = 2\alpha$, $n(a) = \alpha^2 - (x, x)$, and $n(a)$ is
+nondegenerate if and only if $(x, y)$ is.
+
+The algebras of types $\type A$, $\type B$, $\type C$ are special Jordan algebras by definition.
+An algebra of type $\type D$ is a subalgebra of $\A^+$, where $\A$ is the (associative)
+\PG--File: 046.png---\************\*****\********\*******\-----------------
+Clifford algebra of $(x, y)$ (Artin, \textit{Geometric Algebra}, p.~186). But algebras
+of type $\type E$ are not special (as we show below), and are therefore called
+\emph{exceptional} central simple Jordan algebras. Exceptional Jordan division
+algebras exist (over suitable fields $F$; but not, for example, over a finite
+field or the field of all real numbers)~\cite{Ref2}. If an exceptional central
+simple Jordan algebra $\J$ is not a division algebra, then it is reduced, and
+$\J$ is isomorphic to an algebra $\HH(\C_3, \Gamma)$ of self-adjoint elements in $\C_3$ under
+a \emph{canonical involution} $x \to \Gamma^{-1}\overline{x'}\Gamma$ where $\Gamma = \diag\{\gamma_1, \gamma_2, \gamma_3\}$, $\gamma_i \ne 0$ in $F$.
+Isomorphism of reduced exceptional simple Jordan algebras is studied in~\cite{Ref8}.
+
+The unifying feature in the list of central simple Jordan algebras
+above is that, for $t > 2$, a reduced central simple Jordan algebra is
+isomorphic to the algebra $\HH(\D_t, \Gamma)$ defined as follows: $\D$ is an alternative
+algebra (of dimension $1$, $2$, $4$, or $8$) with unity element $u$ and involution
+$d \to\overline d$ satisfying $d + \overline d \in Fu$, $d\overline d = n(d)u$, $n(d)$ nondegenerate on $\D$; $\D_t$ is the
+algebra of all $t \times t$ matrices with elements in $\D$; $\Gamma = \diag\{\gamma_1, \gamma_2, \dots, \gamma_t\}$,
+$\gamma_i \ne 0$ in $F$. Then $x \to \Gamma^{-1}\overline{x'}\Gamma$ is a canonical involution in $\D_t$, and the
+set $\HH(\D_t, \Gamma)$ of all self-adjoint elements in $\D_t$ is a subalgebra of $\D_t^+$ (that
+is, we do not need $\A$ associative to define $\A^+$ by \tagref(23)). If $\D$ is associative,
+then $\D_t = \D \otimes_F F_t$ is associative, and $\J \cong \HH(\D_t, \Gamma)$ is a special Jordan
+algebra. If $\D$ is not associative, then $\J \cong \HH(\D_t, \Gamma)$ is not a Jordan algebra
+unless $t = 3$. Hence we have $\J$ of type $\type B$ if $\D = F1$; $\J$ of type $\type A$ if $\D = \Z$
+(type $\type A_\text{I}$ if $\Z = F \oplus F$; type $\type A_\text{II}$ if $\Z$ is a quadratic field over $F$); $\J$ of type
+$\type C$ if $\D = \Q$; $\J$ of type $\type E$ if $t = 3$ and $\D = \C$. The corresponding dimensions for
+$\J$ are clearly $t + \frac12t(t - 1)(\dim\D)$; that is, $\frac12t(t + 1)$ for type $\type B$, $t^2$ for
+type $\type A$, $2t^2 - t$ for type $\type C$, and $27$ for type $\type E$.
+
+\begin{theorem}[9] Any central simple Jordan algebra $\J$ of type $\type E$ is exceptional
+(that is, is not a special Jordan algebra).
+\PG--File: 047.png---\*************\*****\********\*******\----------------
+\end{theorem}
+
+\begin{proof} It is sufficient to prove that $\HH(\C_3)$ is not special. For, if
+$\J$ were special, then $\J \cong \J'\subseteq \A^+$ with $\A$ associative implies $\J_K = K\otimes \J \cong K\otimes \J'\subseteq
+K\otimes\A^+ = (K\otimes\A)^+ = {\A_K}^+$ so that $\HH((\C_K)_3)\cong\J_K$ is special, a contradiction.
+
+Suppose that $\HH(\C_3)$ is special. There is an associative algebra $\A$ (of
+possibly infinite dimension over $F$) such that $U$ is an isomorphism of $\HH(\C_3)$
+into $\A^+$. For $i = 1,2,3$ define elements $e_i$ in $\A$ and $8$-dimensional subspaces
+\[
+\Ss_i=\{d_i\mid d\in \C\}
+\]
+of $\A$ by
+\begin{myalign}
+\tag{26} &&xU = \xi_1e_1 + \xi_2e_2 + \xi_3e_3 +a_1 + b_2 + c_3
+\intertext{for $x$ in \tagref(24); that is, for $\xi_i$ in $F$ and $a, b, c$ in $\C$. (Note that our
+notation is such that we will never use $e$ for an element of $\C$). Then}
+\tag{27} && \Ss = Fe_1 + Fe_2 + Fe_3 + \Ss_1 + \Ss_2 + \Ss_3 = \HH(\C_3)U
+\intertext{is a $27$-dimensional subspace of $\A$. $\Ss$ is a subalgebra of $\A^+$. The mapping
+$V=U^{-1}$ defined on $\Ss$ (not on all of $\A$) is an isomorphism of $\Ss$ onto $\HH(\C_3)$:}
+\tag{28} && (xU\dotm yU)V = x\dotm y &for all $x,y$ in $\HH(\C_3)$.
+\intertext{For our proof we do not need all of the products in $\A^+$ of elements of $\Ss$.
+However, performing the multiplications in $\HH(\C_3)$, we see that \tagref(28) yields}
+\tag{29} &&{e_i}^2 = e_i~(\ne0),&$i=1,2,3$;\\
+\tag{30} &&e_i\dotm e_j = 0, &$i\ne j$;\\
+\tag{31} &&e_i\dotm a_i = 0, &$a$ in $\C$, $i=1,2,3$;\\
+\tag{32} &&e_i\dotm a_j = \tfrac12 a_j, &$a$ in $\C$, $i\ne j$;\\
+\tag{33} &&{u_i}^2 = e_j+e_k, &$i,j,k$ distinct,
+\end{myalign}
+where $u$ is the unity element in $\C$; and
+\begin{myalign}
+\tag{34}&& 2a_i\dotm b_j=(\overline b\,\overline{\vphantom{b}a})_k,\\ % line broken: too long
+&&&$a,b$ in $\C$, $i,j,k$ a cyclic permutation of $1,2,3$.\\
+\end{myalign}
+Now \tagref(29) and \tagref(30) imply that $e_i$ ($i=1,2,3$) are pairwise orthogonal idempotents.
+For $\A$ is associative, so $e_ie_j+e_je_i=0$ for $i\ne j$ implies ${e_i}^2e_j + e_ie_je_i=
+0 = e_ie_je_i + e_j{e_i}^2$, or $e_ie_j=e_je_i$; hence $e_ie_j=0$ for $i\ne j$. By an
+identical proof it follows from \tagref(31) that
+\PG--File: 048.png---\********\******\****\********\-----------------------
+\begin{myalign}
+\tag{31$'$}  &&e_ia_i = a_ie_i = 0, &$i=1,2,3$.
+\end{myalign}
+For $i, j, k$ distinct, \tagref(32) implies $e_ia_j + a_je_i = a_j = e_ka_j + a_je_k$; then
+$fa_j + a_jf = 2a_j$ for the idempotent $f = e_i + e_k$. Hence $f^2 a_j + fa_jf = 2fa_j$,
+so $fa_jf = fa_j$ and similarly $fa_jf = a_jf$; that is, $fa_j = a_jf = a_j$:
+\begin{myalign}
+\tag{35}    &&(e_i + e_k)a_j = a_j = a_j(e_i + e_k), &$i, j, k$ distinct.
+\intertext{Also \tagref(32) implies $e_ia_j = a_j - a_je_i$, so $e_ia_je_i = a_je_i - a_j{e_i}^2 = 0$:}
+\tag{36}   &&e_ia_je_i = 0,  &$i \ne j$.
+\end{myalign}
+
+For any $a$ in $\C$, define
+\begin{myalign}
+\tag{37}   &&a' = e_1a_3u_3  &in $\A$.
+\intertext{Then $(ab)' = e_1(ab)_3u_3 = e_1(\overline{b_1}\,\overline{\vphantom{b}a_2}
+ + \overline{\vphantom{b}a_2}\:\overline{b_1})u_3
+ = e_1\overline{\vphantom{b}a_2}\:\overline{b_1}u_3$ by \tagref(34) and \tagref(31$'$).
+Also \tagref(34) implies $a_3u_1 + u_1a_3 = (\overline{u}\, \overline{a})_2 = \overline{a_2}$ and}
+\tag{38}      &&u_2b_3 + b_3u_2 = \overline{b_1}.
+\end{myalign}
+Hence $(ab)' = e_1(a_3u_1 + u_1a_3)(u_2b_3 + b_3u_2)u_3 = e_1a_3u_1(u_2b_3 + b_3u_2)u_3$ by \tagref(31$'$).
+Now $b_3u_2u_3 = b_3u_2(e_1 + e_3)u_3 = b_3u_2e_1u_3 = (\overline{b_1} - u_2b_3)e_1u_3 =
+ - u_2b_3e_1u_3 =
+-u_2(e_1 + e_3)b_3e_1u_3 = -u_2e_1b_3e_1u_3 = 0$ by \tagref(35), \tagref(31$'$), \tagref(38), and \tagref(36).
+Also $\allowbreaks u_1u_2b_3 = u_1u_2(e_1 + e_2)b_3 = u_1u_2e_1b_3 = (u_3 - u_2u_1)e_1b_3 = u_3e_1b_3$ by \tagref(35),
+\tagref(31$'$), \tagref(34). Hence $(ab)' = e_1a_3u_1u_2b_3u_3 = e_1a_3u_3e_1b_3u_3 = a'b'$.
+
+Clearly the mapping $a \to a'$ is linear; hence it is a homomorphism of
+$\C$ onto the subalgebra $\C'$ of $\A$ consisting of all $a'$. Since $\C$ is simple, the
+kernel of this homomorphism is either $0$ or $\C$; in the latter case $0 = u' =
+e_1{u_3}^2 = e_1(e_1 + e_2) = e_1 \ne 0$ by \tagref(33), and we have a contradiction. Hence
+$a \to  a'$ is an isomorphism. But $\C'$ is associative, whereas $\C$ is not. Hence
+$\HH(\C_3)$ is an exceptional Jordan algebra.
+\end{proof}
+
+Any central simple exceptional Jordan algebra $\J$ over $F$ is a \emph{cubic algebra}:
+for any $x$ in $\J$,
+\PG--File: 049.png---\********\*****\****\********\------------------------
+\begin{myalign}
+\tag{39}  &&x^3 - T(x)x^2 + Q(x)x - N(x)1 = 0,\\ % break not in original
+&&&$T(x)$, $Q(x)$, $N(x)$ in $F$.\\
+\end{myalign}
+Here $x^2x = xx^2$ ($= x^3$) since $\J$ is commutative. It is sufficient to show \tagref(39)
+for $\HH(\C_3)$. But $x$ in \tagref(24) implies \tagref(39) where
+\begin{myalign}
+\tag{40} &&T(x) = \xi_1 + \xi_2 + \xi_3,\\
+\tag{41}&&\!{\begin{aligned} % there must be a better way to do this
+         Q(x) &= \xi_1\xi_2 + \xi_2\xi_3 + \xi_3\xi_1 - n(a) - n(b) - n(c)\\
+      &= \tfrac{1}{2}\left[\left(T(x)\right)^2 - T(x^2)\right],
+\end{aligned}}\\
+\tag{42} &&N(x) = \xi_1\xi_2\xi_3 - \xi_1n(a) - \xi_2n(b) - \xi_3n(c) + t(abc);
+\intertext{if $F$ has characteristic $\ne 3$ (as well as $\ne 2$), formula \tagref(42) may be written
+as}
+\tag{42$'$}&& N(x) =\tfrac{1}{6}\left[\left(T(x)\right)^3 - 3T(x)T(x^2) + 2T(x^3)\right].
+\end{myalign}
+The cubic norm form \tagref(42) satisfies $N\left(\{xyx\}\right) = \left[N(x)\right]^2 N(y)$; that is, $N(x)$
+permits a new type of composition \cite{Ref35, Ref69}.
+
+In the 1955 Bulletin \hyperlink{cite.Ref64}{article}, only passing mention is made in \S7 of the
+relationships between exceptional central simple Jordan algebras $\J$ and
+exceptional simple Lie algebras and groups; relationships which stem from
+the fact that the derivation algebra $\D(\J)$ is a central simple Lie algebra
+of dimension $52$, called an \emph{exceptional Lie algebra of type $\type F$} (corresponding
+to the $52$-parameter complex exceptional Lie group $F_4$)---a proof of this
+for $F$ of characteristic $\ne 2$ appears in \cite{Ref36} (characteristic $\ne 2,\ 3$ in \cite{Ref70}).
+Since 1955 the relationships, including a characterization of Cayley planes
+by means of $\HH(\C_3,\Gamma)$, have been vigorously exploited \cite{Ref18, Ref19, Ref20, Ref35, Ref36,
+Ref37, Ref39, Ref70, Ref76, Ref78, Ref81, Ref82}.
+\PG--File: 050.png---\********\*****\****\********\------------------------
+
+
+
+\chapter{Power-associative Algebras} % V.
+
+We recall that an algebra $\A$ over $F$ is called \emph{power-associative} in case
+the subalgebra $F[x]$ generated by any element $x$ of $\A$ is associative. We
+have seen that this is equivalent to defining, for any $x$ in $\A$,
+\begin{myalign}
+    &&x^1 = x,\qquad x^{i+1} = xx^i     &for $i = 1,2,3,\ldots$, % NB \dots doesn't work here for some reason
+\intertext{and requiring}
+\tag{1}   &&x^i x^j = x^{i+j}           &for $i,j = 1,2,3,\ldots$
+\end{myalign}
+All algebras mentioned in the \hyperlink{chapter.1}{Introduction} are power-associative (Lie
+algebras trivially, since $x^2 = 0$ implies $x^i = 0$ for $i = 2,3,\dots$). We shall
+encounter in V new examples of power-associative algebras.
+
+The most important tool in the study of noncommutative power-associative
+algebras $\A$ is the passage to the commutative algebra $\A^+$. Let $F$ have
+characteristic $\ne 2$ throughout V; we shall also require that $F$ contains at
+least four distinct elements. The algebra $\A^+$ is the same vector space as $\A$
+over $F$, but multiplication in $\A^+$ is defined by
+\begin{myalign}
+\tag{2}  &&x\dotm y = \tfrac{1}{2} (xy + yx)       &for $x, y$ in $\A$,
+\end{myalign}
+where $xy$ is the (nonassociative) product in $\A$. If $\A$ is power-associative,
+then (as in the \hyperlink{chapter.1}{Introduction}) powers in $\A$ and $\A^+$ coincide, and it follows
+that $\A^+$ is a commutative power-associative algebra.
+
+Let $\A$ be power-associative. Then \tagref(2) implies
+\begin{myalign}
+\tag{3}   &&x^2 x = xx^2              &for all $x$ in $\A$
+\Intertext{and}
+\tag{4}   &&x^2 x^2 = x(xx^2)         &for all $x$ in $\A$.
+\intertext{In terms of associators, we have}
+\tag{3$'$}   &&(x,x,x) = 0            &for all $x$ in $\A$
+\PGx--File: 051.png---\*******\*****\****\********\-------------------------
+\Intertext{and}
+\tag{4$'$}   &&(x,x,x^2) = 0         &for all $x$ in $\A$.
+\intertext{Also \tagref(3) may be written in terms of a commutator as}
+\tag{3$''$}   &&[x^2,x] = 0          &for all $x$ in $\A$.
+\intertext{Using the linearization process employed in \chaplink{4}, we obtain from \tagref(3$''$), by
+way of the intermediate identity}
+\tag{3$'''$}  &&2[x\dotm y,x] + [x^2,y] = 0   &for all $x, y$ in $\A$,
+\end{myalign}
+the multilinear identity
+\begin{myalign}[0em]
+\tag{3m}   &&[x\dotm y,z] + [y\dotm z,x] + [z\dotm x,y] = 0   &for all $x, y, z$ in $\A$.\\
+\intertext{Similarly, assuming four distinct elements in $F$, \tagref(4) is equivalent to}
+\tag{4$''$}   &&2(x,x,x\dotm y) + (x,y,x^2) + (y,x,x^2) = 0 &for all $x, y$ in $\A$,\\
+\Intertext{to}
+\tag{4$'''$}&&{\begin{aligned}
+&2(x,y,x\dotm z) + 2(x,z,x\dotm y) + 2(y,x,x\dotm z) + 2(x,x,y\dotm z)\\
+&\quad+2(z,x,x\dotm y) + (y,z,x^2) + (z,y,x^2) = 0
+\end{aligned}}\\
+&&&for all $x, y, z$ in $\A$,\\
+\Intertext{and finally to the multilinear identity}
+\tag{4m}&&{\begin{aligned}
+&(x,y,z\dotm w) + (z,y,w\dotm x) + (w,y,x\dotm z)\\
+        +&(y,x,z\dotm w) + (z,x,w\dotm y) + (w,x,y\dotm z)\\
+        +&(z,w,x\dotm y) + (x,w,y\dotm z) + (y,w,z\dotm x)\\
+        +&(w,z,x\dotm y) + (x,z,y\dotm w) + (y,z,w\dotm x) = 0\end{aligned}}\\
+ &&&for all $x, y, z, w$ in $\A$,\\
+\end{myalign}
+where in each row of the formula \tagref(4m) we have left one of the four elements
+$x$, $y$, $z$, $w$ fixed in the middle position of the associator and permuted the
+remaining three cyclically.
+
+We omit the proof of the fact that, if $F$ has characteristic $0$, then
+identities \tagref(3) and \tagref(4) are sufficient to insure that an algebra is power-associative;
+the proof involves inductions employing the multilinear
+identities \tagref(3m) and \tagref(4m). We omit similarly a proof of the fact that, if
+$F$ has characteristic $\ne 2,3,5$, the single identity \tagref(4) in a commutative
+algebra implies power-associativity. One consequence of this latter fact
+\PG--File: 052.png---\*************\*****\***********\********\------------
+is that in a number of proofs concerning power-associative algebras separate
+consideration has to be given to the characteristic $3$ or $5$ case by bringing
+in associativity of fifth or sixth powers and an assumption that $F$ contains
+at least 6 distinct elements. We shall omit these details, simply by
+assuming characteristic $\ne 3,5$ upon occasion.
+
+An algebra $\A$ over $F$ is called \emph{strictly power-associative} in case every
+scalar extension $\A_K$ is power-associative. If $\A$ is a commutative power-associative
+algebra over $F$ of characteristic $\ne 2,3,5$, then $\A$ is strictly
+power-associative. The assumption of strict power-associativity is
+employed in the noncommutative case, and in the commutative case of
+characteristic $3$ or $5$, when one wishes to use the method of extension of
+the base field.
+
+\ThoughtBreak
+Let $\A$ be a finite-dimensional power-associative algebra over $F$.
+Just as in the proofs of Propositions \hyperlink{Proposition:1}{1} and \hyperlink{Proposition:2}{2}, one may argue that $\A$ has a
+unique maximal nilideal $\N$, and that $\A/\N$ has maximal nilideal $0$. For if
+$\A$ is a power-associative algebra which contains a nilideal $\I$ such that
+$\A/\I$ is a nilalgebra, then $\A$ is a nilalgebra. [If $x$ is in $\A$, then $\overline{x^s} = \overline{x}^s=0$
+for some $s$, so that $x^s = y\in\I$ and $x^{rs}=(x^s)^r=y^r=0$ for some $r$.] Since
+any homomorphic image of a nilalgebra is a nilalgebra, it follows from the
+second isomorphism theorem that, if $\B$ and $\C$ are nilideals, then so is
+$\B+\C$. For $(\B+\C)/\C\cong \B/(\B\cap\C)$ is a nilalgebra, so $\B+\C$ is. This
+establishes the uniqueness of $\N$. It follows as in the proof of Proposition
+\hyperlink{Proposition:2}{2} that $0$ is the only nilideal of $\A/\N$. $\N$ is called the \emph{nilradical} of $\A$, and
+$\A$ is called \emph{semisimple} in case $\N=0$. Of course any anticommutative algebra
+(for example, any Lie algebra) is a nilalgebra, and hence is its own
+nilradical. Hence this concept of semisimplicity is trivial for anticommutative
+algebras.
+\PG--File: 053.png---\*************\********\********\*******\-------------
+
+For the moment let $\A$ be a commutative power-associative algebra, and
+let $e$ be an idempotent in $\A$. Putting $x=e$ in \tagref(4$''$) and using commutativity,
+we have $y(2{R_e}^3-3{R_e}^2+R_e)=0$ for all $y$ in $\A$, or
+\begin{myalign}
+\tag{5} &&2{R_e}^3 -3{R_e}^2+R_e = 0
+\end{myalign}
+for any idempotent $e$ in a commutative power-associative algebra $\A$. As we
+have seen in the case of Jordan algebras in \chaplink{4}, this gives a Peirce
+decomposition
+\begin{myalign}
+\tag{6} &\A&=\A_1 + \A_{1/2} + \A_0
+\intertext{of $\A$ as a vector space direct sum of subspaces $\A_i$ defined by}
+\tag{7} &\A_i&=\{x_i \mid x_i e = ix_i\},&$i=1, 1/2, 0$; $\A$ commutative.\\
+\end{myalign}
+Now if $\A$ is any power-associative algebra, the algebra $\A^+$ is a commutative
+power-associative algebra. Hence we have the Peirce decomposition \tagref(6) where
+\begin{myalign}
+\tag{7$'$}&& \A_i = \{x_i \mid ex_i+x_i e = 2ix_i\},&$i = 1, 1/2, 0$.
+\end{myalign}
+Put $y = z = e$ in \tagref(3m) and let $x=x_i \in\A_i$ as in \tagref(7$'$) to obtain $(2i-1)[x_i,e]=
+0$; that is, $x_i e=ex_i$ if $i\ne 1/2$. Hence \tagref(7$'$) becomes
+\begin{myalign}
+\tag{7$''$}&&{\begin{aligned} % not in original
+ \A_i&=\{x_i \mid ex_i = x_i e = ix_i\},\qquad i = 1,0;\\
+ \A_{1/2} &=\{x_{1/2}\mid ex_{1/2}+x_{1/2}e=x_{1/2}\}
+ \end{aligned}}
+\end{myalign}
+in the Peirce decomposition \tagref(6) of any power-associative algebra $\A$. As
+we have just seen, the properties of commutative power-associative algebras
+may be used (via $\A^+$) to obtain properties of arbitrary power-associative
+algebras.
+
+Let $\A$ be a commutative power-associative algebra with Peirce decomposition
+\tagref(6), \tagref(7) relative to an idempotent $e$. Then $\A_1$ and $\A_0$ are orthogonal subalgebras
+of $\A$ which are related to $\A_{1/2}$ as follows:
+\begin{myalign}
+&\A_{1/2}\A_{1/2}&\subseteq \A_1+\A_0,\\
+\tag{8}& \A_1\A_{1/2}&\subseteq \A_{1/2}+\A_0,\\
+&\A_0\A_{1/2}&\subseteq \A_1+\A_{1/2}.
+\end{myalign}
+\PG--File: 054.png---\*************\*****\********\*******\----------------
+Note that the last two inclusion relations of \tagref(8) are weaker than for
+Jordan algebras in \chaplink{4}. The proofs are similar to those in \chaplink{4}, and are
+given by putting $x = e$, $y=y_j\in\A_j$, $z=x_i\in\A_i$ in \tagref(4$'''$). We omit the
+details except to note that the characteristic $3$ case of orthogonality
+requires associativity of fifth powers.
+
+For $x\in\A_1$, $w\in\A_{1/2}$, we have $wx =(wx)_{1/2} + (wx)_0 \in\A_{1/2}+\A_0$ by \tagref(8).
+Then $w\to(wx)_{1/2}$ is a linear operator on $\A_{1/2}$ which we denote by $S_x$:
+\begin{myalign}
+\tag{9} &&wS_x = (wx)_{1/2} &for $x\in\A_1$, $w\in\A_{1/2}$.\\
+\intertext{If $\HH$ is the (associative) algebra of all linear operators on $\A_{1/2}$, then
+$x\to2S_x$ is a homomorphism of $\A_1$ into the special Jordan algebra $\HH^+$,
+for $x\to S_x$ is clearly linear and we verify}
+\tag{10} &&S_{xy} = S_x S_y + S_y S_x &for all $x,y$ in $\A_1$
+\end{myalign}
+as follows: put $x\in\A_1$, $y\in\A_1$, $z = e$, $w\in\A_{1/2}$ in \tagref(4m) to obtain
+\begin{myalign}(0em)
+\tag{11}&& e\left[y(wx) + x(wy) + w(xy)\right] + w(xy) - 2x(yw) - 2y(xw) = 0,
+\end{myalign}
+since $e(yw)=\frac12(yw)_{1/2}$ implies $x\left[e(yw)\right]=\frac12x(yw)_{1/2}=\frac12x(yw)$ and $y\left[e(xw)\right]=
+\frac12y(xw)$ by interchange of $x$ and $y$. By taking the $\A_{1/2}$ component in \tagref(11),
+we have \tagref(10) after dividing by $3$. Similarly, defining the linear operator
+$T_z$ on $\A_{1/2}$ for any $z$ in $\A_0$ by
+\begin{myalign}
+\tag{12}& wT_z &= (wz)_{1/2} &for $z\in\A_0$, $w\in\A_{1/2}$,
+\Intertext{we have}
+\tag{13}& T_{zy} &= T_z T_y + T_y T_z &for all $z,y$ in $\A_0$,
+\Intertext{and}
+\tag{14}& S_x T_z &= T_z S_x &for all $x$ in $\A_1$, $z$ in $\A_0$.\\
+\end{myalign}
+This is part of the basic machinery used in developing the structure
+theory for commutative power-associative algebras as reported in the 1955 Bulletin
+\hyperlink{cite.Ref64}{article}. The result that simple algebras (actually rings) of degree greater
+\PG--File: 055.png---\*******\********\****\********\----------------------
+than $2$ are Jordan has been extended by the same technique to flexible power-associative
+rings (the conclusion being that $\A^+$ is Jordan) \cite{Ref58}. All
+semisimple commutative power-associative algebras of characteristic $0$ are
+Jordan algebras \cite{Ref51}. The determination of all simple commutative power-associative
+algebras of degree $2$ and characteristic $p > 0$ is still not
+complete \cite{Ref1, Ref24}.
+
+Here we shall develop only as much of the technique as will be required
+in the proof of the following generalization of Wedderburn's theorem that
+every finite associative division ring is a field (Artin, Geometric Algebra,
+p.~37). In \chaplink{4} it was mentioned that exceptional Jordan division algebras do
+exist over suitable fields $F$; however, $F$ cannot be finite in that event and
+we assume this particular case of the following theorem (as well as
+Wedderburn's theorem) in the proof of
+
+\begin{theorem}[10] Let $\D$ be a finite power-associative division ring of
+characteristic $\ne 2,3,5$. Then $\D$ is a field.
+\end{theorem}
+
+For the proof we require an exercise and a lemma.
+
+\begin{exercise} If $u$ and $v$ are orthogonal idempotents in a commutative
+power-associative algebra $\A$, then
+\begin{myalign}
+\tag{15}&&  R_u R_v = R_v R_u.
+\end{myalign}
+(Hint: put $x = u$, $y = v$ in \tagref(4$'''$) and use \tagref(5). After considerable
+manipulation, and use of characteristic $\ne 3,5$, this (together with what one
+obtains by interchanging $u$ and $v$) will yield \tagref(15).)
+\end{exercise}
+
+\begin{lemma} Let $e$ be a principal idempotent in a commutative power-associative
+algebra $\A$ (that is, $\A_0$ in \tagref(7) is a nilalgebra). Then every
+element in $\A_{1/2}$ is nilpotent.
+\PG--File: 056.png---\*******\********\****\********\----------------------
+\end{lemma}
+
+\begin{proof} To obtain \tagref(18$'$) below, one does not need to assume that $e$ is
+principal. For any $w \in\A_{1/2}$ put $x = w$, $y = e$ in \tagref(4$''$) to obtain
+\begin{myalign}
+\tag{16}  && w^3 - w(ew^2)-ew^3 = 0     &for $w$ in $\A_{1/2}$
+\end{myalign}
+Let $x = (w^2)_1 \in\A_1$, $z = (w^2)_0 \in\A_0$, so that $w^2 = x + z$, $ew^2 = x$, $w(ew^2) =
+wx = (wx)_{1/2} + (wx)_0$. Also $\allowbreaks w^3 = wx + wz = (wx)_{1/2} + (wx)_0 + (wz)_{1/2} + (wz)_1$
+so that $ew^3 =  \frac{1}{2}(wx)_{1/2} +  \frac{1}{2}(wz)_{1/2} + (wz)_1$. Then \tagref(16) implies $(wx)_{1/2} =
+(wz)_{1/2}$, or
+\begin{myalign}
+\tag{17}  && wS_x = wT_z  &for any $w$ in $\A_{1/2}$
+\intertext{where $w^2 = x + z$, $x = (w^2)_1$, $z = (w^2)_0$. Now}
+\tag{18}  &&   w{S_x}^k = w{T_z}^k  &for $k = 1,2,3,\ldots$
+\intertext{For \tagref(17) is the case $k = 1$ of \tagref(18) and, assuming \tagref(18), we have $w{S_x}^{k+1} =
+w{S_x}^kS_x = w{T_z}^kS_x = wS_x{T_z}^{k} = w{T_z}^{k+1}$ by \tagref(14). But \tagref(10) and \tagref(13) imply
+$S_{x^k} = 2^{k-1}{S_x}^k$, $T_{z^k} = 2^{k-1}{T_z}^k$, so \tagref(18) yields}
+\tag{18$'$} &&     wS_{x^k} = wT_{z^k}    &for $k = 1,2,3,\ldots$
+\end{myalign}
+where $w$ is any element of $\A_{1/2}$ and $w^2 = x + z$, $x \in\A_1$, $z \in\A_0$.
+
+Now let $e$ be a principal idempotent in $\A$. Then every element $w$ in
+$\A_{1/2}$ is nilpotent. For $z = (w^2)_0$ is nilpotent, and $z^k=0$ for some $k$.
+By \tagref(18$'$) we have $w^{2k+1} = w(w^2)^k = w(x+z)^k = w(x^k + z^k) = wx^k = (wx^k)_{1/2} +
+(wx^k)_0 = wS_{x^k} + (wx^k)_0 = (wx^k)_0$ in $\A_0$ is nilpotent; hence $w$ is
+nilpotent. % possible missing text?
+\end{proof}
+
+We use the Lemma to prove that any finite-dimensional power-associative
+division algebra $\D$ over a field $F$ has a unity element $1$. For $\D^+$ is a
+(finite-dimensional) commutative power-associative algebra without nilpotent
+elements $\ne 0$, so $\D^+$ contains a principal idempotent $e$ (as in \chaplink{4}, $\D^+$ contains
+an idempotent $e$; if $e$ is not principal, there is an idempotent $u \in \D_0^+ = \D_{0,e}^+$,
+\PG--File: 057.png---\*************\*****\****\********\-------------------
+$e' = e + u$ is idempotent, $\dim\D_{1,e}^+ < \dim\D_{1,e'}^+$, and the increasing dimensions
+must terminate). By the Lemma, since $0$ is the only nilpotent element in
+$\D^+$, we have $\D_{1/2}^+ = \D_0^+ = 0,$ $\D^+ = \D_1^+$, $e$ is a unity element for $\D^+$. By \tagref(7$''$) $e$
+is a unity element for $\D$.
+
+\begin{proof}[Proof of Theorem 10]
+We are assuming that $\D$ is a finite power-associative
+division ring. We have seen in \chaplink{2} that this means that $\D$ is
+a (finite-dimensional) division algebra over a (finite) field. Hence we
+have just seen that $\D$ has a unity element $1$, so that $\D$ is an algebra over
+its center. Thus we may as well take $\D$ to be an algebra over its center $F$,
+a finite field. Hence $F$ is perfect (Zariski-Samuel, Commutative Algebra,
+vol.~I, p.~65).
+
+Now $\D^+$ is a Jordan algebra over $F$. For let $x, y$ be any elements of $\D^+$.
+If $x \in F1$, the Jordan identity
+\begin{myalign}
+\tag{19} &&(x \dotm y)\dotm x^2 = x\dotm (y\dotm x^2) &for all $x, y$ in $\D^+$
+\end{myalign}
+holds trivially. Otherwise the (commutative associative) subalgebra $F[x]$
+of $\D^+$ is a finite (necessarily separable) extension of $F$, so there is an
+extension $K$ of $F$ such that $F[x]_K= K \oplus K  \oplus \dotsb  \oplus K$, $x$ is a linear combination
+$x= \xi_1e_1 +\xi_2e_2 + \dotsb +\xi_ne_n$ of pairwise orthogonal idempotents $e_i$ in
+$F[x]_K \subseteq (\D^+)_K$ with coefficients in $K$. In order to establish \tagref(19), it is
+sufficient to establish
+\begin{myalign}[0em]
+\tag{19$'$} &&(e_i\dotm y)\dotm (e_j\dotm e_k) = e_i\dotm \left[y\dotm (e_j\dotm e_k)\right], &$i,j,k = 1,\ldots,n$.\\
+\end{myalign}
+For $j \ne k$, \tagref(19$'$) is obvious; for $j = k$, \tagref(19$'$) reduces to \tagref(15).
+
+Now the radical of $\D^+$ (consisting of nilpotent elements) is $0$.
+Although our proof of the \hyperlink{cor:thm:7}{Corollary} to Theorem 7 is valid only for
+characteristic $0$, we remarked in \chaplink{4} that the conclusion is valid for
+characteristic $\ne 0$. Hence $\D^+$ is a direct sum $\Ss_1 \oplus \dotsb \oplus \Ss_r$ of $r$ simple
+ideals $\Ss_i$, each with unity element $e_i$. The existence of an idempotent
+\PG--File: 058.png---\*************\*****\****\********\-------------------
+$e \ne 1$ in $\D^+$ is sufficient to give zero divisors in $\D$, a contradiction, since
+the product $e(1-e) = 0$ in $\D$. Hence $r = 1$ and $\D^+$ is a simple Jordan algebra
+over $F$. Let $C$ be the center of $\D^+$. Then $C$ is a finite separable extension
+of $F$, $C = F[z]$, $z \in C$ (Zariski-Samuel, ibid, p.~84). If $\D^+ = C = F[z]$, then
+$\D = F[z]$ is a field, and the theorem is established. Hence we may assume
+that $\D^+ \ne C$, so $\D^+$ is a central simple Jordan algebra of degree $t \ge 2$ over
+the finite field $C$ and is of one of the types $\type A$--$\type E$ listed in \chaplink{4}. We are
+assuming that type $\type E$ is known not to occur. The other types are eliminated
+as follows.
+
+Wedderburn's theorem implies that, over any finite field, there are no
+associative central division algebras of dimension $> 1$. Hence, by Wedderburn's
+theorem on simple associative algebras, every associative central simple
+algebra over a finite field is a total matric algebra. Thus we have the
+following possibilities:
+
+\begin{IItemize}
+\item[$\type A_\text{I}$.] $\D^+ \cong {C_t}^+$, $t \ge 2$. Then ${C_t}^+$ contains an idempotent $e_{11} \ne 1$,
+a contradiction.
+
+\item[$\type A_\text{II}$.] $\D^+$ is the set $\HH(\Z_t)$ of self-adjoint elements in $\Z_t$, $\Z$ a quadratic
+extension of $C$, where the involution may be taken to be $a \to g^{-1}\overline {a'}g$ with $g$
+a diagonal matrix. Hence $\HH(\Z_t)$ contains $e_{11} \ne 1$, a contradiction.
+
+\item[$\type B$.] $\D^+ \cong\HH(C_t)$, the involution being $a \to g^{-1} a' g$ with $g$ diagonal;
+hence $\HH(C_t)$ contains $e_{11} \ne 1$, a contradiction.
+
+\item[$\type C$.] $\D^+ \cong\HH(C_{2t})$, the involution being $a \to g^{-1} a' g$, $g = \begin{pmatrix} 0 &1_t\\ -1_t &0 \end{pmatrix}$;
+$\HH(C_{2t})$ contains the idempotent $e_{11}+ e_{t+1, t+1} \ne 1$, a contradiction.
+\end{IItemize}
+
+There remains the possibility that $\D^+$ might be of type $\type D$ (where the
+dimension is necessarily $\ge 3$). The basis $u_1, \dotsc, u_n$ for $\M$ may be normalized
+so that $(u_i, u_j)=0$ for $i \ne j$, $(u_i, u_i) = \alpha_i \ne 0$ in $C$; that is, ${u_i}^2 = \alpha_i 1$,
+$u_i \dotm u_j=0$ for $i \ne j$. Each of the fields $C[u_i]$ is a quadratic extension of
+\PG--File: 059.png---\*******\********\****\********\----------------------
+$C$. But in the sense of isomorphism there is only one quadratic extension of
+the finite field $C$ (Zariski-Samuel, ibid, pp.~73, 83); hence all $\alpha_i$ may be
+taken to be the same nonsquare $\alpha$ in $C$. Let $\beta$ be any element of $C$. Then
+$w = \beta u_1 + u_2 \notin F1$ implies $F[w]$ is isomorphic to $F[u_1]$, so $w^2 = (\beta^2 + 1)\alpha 1 =
+\gamma^2 \alpha 1$, $\gamma$ in $C$; that is, for any $\beta \in C$, there is $\gamma \in C$ satisfying
+\begin{myalign}
+\tag{20}&&  \gamma^2 = \beta^2 + 1.
+\end{myalign}
+Now let $P$ be the prime field of characteristic $p$ contained in $C$. It
+follows from \tagref(20) that all elements in $P$ are squares of elements in $C$. For
+$1$ (also $0$) satisfies this condition, and it follows by induction from \tagref(20)
+that all elements in $P$ do. In particular, $-1 = \beta^2$ for some $\beta \in C$. Then
+$w^2 = (\beta u_1 + u_2)^2 = 0$, a contradiction.
+\end{proof}
+
+\begin{theorem}[11] Let $\A$ be a finite-dimensional power-associative algebra
+over $F$ satisfying the following conditions:
+\begin{myalign}[13em]
+\text{(i)}\DPanchor{thm11:i}&&\text{there is an (associative) trace form $(x,y)$  defined on $\A$;}\\
+\text{(ii)}\DPanchor{thm11:ii}&& (e,e) \ne 0  &for any idempotent $e$ in $\A$;\\
+\text{(iii)}\DPanchor{thm11:iii}&& (x,y) = 0    &if $x \dotm y$ is nilpotent, $x, y$ in $\A$.\\
+\end{myalign}
+Then the nilradical $\N$ of $\A$ coincides with the nilradical of $\A^+$, and is the
+radical $\A^\perp$ of the trace form $(x,y)$. The semisimple power-associative algebra
+$\Ss = \A/\N$ satisfies (i)--(iii) with $(x,y)$ nondegenerate. For any such $\Ss$ we
+have
+\begin{myalign}
+\text{(a)}\DPanchor{thm11:a}&& \text{$\Ss = \Ss_1 \oplus \dotsb \oplus \Ss_t$ for simple $\Ss_i$;}\\
+\text{(b)}\DPanchor{thm11:b}&& \text{$\Ss$ is flexible.}
+\intertext{If $F$ has characteristic $\ne 5$, then}
+\text{(c)}\DPanchor{thm11:c}&& \text{$\Ss^+$ is a semisimple Jordan algebra;}\\
+\text{(d)}\DPanchor{thm11:d}&& \text{$\Ss_i^+$ is a simple (Jordan) algebra, $i = 1,\dotsc,t$.}
+\end{myalign}
+\end{theorem}
+
+\begin{proof} By (\hyperlink{thm11:i}{i}) we know from \chaplink{4} that $\A^\perp$ is an ideal of $\A$. If there
+were an idempotent $e$ in $\A^\perp$, then (\hyperlink{thm11:ii}{ii}) would imply $(e,e) \ne 0$, a contradiction.
+\PG--File: 060.png---\*******\*****\******\********\-----------------------
+Hence $\A^\perp$ is a nilideal: $\A^\perp \subseteq\N$. Conversely, $x$ in $\N$ implies $x\dotm y$ is in $\N$
+for all $y$ in $\A$, so that $(x,y) = 0$ for all $y$ in $\N$ by (\hyperlink{thm11:iii}{iii}), or $x$ is in $\A^\perp$.
+Hence $\N \subseteq\A^\perp$, $\N = \A^\perp$. Any ideal of $\A$ is clearly an ideal of $\A^+$; hence
+any nilideal of $\A$ is a nilideal of $\A^+$, and $\N$ is contained in the nilradical
+$\N_1$ of $\A^+$. But $x$ in $\N_1$ implies $x \dotm y$ is in $\N_1$ for all $y$ in $\A^+$, or $(x,y) = 0$
+by (\hyperlink{thm11:iii}{iii}) and we have $\N_1 \subseteq\A^\perp = \N$.
+
+Now $(x,y)$ induces a nondegenerate symmetric bilinear form $(\overline x, \overline y)$ on
+$\A/\A^\perp = \A/\N$ where $\overline x = x + \N$, etc.; that is, $(\overline x, \overline y) = (x,y)$. Then $(\overline x\:\overline y, \overline z) =
+(\overline{xy}, \overline z) = (xy,z) = (x,yz) = (\overline x,\overline y\:\overline z)$, so $(\overline x,\overline y)$ is a trace form. To show (ii)
+we take any idempotent $\overline e$ in $\A/\N$ and use the power-associativity of $\A$ to
+``lift'' the idempotent to $\A$: $F[e]$ is a subalgebra of $\A$ which is not a nilalgebra,
+so there is an idempotent $u \in F[e] \subseteq Fe + \N$, and $\overline u = \overline e$. Then $(\overline e,\overline e) = (\overline u,\overline u) =
+(u,u) \ne 0$. Suppose $\overline x \dotm \overline y = \overline{x \dotm y}$ is nilpotent. Then some power of $x \dotm y$ is in $\N$,
+$x \dotm y$ is nilpotent, and $(\overline x,\overline y) = (x,y) = 0$, establishing (\hyperlink{thm11:iii}{iii}).
+
+Now let $\Ss$ satisfy (\hyperlink{thm11:i}{i})--(\hyperlink{thm11:iii}{iii}) with $(x,y)$ nondegenerate. Then the
+nilradical of $\Ss$ is $0$, and the hypotheses of \hyperlink{Theorem:7}{Theorem~7} apply. For if $\I^2 = 0$
+for an ideal $\I$ of $\Ss$, then $\I$ is a nilideal, $\I = 0$. We have $\Ss = \Ss_1 \oplus \dotsb \oplus \Ss_t$
+for simple $\Ss_i$; also we know that the $\Ss_i$ are not nilalgebras (for then they
+would be nilideals of $\Ss$), but this will follow from (\hyperlink{thm11:d}{d}).
+
+Now \tagref(3$'''$) implies that $a \dotm z = 0$ where $a = 2[x \dotm y,x] + [x^2,y]$. Since
+$a \dotm z$ is nilpotent, (\hyperlink{thm11:iii}{iii}) implies $\allowbreaks(a,z) = \left((xy)x,z\right) + \left((yx)x,z\right) - \left(x(xy),z\right)
+- \left(x(yx),z\right) + (x^2y,z) - (yx^2,z) = 0$ for all $x, y, z$ in $\Ss$. The properties of
+a trace form imply that
+\begin{myalign}
+\tag{21}&&
+  (xy + yx,xz - zx) = (x^2,zy - yz).
+\end{myalign}
+Interchange $z$ and $y$ to obtain $(xz + zx,xy - yx) = (x^2,yz - zy) = (xy + yx,
+zx - xz)$ by \tagref(21). Add $(xy + yx,xz + zx)$ to both sides of this to obtain
+$(xy,xz + zx) = (xy + yx,zx)$. Then $(xy,xz) = (yx,zx)$, so that
+\begin{myalign}
+\tag{22}
+  &&((xy)x,z) = (x(yx),z)   &for all $x,y,z$ in $\Ss$.
+\end{myalign}
+\PG--File: 061.png---\*******\********\********\*******\-------------------
+Since $(x,y)$ is nondegenerate on $\Ss$, \tagref(22) implies $(xy)x = x(yx)$; that is, $\Ss$ is
+flexible.
+
+To prove (\hyperlink{thm11:c}{c}) we note first that $(x,y)$ is a trace form on $\Ss^+$:
+\begin{myalign}
+\tag{23}  &&   (x\dotm y,z) = (x,y\dotm z)  &   for all $x,y,z$ in $\Ss$.
+\end{myalign}
+Also it follows from \tagref(23), just as in formula \tagref[IV.](14) of \chaplink{4}, that
+\begin{myalign}
+\tag{24}  && (yS_1S_2\dotsm S_h,z) = (y,zS_h\dotsm S_2S_1)
+\end{myalign}
+where $S_i$ are right multiplications of the commutative algebra $\Ss^+$. In the
+commutative power-associative algebra $\Ss^+$ formula \tagref(4$''$) becomes
+\begin{myalign}
+\tag{25} && 4x^2\dotm(x\dotm y) - 2x\dotm\left[x\dotm(x\dotm y)\right] - x\dotm(y\dotm x^2) - y\dotm x^3 = 0.
+\end{myalign}
+Applying the same procedure as above, we write $a$ for the left side of \tagref(25),
+have $a\dotm z = 0$ for all $z$ in $\Ss^+$, so (\hyperlink{thm11:iii}{iii}) implies $\allowbreaks 4\left(x^2\dotm(x\dotm y),z\right) - 2\left(x\dotm\left[x\dotm(x\dotm y)\right],z\right)
+-\left(x\dotm(y\dotm x^2),z\right) - (y\dotm x^3,z) = 0$ or
+\begin{myalign}(0em)
+\tag{26} && (y\dotm z,x^3) + 2\left(x\dotm[x\dotm(x\dotm y)],z\right) = 4(x\dotm y,x^2\dotm z)-(y\dotm x^2,x\dotm z).
+\end{myalign}
+By \tagref(24) the left-hand side of \tagref(26) is unaltered by interchange of $y$ and $z$.
+Hence $4(x\dotm y, x^2\dotm z)-(y\dotm x^2,x\dotm z) = 4(x\dotm z,x^2\dotm y) - (z\dotm x^2,x\dotm y)$ so that (after
+dividing by $5$) we have $(x\dotm y,x^2\dotm z) = (y\dotm x^2, x\dotm z)$. Hence $\left((x\dotm y)\dotm x^2,z\right) =
+\left(x\dotm(y\dotm x^2),z\right)$ for all $z$ in $\Ss$, or $(x\dotm y)\dotm x^2 = x\dotm(y\dotm x^2)$, $\Ss^+$ is a Jordan algebra.
+We know from \chaplink{4} that, since the nilradical of $\Ss^+$ is $0$, $\Ss^+$ is a direct sum
+of simple ideals, but it is conceivable that these are not the ${\Ss_i}^+$ given by
+(\hyperlink{thm11:a}{a}). To see that the simple components of $\Ss^+$ are the ${\Ss_i}^+$ given by (\hyperlink{thm11:a}{a}), we
+need to establish (\hyperlink{thm11:d}{d}).
+
+Let $\T$ be an ideal of ${\Ss_i}^+$; we need to show that $\T$ is an ideal of $\Ss_i$.
+It follows from (\hyperlink{thm11:a}{a}) that $\T$ is an ideal of $\Ss^+$, and is therefore by (\hyperlink{thm11:c}{c}) a
+direct sum of simple ideals of $\Ss^+$ each of which has a unity element. The
+sum of these pairwise orthogonal idempotents in $\Ss^+$ is the unity element $e$ of
+$\T$. Now $e$ is an idempotent in $\Ss^+$ (and $\Ss$), and the Peirce decomposition \tagref(7$''$)
+characterizes $\T$ as
+\begin{myalign}
+\tag{27} &&  \T = \Ss_{1,e} = \{t \in \Ss \mid et = te = t\}.
+\end{myalign}
+\PG--File: 062.png---\*******\*****\********\*******\----------------------
+Let $s$ be any element of $\Ss$. Then flexibility implies $(s,t,e) + (e,t,s) = 0$,
+or
+\begin{myalign}
+\tag{28}  && (st)e - st + ts = e(ts) &for all $t \in\T$, $s \in\Ss$.
+\end{myalign}
+But $\T$ an ideal of $\Ss^+$ implies that $s\dotm t \in\T$, so that $st + ts = e(st + ts) =
+e(st) + (st)e - st + ts$ by \tagref(27) and \tagref(28); that is, $e(st) + (st)e = 2st$, and
+$st$ is in $\T=\Ss_{1,e}$ by \tagref(7$'$). But then $s\dotm t$ in $\T$ implies $ts$ is in $\T$ also;
+$\T$ is an ideal of $\Ss$. Then $\T\subseteq\Ss_i$ is an ideal of $\Ss_i$. Hence the only ideals
+of ${\Ss_i}^+$ are $0$ and ${\Ss_i}^+$. Since ${\Ss_i}^+$ cannot be a zero algebra, ${\Ss_i}^+$ is simple.
+\end{proof}
+
+We list without proof the central simple flexible algebras $\A$ over $F$
+which are such that $\A^+$ is a (central) simple Jordan algebra. These are the
+algebras which (over their centers) can appear as the simple components $\Ss_i$
+in (\hyperlink{thm11:a}{a}) above:
+
+\begin{Itemize}\let\LeftBracket\empty\let\RightBracket.
+\item[1] $\A$ is a central simple (commutative) Jordan algebra.
+
+\item[2] $\A$ is a \emph{quasiassociative} central simple algebra. That is, there is
+a scalar extension $\A_K$, $K$ a quadratic extension of $F$, such that $\A_K$ is
+isomorphic to an algebra $\B(\lambda)$ defined as follows: $\B$ is a central simple
+associative algebra over $K$, $\lambda \ne \frac12$ is a fixed element of $K$, and $\B(\lambda)$ is the
+same vector space over $K$ as $\B$ but multiplication in $\B(\lambda)$ is defined by
+\begin{myalign}
+\tag{29} &&  x \ast y = \lambda xy + (1-\lambda)yx &for all $x,y$ in $\B$
+\end{myalign}
+where $xy$ is the (associative) product in $\B$.
+
+\item[3] $\A$ is a flexible quadratic algebra over $F$ with nondegenerate norm form.
+\end{Itemize}
+
+Note that, except for Lie algebras, all of the central simple algebras
+which we have mentioned in these notes are listed here (associative algebras
+are the case $\lambda=1$ (also $\lambda=0$) in 2; Cayley algebras are included in 3).
+
+\ThoughtBreak
+We should remark that, if an algebra $\A$ contains $1$, any trace form $(x,y)$
+on $\A$ may be expressed in terms of a linear form $T(x)$. That is, we write
+\begin{myalign}
+\tag{30} &  T(x) &= (1,x)     &for all $x$ in $\A$,
+\PGx--File: 063.png---\*******\********\********\*******\-------------------
+\Intertext{and have}
+\tag{31} &  (x,y) &= T(xy)    &for all $x,y$ in $\A$
+\end{myalign}
+since $(x1,y) = (1,xy)$. The symmetry and the associativity of the trace form
+$(x,y)$ are equivalent to the vanishing of $T(x)$ on commutators and associators:
+\begin{myalign}
+\tag{32} &&{\begin{aligned} &T(xy) = T(yx),\\ &T((xy)z) = T(x(yz))
+  \end{aligned}} &for all $x,y,z$ in $\A$. % not aligned in original
+\end{myalign}
+If $\A$ is power-associative, hypotheses (ii) and (iii) of \hyperlink{Theorem:11}{Theorem} 11 become
+\begin{myalign}
+\tag{33} &  T(e) &\ne 0       &for any idempotent $e$ in $\A$,\\
+\Intertext{and}
+\tag{34} &  T(z) &= 0         &for any nilpotent $z$ in $\A$,\\
+\end{myalign}
+the latter being evident as follows: \tagref(34) implies that, if $x\dotm y$ is nilpotent,
+then $0=T(x\dotm y)=(1,x\dotm y)= \frac12(1,xy) + \frac12(1,yx) = \frac12(x,y) + \frac12(y,x) = (x,y)$
+and, conversely, if $z = 1\dotm z$ is nilpotent, then (\hyperlink{thm11:iii}{iii}) implies $T(z) = (1,z) = 0$.
+
+\ThoughtBreak
+A natural generalization to noncommutative algebras of the class of
+(commutative) Jordan algebras is the class of algebras $\J$ satisfying the
+Jordan identity
+\begin{myalign}
+\tag{35} &&  (xy)x^2 = x(yx^2)    &for all $x,y$ in $\J$.
+\end{myalign}
+As in \chaplink{4}, we can linearize \tagref(35) to obtain
+\begin{myalign}
+\tag{35$'$} && (x,y,w\dotm z) + (w,y,z\dotm x) + (z,y,x\dotm w) = 0\\ % break not in original
+&& &for all $w,x,y,z$ in $\J$.
+\end{myalign}
+If $\J$ contains $1$, then $w = 1$ in \tagref(35$'$) implies
+\begin{myalign}
+\tag{36}  &&(x,y,z) + (z,y,x) = 0 &for all $x,y,z$ in $\J$;
+\intertext{that is, $\J$ is \emph{flexible}:}
+\tag{37}  &&(xy)x = x(yx)    &for all $x,y$ in $\J$.
+\end{myalign}
+If a unity element $1$ is adjoined to $\J$ as in \chaplink{2}, then a necessary and
+sufficient condition that \tagref(35$'$) be satisfied in the algebra with $1$ adjoined
+is that both \tagref(35$'$) and \tagref(36) be satisfied in $\J$. Hence we define a \emph{noncommutative
+Jordan algebra} to be an algebra satisfying both \tagref(35) and \tagref(37).
+
+\begin{exercise} Prove: A flexible algebra $\J$ is a noncommutative Jordan
+\PG--File: 064.png---\*******\********\********\*******\-------------------
+algebra if and only if any one of the following is satisfied:
+\begin{myalign}
+\tag{38}  &&(x^2y)x = x^2(yx)    &for all $x,y$ in $\J$;\\
+\tag{39}  &&x^2(xy) = x(x^2y)    &for all $x,y$ in $\J$;\\
+\tag{40}  &&(yx)x^2 = (yx^2)x    &for all $x,y$ in $\J$;\\
+\tag{41}  &&\J^+ \rlap{ is a (commutative) Jordan algebra.}
+\end{myalign}
+\end{exercise}
+
+We see from \tagref(41) that any semisimple algebra (of characteristic $\ne 5$)
+satisfying the hypotheses of \hyperlink{Theorem:11}{Theorem} 11 is a noncommutative Jordan algebra.
+
+Since \tagref(35$'$) and \tagref(36) are multilinear, any scalar extension $\A_K$ of a
+noncommutative Jordan algebra is a noncommutative Jordan algebra. It may
+be verified directly that any noncommutative Jordan algebra is power-associative
+(hence strictly power-associative).
+
+Let $\J$ be any noncommutative Jordan algebra. By \tagref(41) $\J^+$ is a (commutative)
+Jordan algebra, and we have seen in \chaplink{4} that a trace form on $\J^+$ may be given
+in terms of right multiplications of $\J^+$. Our application of this to the
+situation here works more smoothly if there is a unity element $1$ in $\J$, so
+(if necessary) we adjoin one to $\J$ to obtain a noncommutative Jordan algebra
+$\J_1$ with $1$ and having $\J$ as a subalgebra (actually ideal). Then by the proof
+of \hyperlink{Theorem:6}{Theorem} 6 we know that
+\begin{myalign}
+\tag{42} && (x,y) = \trace R^+_{x\dotm y} = \tfrac12\trace (R_{x\dotm y} + L_{x\dotm y})\\ % break not in original
+&& &for all $x,y$ in $\J_1$
+\end{myalign}
+is a trace form on ${\J_1}^+$ where $R^+$ indicates the right multiplication in ${\J_1}^+$;
+hence \tagref(23) holds for all $x,y,z$ in $\J_1$, where $(x,y)$ is the symmetric bilinear
+form \tagref(42). In terms of $T(x)$ defined by \tagref(30), we see that \tagref(23) is equivalent
+to
+\begin{myalign}
+\tag{43}   &&T\left((x\dotm y)\dotm z\right)=T\left(x\dotm(y\dotm z)\right) &for all $x,y,z$ in $\J_1$.
+\intertext{Now \tagref(36) implies}
+\tag{44}   &&L_{xy} - L_yL_x + R_yR_x - R_{yx} = 0  &for all $x,y$ in $\J_1$.
+\end{myalign}
+Interchanging $x$ and $y$ in \tagref(44), and subtracting, we have
+\PG--File: 065.png---\*******\********\********\*******\-------------------
+\begin{myalign}[0em]
+\tag{45}    && R_{[x,y]} + L_{[x,y]} = [R_y,R_x] + [L_x,L_y] &for all $x,y$ in $\J_1$.\\
+\end{myalign}
+Hence $T\left([x,y]\right) = \left(1,[x,y]\right) = \frac12\trace (R_{[x,y]} + L_{[x,y]}) = 0$ by \tagref(42) and \tagref(45).
+Then $xy = x\dotm y + \frac12[x,y]$ implies $T(xy) = T(x\dotm y) = \frac12T(xy) + \frac12T(yx)$, or
+\begin{myalign}
+\tag{46}      &&T(xy) = T(yx) = (x,y)  &for all $x,y$ in $\J_1$
+\end{myalign}
+since $T(x\dotm y) = (1,x\dotm y) = (x,y)$ by \tagref(23). Now \tagref(43) and \tagref(46) imply that $(x,y)$
+is a trace form on $\J_1$. For $0 = 4T\left[(x\dotm y)\dotm z-x\dotm(y\dotm z)\right] = T\left[(xy)z+(yx)z+
+z(xy) + z(yx) - x(yz)-x(zy)-(yz)x-(zy)x\right] = 2T\left[(xy)z-x(yz)-(zy)x +
+z(yx)\right] = 4T\left[(xy)z-x(yz)\right]$ by \tagref(36), so $T\left((xy)z\right) = T\left(x(yz)\right)$, or $(xy,z) = (x,yz)$
+as desired. Then \tagref(42) induces a trace form on the subalgebra $\J$ of $\J_1$.
+
+\begin{corollary}[to Theorem 11] Modulo its nilradical, any finite-dimensional
+noncommutative Jordan algebra of characteristic $0$ is (uniquely) expressible
+as a direct sum $\Ss_1\oplus\dotsb\oplus\Ss_t$ of simple ideals $\Ss_i$. Over their centers these
+$\Ss_i$ are central simple algebras of the following types: (commutative) Jordan,
+quasiassociative, or flexible quadratic.
+\end{corollary}
+
+\begin{proof} Only the verification for $(x,y)$ in \tagref(42) of hypotheses (\hyperlink{thm11:ii}{ii}) and
+(\hyperlink{thm11:iii}{iii}) remains. But these are \tagref[IV.](12) and \tagref[IV.](8) of \chaplink{4}.
+\end{proof}
+
+It was remarked in \chaplink{4} that, although proof was given only for commutative
+Jordan algebras of characteristic $0$, the results were valid for arbitrary
+characteristic ($\ne 2$). The same statement cannot be made here. The trace
+argument in \hyperlink{Theorem:11}{Theorem} 11 can be modified to give the direct sum decomposition
+for semisimple algebras \cite{Ref58}. But new central simple algebras occur for
+characteristic $p$ \cite{Ref52, Ref55}; central simple algebras which are not listed in
+the Corollary above are necessarily of degree one \cite{Ref58} and are \emph{ramified} in
+the sense of \cite{Ref35}.
+
+A finite-dimensional power-associative algebra $\A$ with $1$ over $F$ is
+called a \emph{nodal algebra} in case every element of $\A$ is of the form $\alpha 1 + z$
+\PG--File: 066.png---\*******\************\****\********\------------------
+where $\alpha \in F$ and $z$ is nilpotent, and $\A$ is not of the form $\A = F1 + \N$ for
+$\N$ a nil subalgebra of $\A$. There are no such algebras which are alternative
+(of arbitrary characteristic), commutative Jordan (of characteristic $\ne 2$)
+\cite{Ref32}, or noncommutative Jordan of characteristic $0$. But nodal noncommutative
+Jordan algebras of characteristic $p > 0$ do exist. Any nodal algebra has a
+homomorphic image which is a simple nodal algebra.
+
+Let $\J$ be a nodal noncommutative Jordan algebra over $F$. Since the
+commutative Jordan algebra $\J^+$ is not a nodal algebra, $\J^+ =F1 + \N^+$ where
+$\N^+$ is a nil subalgebra of $\J^+$; that is, $\J= F1 + \N$, where $\N$ is a subspace
+of $\J$ consisting of all nilpotent elements of $\J$, and $x\dotm y \in\N$ for all $x,y \in\N$.
+For any elements $x,y \in\N$ we have
+\begin{myalign}
+\tag{47}  && xy = \lambda 1 + z,  &$ \lambda \in F$, $z \in\N$.
+\end{myalign}
+There must exist $x, y$ in $\N$ with $\lambda\ne 0$ in \tagref(47). Since $\N^+$ is a nilpotent
+commutative Jordan algebra, the powers of $\N^+$ lead to $0$; equivalently,
+the subalgebra $(\N^+)^*$ of $\M(\J^+)$ is nilpotent. Now \tagref(47) implies $yx = -\lambda 1+
+(2x\dotm y-z)$ and $(xy)x = \lambda x + zx = x(yx) = -\lambda x+2x(x\dotm y)-xz$, or
+\begin{myalign}
+\tag{48} &&  x(x\dotm y) = \lambda x + x\dotm z.
+\end{myalign}
+Now $\allowbreaks0=(x,x,y)+(y,x,x)=x^2y-x(\lambda 1+z)+(-\lambda 1+2x\dotm y-z)x-yx^2=
+2x^2y-2\lambda x-2x\dotm z+4(x\dotm y)\dotm x-2x(x\dotm y)-2x^2\dotm y$ implies
+\begin{myalign}
+\tag{49} &&  x^2y=2\lambda x+2x\dotm z-2(x\dotm y)\dotm x+x^2\dotm y
+\end{myalign}
+by \tagref(48). Defining
+\begin{myalign}
+\tag{47$'$}&&   x_iy=\lambda_i1+z_i, &$\lambda_i\in F$, $z_i\in\N$,
+\end{myalign}
+linearization of \tagref(49) gives
+\begin{myalign}
+\tag{49$'$}&
+   (x_1\dotm x_2)y&=\lambda_1x_2+\lambda_2x_1+x_1\dotm z_2+x_2\dotm z_1\\
+&&\qquad -(x_1\dotm y)\dotm x_2-(x_2\dotm y)\dotm x_1+(x_1\dotm x_2)\dotm y.
+\PGx--File: 067.png---\*************\*****\****\********\-------------------
+\end{myalign}
+
+\begin{theorem}[12] Let $\J$ be a simple nodal noncommutative Jordan algebra
+over $F$. Then $F$ has characteristic $p$, $\J^+$ is the $p^n$-dimensional (commutative)
+associative algebra $\J^+ = F[1, x_1,\dots,x_n]$, $x_i^p = 0$, $n \ge 2$, and multiplication
+in $\J$ is given by
+\begin{myalign}
+\tag{50} && fg=f \dotm g+\sum_{i,j=1}^n\frac{\partial f}{\partial x_i}\dotm
+ \frac{\partial g}{\partial x_j}\dotm c_{ij},&$c_{ij} = -c_{ji}$,
+\end{myalign}
+where at least one of the $c_{ij}$ ($= -c_{ji}$) has an inverse.
+\end{theorem}
+
+\begin{proof} Since $\J = F1 + \N$, every element $a$ in $\J$ is of the form
+\begin{myalign}
+\tag{51} &&  a=\alpha 1 + x,  &$\alpha \in F$, $x \in\N$.
+\end{myalign}
+By \tagref(51) every associator relative to the multiplication in $\J^+$ is an
+associator
+\begin{myalign}[0em]
+\tag{52}&& [x_1,x_2,x_3] = (x_1 \dotm x_2)\dotm x_3 - x_1\dotm (x_2\dotm x_3), &$x_i \in\N$.\\
+\end{myalign}
+We shall first show that $\J^+$ is associative by showing that the subspace $\B$
+spanned by all of the associators \tagref(52) is $0$. For any $y$ in $\N$, \tagref(49$'$) implies
+that $(x_1\dotm x_2)y$ is in $\N$, so $\allowbreaks\left[(x_1 \dotm x_2 )\dotm x_3\right]y
+ =\lambda_3x_1 \dotm x_2 + (x_1 \dotm x_2)\dotm z_3 +x_3\dotm \left[\lambda_1x_2\right. +
+\lambda_2x_1 +x_1 \dotm z_2 + x_2\dotm z_1 -(x_1 \dotm y)\dotm x_2 -(x_2 \dotm y)\dotm x_1
+ + \left.(x_1 \dotm x_2)\dotm y\right]-\left[(x_1 \dotm x_2)\dotm y\right]\dotm x_3 -
+(x_3 \dotm y)\dotm (x_1 \dotm x_2) + \left[(x_1 \dotm x_2)\dotm x_3\right]\dotm y$ by \tagref(47$'$) and \tagref(49$'$).
+ Interchange subscripts
+$1$ and $3$, and subtract, to obtain $\allowbreaks[x_1,x_2,x_3]y = [x_1,x_2,x_3] + [x_1,z_2,x_3] +
+[z_1,x_2,x_3] - [x_1\dotm y,x_2,x_3] - [x_1,x_2\dotm y,x_3] + [x_3\dotm y,x_2,x_1] + [x_1,x_2,x_3]\dotm y$, so
+that we have the first inclusion in
+\begin{myalign}
+\tag{53} &&\B\N \subseteq\B + \B\dotm\N, \qquad \N\B \subseteq\B + \B\dotm\N.
+\end{myalign}
+The second part of \tagref(53) follows from $nb = -bn + 2b\dotm n$ for $b$ in $\B$, $n$ in $\N$.
+
+Define an ascending series $\C_0 \subseteq\C_1 \subseteq\C_2 \dotsb$ of subspaces $\C_i$ of $\J$ by
+\begin{myalign}
+\tag{54}&& \C_0 = \B, \qquad \C_{i+1} = \C_i + \C_i\dotm\N.
+\end{myalign}
+Note that all the $\C_i$ are contained in $\N$ (actually in $\N \dotm\N \dotm\N$, since $\B$ is).
+Since $(\N^+)^*$ is nilpotent, there is a positive integer $k$ such that $\C_{k+1} = \C_k$.
+We prove by induction on $i$ that
+\PG--File: 068.png---\*************\******\****\********\------------------
+\begin{myalign}
+\tag{55}&& \C_i\N \subseteq\C_{i+1}, \qquad \N\C_i \subseteq\C_{i+1}.
+\end{myalign}
+The case $i = 0$ of \tagref(55) is \tagref(53). We assume \tagref(55) and prove that $\C_{i+1}\N \subseteq\C_{i+2}$
+as follows: by the assumption of the induction it is sufficient to show
+\begin{myalign}
+\tag{56}&&(\C_i \dotm\N)\N \subseteq\C_{i+1}+\C_{i+1}\dotm\N.
+\end{myalign}
+Now the flexible law \tagref(36) is equivalent to
+\begin{myalign}[0em]
+\tag{57} &&(x\dotm y)z = (yz)\dotm x + (yx)\dotm z - (y\dotm z)x &for all $x, y, z$ in $\J$.\\
+\end{myalign}
+Put $x$ in $\C_i$, $y$ and $z$ in $\N$ into \tagref(57), and use $yz=\mu1+w$, $\mu \in F$, $w \in\N$, to
+see that each term of the right-hand side of \tagref(57) is in $\C_{i+1} + \C_{i+1}\dotm\N$ by the
+assumption \tagref(55) of the induction. We have established \tagref(56), and therefore
+$\C_{i+1}\N \subseteq\C_{i+2}$. Then, as above, $\N\C_{i+1} \subseteq\C_{i+1}\N + \C_{i+1}\dotm\N \subseteq\C_{i+2}$, and we have
+established \tagref(55). For the positive integer $k$ such that $\C_{k+1} = \C_k$, we have
+$\C_k$ an ideal of $\J$. For $\C_k\J=\C_k(F1 +\N) \subseteq\C_k$ by \tagref(55), and similarly $\J\C_k \subseteq\C_k$.
+The ideal $\C_k$, being contained in $\N$, is not $\J$. Hence $\C_k = 0$, since $\J$ is
+simple. But $\B \subseteq\C_k$, so $\B = 0$, $\J^+$ is associative.
+
+An ideal $\I$ of an algebra $\A$ is called a \emph{characteristic} ideal (or $\D$-ideal)
+in case $\I$ is mapped into itself by every derivation of $\A$. $\A$ is called \emph{$\D$-simple}
+if $0$ and $\A$ are the only characteristic ideals of $\A$.
+
+We show next that the commutative associative algebra $\J^+$ is $\D$-simple.
+Interchange $x$ and $y$ in \tagref(36) to obtain
+\begin{myalign}
+\tag{36$'$} &&(y, x, z) + (z, x, y) = 0  &for all $x, y, z$ in $\J$;
+\intertext{interchange $y$ and $z$ in \tagref(36) to obtain}
+\tag{36$''$}  &&(x, z, y) + (y, z, x) = 0 &for all $x, y, z$ in $\J$;
+\intertext{adding \tagref(36) and \tagref(36$'$), and subtracting \tagref(36$''$), we obtain the identity}
+\tag{58} &&[x\dotm y, z] = [x, z]\dotm y + x\dotm [y, z] &for all $x, y, z$ in $\J$,
+\intertext{which is valid in any flexible algebra. Identity \tagref(58) is equivalent to the
+statement that}
+\tag{59} &&D = R_z - L_z &for any $z$ in $\J$
+\end{myalign}
+\PG--File: 069.png---\*******\*******\****\********\-----------------------
+is a derivation of $\J^+$. If $\I$ is an ideal of $\J^+$, then $x\dotm z$ is in $\I$ for
+all $x$ in $\I$, $z$ in $\J$. If, furthermore, $\I$ is characteristic, then $[x,z] = xD$
+is in $\I$, since $D$ in \tagref(59) is a derivation of $\J^+$. Hence $xz = x\dotm z + \frac{1}{2}[x,z]$
+and $zx = x\dotm z - \frac{1}{2}[x,z]$ are in $\I$ for all $x$ in $\I$, $z$ in $\J$; that is, $\I$ is an
+ideal of $\J$. Hence $\J$ simple implies that the commutative associative algebra
+$\J^+$ is $\D$-simple.
+
+It is a recently proved result in the theory of commutative associative
+algebras (see \cite{Ref24}) that, if $\A$ is a finite-dimensional $\D$-simple commutative
+associative algebra of the form $\A = F1 + \R$ where $\R$ is the radical of $\A$, then
+(except for the trivial case $\A = F1$ which may occur at characteristic $0$,
+and which does not give a nodal algebra) $F$ has characteristic $p$ and $\A$ is the
+$p^n$-dimensional algebra $\A = F[1,x_1,\dots,x_n]$,  ${x_i}^p = 0$.
+
+Now any derivation $D$ of such an algebra has the form
+\begin{myalign}
+\tag{60} &&f \to fD= \sum_{i=1}^n \frac{\partial f}{\partial x_i}\dotm a_i,  &$a_i \in\A$,
+\intertext{where the $a_i$ of course depend on the derivation $D$. Then \tagref(59) implies that
+$f \to [f,g]$ is a derivation of $\J^+$ for any $g$ in $\J$. By \tagref(60) we have}
+\tag{61} &&[f,g]= \sum_{i=1}^n \frac{\partial f}{\partial x_i}\dotm a_i(g),  &$a_i(g) \in\J$.
+\intertext{To evaluate the $a_i(g)$, note that $x_iD = [x_i,g] = a_i(g)$ and}
+\tag{62} &&[g,x_i]= \sum_{j=1}^n \frac{\partial g}{\partial x_j}\dotm a_j(x_i).
+\intertext{Then $a_j(x_i) = [x_j,x_i]$ implies
+ $a_i(g) = -[g, x_i] = \displaystyle-\sum_{j=1}^n \frac{\partial g}{\partial x_j}\dotm [x_j, x_i]$, or}
+\tag{63}   &&[f,g] = \sum_{i,j=1}^n \frac{\partial f}{\partial x_i}\dotm \frac{\partial g}{\partial x_j}\dotm [x_i,x_j]
+\end{myalign}
+by \tagref(61), so that $fg = f\dotm g + \frac{1}{2} [f,g]$ implies \tagref(50) where $c_{ij} = \frac{1}{2} [x_i,x_j]$.
+If every $c_{ij}$ were in $\N$, then $\N$ would be a subalgebra of $\A$, a contradiction.
+Hence at least one of the $c_{ij}$ is of the form \tagref(51) with $\alpha \ne 0$, so it has an
+\PG--File: 070.png---\*******\*****\********\*******\----------------------
+inverse, and $n \ge 2$.
+\end{proof}
+
+Not every algebra described in the conclusion of \hyperlink{Theorem:12}{Theorem} 12 is simple
+(see \cite{Ref55}). However, all such algebras of dimension $p^2$ are, and for every
+even $n$ there are simple algebras of dimension $p^n$. There are relationships
+between the derivation algebras of nodal noncommutative Jordan algebras and
+recently discovered (non-classical) simple Lie algebras of characteristic
+$p$ \cite{Ref7, Ref11, Ref17, Ref68}. For a general discussion of Lie algebras of characteristic
+$p$, see \cite{Ref61}.
+\PG--File: 071.png---\*******\*****\********\*******\----------------------
+
+\clearpage
+The following list of papers on nonassociative algebras is intended
+to bring up to date (May 1961) the selective bibliography which appears
+at the end of the expository article \cite{Ref64}.
+
+
+\begin{thebibliography}{99}
+
+\bibitem{Ref1} %1.
+ A.\,A. Albert, \textit{On partially stable algebras}, Trans.\ Amer.\ Math.\ Soc.\
+      vol.~84 (1957), pp.~430--443; \textit{Addendum to the paper on partially
+      stable algebras}, ibid, vol.~87 (1958), pp.~57--62.
+
+\bibitem{Ref2} %2.
+\bysame, \textit{A construction of exceptional Jordan division algebras},
+      Ann.\ of Math.\ (2) vol.~67 (1958), pp.~1--28.
+
+\bibitem{Ref3} %3.
+\bysame, \textit{On the orthogonal equivalence of sets of real symmetric
+      matrices}, J. Math.\ Mech.\ vol.~7 (1958), pp.~219--236.
+
+\bibitem{Ref4} %4.
+\bysame, \textit{Finite noncommutative division algebras}, Proc.\ Amer.\
+      Math.\ Soc.\ vol.~9 (1958), pp.~928--932.
+
+\bibitem{Ref5} %5.
+\bysame, \textit{Finite division algebras and finite planes}, Proc.\
+      Symposia Applied Math., vol.~X, Combinatorial Analysis, Amer.\ Math.\
+      Soc.\ 1960, pp.~53--70.
+
+\bibitem{Ref6} %6.
+\bysame, \textit{A solvable exceptional Jordan algebra}, J. Math.\ Mech.\
+      vol.~8 (1959), pp.~331--337. % [**period after "vol" missing][F1: inserted period]
+
+\bibitem{Ref7} %7.
+A.\,A. Albert and M.\,S. Frank, \textit{Simple Lie algebras of characteristic $p$},
+      Univ.\ e Politec.\ Torino.\ Rend.\ Sem.\ Mat.\ vol.~14 (1954--5), pp.~117--139.
+
+\bibitem{Ref8} %8.
+A.\,A. Albert and N.~Jacobson, \textit{On reduced exceptional simple Jordan
+      algebras}, Ann.\ of Math.\ (2) vol.~66 (1957), pp.~400--417.
+
+\bibitem{Ref9} %9.
+A.\,A. Albert and L.\,J. Paige, \textit{On a homomorphism property of certain
+     Jordan algebras}, Trans.\ Amer.\ Math.\ Soc.\ vol.~93 (1959), pp.~20--29.
+
+\bibitem{Ref10} %10.
+S.\,A. Amitsur, \textit{A general theory of radicals. II.\ Radicals in rings
+     and bicategories}, Amer.\ J. Math.\ vol.~76 (1954), pp.~100--125; \textit{III.\
+     Applications}, ibid, pp.~126--136.
+
+\bibitem{Ref11} %11.
+Richard Block, \textit{New simple Lie algebras of prime characteristic}, Trans.\
+    Amer.\ Math.\ Soc.\ vol.~89 (1958), pp.~421--449.
+
+\bibitem{Ref12} %12.
+R.~Bott and J.~Milnor, \textit{On the parallelizability of the spheres}, Bull.\
+    Amer.\ Math.\ Soc.\ vol.~64 (1958), pp. 87--89.
+
+\bibitem{Ref13} %13.
+Bailey Brown and N.\,H. McCoy, \textit{Prime ideals in nonassociative rings},
+    Trans.\ Amer.\ Math.\ Soc.\ vol.~89 (1958), pp.~245--255.
+
+\PG--File: 072.png---\*******\********\********\*******\-------------------
+\bibitem{Ref14} %14.
+R.\,H. Bruck, \textit{Recent advances in the foundations of euclidean plane
+       geometry}, Amer.\ Math.\ Monthly vol.~62 (1955), No.~7, part~II,
+       pp.~2--17.
+
+\bibitem{Ref15} %15.
+P.\,M. Cohn, \textit{Two embedding theorems for Jordan algebras}, Proc.\ London
+       Math.\ Soc.\ (3) vol.~9 (1959), pp.~503--524.
+
+\bibitem{Ref16} %16.
+J.~Dieudonn\'e, \textit{On semi-simple Lie algebras}, Proc.\ Amer.\ Math.\ Soc.\
+       vol.~4 (1953), pp.~931--932.
+
+\bibitem{Ref17} %17.
+M.\,S. Frank, \textit{A new class of simple Lie algebras}, Proc.\ Nat.\ Acad.\
+       Sci.\ U.S.A. vol.~40 (1954), pp.~713--719.
+
+\bibitem{Ref18} %18.
+Hans Freudenthal, \textit{Sur le groupe exceptionnel $E_7$}, Nederl.\ Akad.\
+       Wet\-ensch.\ Proc.\ Ser.~A. vol.~56 (1953), pp.~81--89; \textit{Sur des invariants
+       caract\'eristiques des groupes semi-simples}, ibid, pp.~90--94; \textit{Sur le
+       groupe exceptionnel $E_8$}, ibid, pp.~95-98; \textit{Zur ebenen Oktavengeometrie},
+       ibid, pp.~195--200.
+
+\bibitem{Ref19} %19.
+\bysame, \textit{Beziehungen der $E_7$ und $E_8$ zur Oktavenebene I.}
+       Nederl.\ Akad.\ Wetensch.\ Proc.\ Ser.~A. vol.~57 (1954), pp.~218--230;
+       \textit{II.} ibid, pp.~363--368; \textit{III.} ibid, vol.~58 (1955), pp.~151--157; \textit{IV.}
+       ibid, pp.~277--285; \textit{V.} ibid, vol.~62 (1959), pp.~165--179; \textit{VI.} ibid,
+       pp.~180--191; \textit{VII.} ibid, pp.~192--201; \textit{VIII.} ibid, pp.~447--465; \textit{IX.}
+       ibid, pp.~466--474.
+
+\bibitem{Ref20} %20.
+\bysame, \textit{Lie groups and foundations of geometry}, Canadian
+       Mathematical Congress Seminar, 1959 (mimeographed).
+
+\bibitem{Ref21} %21.
+Marshall Hall, Jr., \textit{Projective planes and related topics}, Calif.\ Inst.\
+       of Tech, 1954, vi + 77pp.
+
+\bibitem{Ref22} %22.
+\bysame, \textit{An identity in Jordan rings}, Proc.\ Amer.\ Math.\ Soc.\
+       vol.~7 (1956), pp.~990--998.
+
+\bibitem{Ref23} %23.
+L.\,R. Harper, Jr., \textit{Proof of an Identity on Jordan algebras}, Proc.\ Nat.\
+       Acad.\ Sci.\ U.S.A. vol.~42 (1956), pp.~137--139.
+
+\bibitem{Ref24} %24.
+\bysame, \textit{On differentiably simple algebras}, Trans.\ Amer.\
+       Math.\ Soc.\ vol.~100 (1961), pp.~63--72.
+
+\bibitem{Ref25} %25.
+Bruno Harris, \textit{Centralizers in Jordan algebras}, Pacific J. Math, vol.~8
+       (1958), pp.~757--790.
+
+\PG--File: 073.png---\*******\*******\********\*******\--------------------
+
+\bibitem{Ref26}% 26.
+\bysame, \textit{Derivations of Jordan algebras}, Pacific J. Math.\ vol.~9
+(1959), pp.~495--512.
+
+\bibitem{Ref27} %27.
+I.\,N. Herstein, \textit{On the Lie and Jordan rings of a simple associative ring},
+Amer.\ J. Math., vol.~77 (1955), pp.~279--285.
+
+\bibitem{Ref28} %28.
+\bysame, \textit{The Lie ring of a simple associative ring}, Duke Math.\ J.
+vol.~22 (1955), pp.~471--476.
+
+\bibitem{Ref29} %29.
+\bysame, \textit{Jordan homomorphisms}, Trans.\ Amer.\ Math.\ Soc.\ vol~81
+(1956), pp.~331--341.
+
+\bibitem{Ref30} %30.
+\bysame, \textit{Lie and Jordan systems in simple rings with involution},
+Amer.\ J. Math.\ vol.~78 (1956), pp.~629--649.
+
+\bibitem{Ref31} %31.
+N.~Jacobson, \textit{Some aspects of the theory of representations of Jordan
+algebras}, Proc.\ Internat.\ Congress of Math., 1954, Amsterdam, vol.~III,
+pp.~28--33.
+
+\bibitem{Ref32} %32.
+\bysame, \textit{A theorem on the structure of Jordan algebras}, Proc.\ Nat.\
+Acad.\ Sci.\ U.S.A. vol.~42 (1956), pp.~140--147.
+
+\bibitem{Ref33} %33.
+\bysame, \textit{Composition algebras and their automorphisms}, Rend.\ Circ.\
+Mat.\ Palermo (2) vol.~7 (1958), pp.~55--80.
+
+\bibitem{Ref34} %34.
+\bysame, \textit{Nilpotent elements in semi-simple Jordan algebras}, Math.\
+Ann.\ vol.~136 (1958), pp.~375--386.
+
+\bibitem{Ref35} %35.
+\bysame, \textit{Some groups of transformations defined by Jordan algebras.
+I.}, J. Reine Angew.\ Math.\ vol.~201 (1959), pp.~178--195; \textit{II.}, ibid, %[**or , see next line][F1: changed "ibid." to "ibid,"]
+vol.~204 (1960), pp.~74--98; \textit{III.}, ibid, vol.~207 (1961), pp.~61--85.
+
+\bibitem{Ref36} %36.
+\bysame, \textit{Exceptional Lie algebras}, dittoed, 57 pp.
+
+\bibitem{Ref37} %37.
+\bysame, \textit{Cayley planes}, dittoed, 28 pp.
+
+\bibitem{Ref38} %38.
+N.~Jacobson and L.\,J. Paige, \textit{On Jordan algebras with two generators},
+J. Math.\ Mech.\ vol.~6 (1957), pp.~895--906.
+
+\bibitem{Ref39} %39.
+P.~Jordan, \textit{\"Uber eine nicht-desarguessche ebene projecktive Geometrie},
+Abh.\ Math.\ Sem.\ Hamb.\ Univ.\ vol.~16 (1949), pp.~74--76.
+
+\bibitem{Ref40} %40.
+Irving Kaplansky, \textit{Lie algebras of characteristic $p$}, Trans.\ Amer.\ Math.\
+Soc.\ vol.~89 (1958), pp.~149--183.
+
+\bibitem{Ref41} %41.
+Erwin Kleinfeld, \textit{Primitive alternative rings and semisimplicity}, Amer.\
+J. Math.\ vol.~77 (1955), pp.~725--730.
+
+\PG--File: 074.png---\*******\********\********\*******\-------------------
+\bibitem{Ref42} %42.
+\bysame, \textit{Generalization of a theorem on simple alternative rings},
+       Portugal.\ Math.\ vol.~14 (1956), pp.~91--94.
+
+\bibitem{Ref43} %43.
+\bysame, \textit{Standard and accessible rings}, Canad.\ J. Math.\ vol.~8
+       (1956), pp.~335--340.
+
+\bibitem{Ref44} %44.
+\bysame, \textit{Alternative nil rings}, Ann.\ of Math.\ (2) vol.~66 (1957),
+       pp.~395--399.
+
+\bibitem{Ref45} %45.
+\bysame, \textit{Assosymmetric rings}, Proc.\ Amer.\ Math.\ Soc.\ vol~8 (1957),
+       pp.~983--986.
+
+\bibitem{Ref46} %46.
+\bysame, \textit{A note on Moufang-Lie rings}, Proc.\ Amer.\ Math.\ Soc.\
+       vol.~9 (1958), pp.~72--74.
+
+\bibitem{Ref47} %47.
+\bysame, \textit{Quasi-nil rings}, Proc.\ Amer.\ Math.\ Soc.\ vol.~10 (1959),
+       pp.~477--479.
+
+\bibitem{Ref48} %48.
+\bysame, \textit{Simple algebras of type $(1,1)$ %[** I checked on Internet, it seems to be (1,1) and not (l,l)]
+ are associative}, Canadian
+       J. Math.\ vol.~13 (1961), pp.~129--148.
+
+\bibitem{Ref49} %49.
+Max Koecher, \textit{Analysis in reelen Jordan-Algebren}, Nachr.\ Acad.\ Wiss.\
+       G\"ottingen Math.-Phys.\ Kl.\ IIa 1958, pp.~67--74.
+
+\bibitem{Ref50} %50.
+L.\,A. Kokoris, \textit{Power-associative rings of characteristic two}, Proc.\ Amer.\
+       Math.\ Soc.\ vol.~6 (1955), pp.~705--710.
+
+\bibitem{Ref51} %51.
+\bysame, \textit{Simple power-associative algebras of degree two}, Ann.\ of
+       Math.\ (2) vol.~64 (1956), pp.~544--550.
+
+\bibitem{Ref52} %52.
+\bysame, \textit{Simple nodal noncommutative Jordan algebras}, Proc.\ Amer.\
+       Math.\ Soc.\ vol.~9 (1958), pp.~652--654.
+
+\bibitem{Ref53} %53.
+\bysame, \textit{On nilstable algebras}, Proc.\ Amer.\ Math.\ Soc.\ vol.~9
+      (1958), pp.~697--701.
+
+\bibitem{Ref54} %54.
+\bysame, \textit{On rings of $(\gamma,\delta)$-type}, Proc.\ Amer.\ Math.\ Soc.\ vol.~9
+      (1958), pp.~897--904.
+
+\bibitem{Ref55} %55.
+\bysame, \textit{Nodal noncommutative Jordan algebras}, Canadian J. Math.\
+      vol.~12 (1960), pp.~488--492.
+
+\bibitem{Ref56} %56.
+I.\,G. Macdonald, \textit{Jordan algebras with three generators}, Proc.\ London
+      Math.\ Soc.\ (3), vol.~10 (1960), pp.~395--408.
+
+\bibitem{Ref57} %57.
+R.\,H. Oehmke, \textit{A class of noncommutative power-associative algebras},
+      Trans.\ Amer.\ Math.\ Soc.\ vol.~87 (1958), pp.~226--236.
+
+\PG--File: 075.png---\*******\************\********\*******\---------------
+\bibitem{Ref58} %58.
+\bysame, \textit{On flexible algebras}, Ann.\ of Math.\ (2) vol.~68 (1958),
+       pp.~221--230.
+
+\bibitem{Ref59} %59.
+\bysame, \textit{On flexible power-associative algebras of degree two},
+       Proc.\ Amer.\ Math.\ Soc.\ vol.~12 (1961), pp.~151--158.
+
+\bibitem{Ref60} %60.
+Gunter Pickert, \textit{Projektive Ebenen}, Berlin, 1955.
+
+\bibitem{Ref61} %6l.
+Report of a Conference on Linear Algebras, Nat.\ Acad.\ of Sci.\ -- Nat.\
+       Res.\ Council, Publ.\ 502, v + 60 pp. (1957) (N.~Jacobson, \textit{Jordan
+       algebras}, pp.~12--19; Erwin Kleinfeld, \textit{On alternative and right
+       alternative rings}, pp.~20--23; G.\,B. Seligman, \textit{A survey of Lie
+       algebras of characteristic $p$}, pp.~24--32.)
+
+\bibitem{Ref62} %62.
+R.\,L. San Soucie, \textit{Right alternative rings of characteristic two},
+       Proc.\ Amer.\ Math.\ Soc.\ vol.~6 (1955), pp.~716--719.
+
+\bibitem{Ref63} %63.
+\bysame, \textit{Weakly standard rings}, Amer.\ J. Math.\ vol.~79 (1957),
+       pp.~80--86.
+
+\bibitem{Ref64} %64.
+R.\,D. Schafer, \textit{Structure and representation of nonassociative algebras},
+       Bull.\ Amer.\ Math.\ Soc.\ vol.~61 (1955), pp.~469--484.
+
+\bibitem{Ref65} %65.
+\bysame, \textit{On noncommutative Jordan algebras}, Proc.\ Amer.\ Math.\ Soc.\ %, [F1: inconsistent comma removed]
+       vol.~9 (1958), pp.~110--117.
+
+\bibitem{Ref66} %66.
+\bysame, \textit{Restricted noncommutative Jordan algebras of characteristic
+       $p$}, Proc.\ Amer.\ Math.\ Soc.\ vol.~9 (1958), pp.~141--144.
+
+\bibitem{Ref67} %67.
+\bysame, \textit{On cubic forms permitting composition}, Proc.\ Amer.\ Math.\
+       Soc.\ vol.~10 (1959), pp.~917--925.
+
+\bibitem{Ref68} %68.
+\bysame, \textit{Nodal noncommutative Jordan algebras and simple Lie
+       algebras of characteristic $p$}, Trans.\ Amer.\ Math.\ Soc.\ vol.~94 (1960),
+       pp.~310--326.
+
+\bibitem{Ref69} %69.
+\bysame, \textit{Cubic forms permitting a new type of composition}, J. Math.\
+       Mech.\ vol.~10 (1961), pp.~159--174.
+
+\bibitem{Ref70} %70.
+G.\,B. Seligman, \textit{On automorphisms of Lie algebras of classical type},
+       Trans.\ Amer.\ Math.\ Soc.\ vol.~92 (1959), pp.~430--448; \textit{II}, ibid, vol.~94
+       (1960), pp.~452--482; \textit{III}, ibid, vol.~97 (1960), pp.~286--316.
+
+\bibitem{Ref71} %71.
+A.\,I. Shirshov, \textit{On special $J$-rings}, Mat.\ Sbornik N.~S. vol.~38 (80)
+       (1956), pp.~149--166 (Russian).
+
+\bibitem{Ref72} %72.
+\bysame, \textit{Some questions in the theory of rings close to associative},
+       Uspehi Mat.\ Nauk, vol.~13 (1958), no.~6 (84), pp.~3--20 (Russian).
+
+\PG--File: 076.png---\*******\************\********\*******\---------------
+\bibitem{Ref73} %73.
+M.\,F. Smiley, \textit{Jordan homomorphisms onto prime rings}, Trans.\ Amer.\ Math.\
+       Soc.\ vol.~84 (1957), pp.~426--429.
+
+\bibitem{Ref74} %74.
+\bysame, \textit{Jordan homomorphisms and right alternative rings}, Proc.\
+       Amer.\ Math.\ Soc.\ vol.~8 (1957), pp.~668--671.
+
+\bibitem{Ref75} %75.
+T.\,A. Springer, \textit{On a class of Jordan algebras}, Nederl.\ Akad.\ Wetensch.\
+       Proc.\ Ser.~A. vol.~62 (1959), pp.~254--264.
+
+\bibitem{Ref76} %76.
+\bysame, \textit{The projective octave plane}, Nederl.\ Akad.\ Wetensch.\
+       Proc.\ Ser.~A. vol.~63 (1960), pp.~74--101.
+
+\bibitem{Ref77} %77.
+\bysame, \textit{The classification of reduced exceptional simple
+       Jordan algebras}, Nederl.\ Akad.\ Wetensch.\ Proc.\ Ser.~A. vol.~63
+       (1960), pp.~414--422.
+
+\bibitem{Ref78} %78.
+Taeil Suh, \textit{On isomorphisms of little projective groups of Cayley planes},
+       Yale dissertation, 1960.
+
+\bibitem{Ref79} %79.
+E.\,J. Taft, \textit{Invariant Wedderburn factors}, Illinois J. Math.\ vol.~1
+       (1957), pp.~565--573.
+
+\bibitem{Ref80} %80.
+\bysame, \textit{The Whitehead first lemma for alternative algebras},
+       Proc.\ Amer.\ Math.\ Soc.\ vol.~8 (1957), pp.~950--956.
+
+\bibitem{Ref81} %81.
+J.~Tits, \textit{Le plan projectif des octaves et les groupes de Lie exceptionnels},
+       Acad.\ Roy.\ Belgique Bull.\ Cl.\ Sci.\ (5) vol.~39 (1953), pp.~309--329;
+       \textit{Le plan projectif des octaves et les groupes exceptionnels $E_6$ et $E_7$},
+       ibid, vol.~40 (1954), pp.~29--40.
+
+\bibitem{Ref82} %82.
+\bysame, \textit{Sur la trialit\'e et les alg\`ebres d'octaves}, Acad.\ Roy.\ Belg.\
+       Bull.\ Cl.\ Sci.\ (5) vol.~44 (1958), pp.~332--350.
+
+\bibitem{Ref83} %83.
+M.\,L. Tomber, \textit{Lie algebras of types $A$, $B$, $C$, $D$, and $F$}, Trans.\ Amer.\
+       Math.\ Soc.\ vol.~88 (1958), pp.~99--106.
+
+\bibitem{Ref84} %84.
+F.~van der Blij and T.\,A. Springer, \textit{The arithmetics of octaves and of the
+       group $G_2$}, Nederl.\ Akad.\ Wetensch.\ Proc.\ Ser.~A. vol.~62 (1959),
+       pp.~406--418.
+
+\bibitem{Ref85} %85.
+L.\,M. Weiner, \textit{Lie admissible algebras}, Univ.\ Nac.\ Tucum\'an.\ Rev.\ Ser.~A.
+       vol.~II (1957), pp.~10--24.
+
+\end{thebibliography}
+
+% we *do* want the licence to start recto, to emphasise it is an addition
+\cleartorecto
+\pagestyle{licence}
+\setlength\parskip{0pt}\raggedbottom
+\phantomsection
+\par
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+Algebras, by R. D. Schafer
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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% End of the Project Gutenberg EBook of An Introduction to Nonassociative %
+% Algebras, by R. D. Schafer                                              %
+%                                                                         %
+% *** END OF THIS PROJECT GUTENBERG EBOOK NONASSOCIATIVE ALGEBRAS ***     %
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diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
diff --git a/README.md b/README.md
new file mode 100644
index 0000000..f9fbef2
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #25156 (https://www.gutenberg.org/ebooks/25156)