<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">%% Copyright (C) 2003 by Markus Rost.
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\title{Quadratic elements in a central simple algebra of degree four}
\author{Markus Rost}
\address{Department of Mathematics,
  The Ohio State University,
  231 W 18th Avenue,
  Columbus, OH 43210, USA}
%\curraddr{}
\email{rost@math.ohio-state.edu}
\urladdr{http://www.math.ohio-state.edu/\~{}rost}
%\thanks{}
%\subjclass{}
\date{\today}
\date{May 30, 2003}

\begin{document}

%\begin{abstract}
%\end{abstract}

\maketitle

%\notfinished\bigskip
\tableofcontents

\section*{Introduction}

By a theorem of Albert \cite[Theorem~11.9]{Albert:39} a central simple
algebra~$A$ of degree~$4$ contains a biquadratic subalgebra.  The
major point is to prove the existence of a quadratic subalgebra.  See
\cite{Haile:89} for a short proof.

One can show this also using the fact from projective geometry that,
given $4$ generic lines in $3$-space, there are exactly $2$ lines
which meet them all.  Namely, for any (possibly non-split)~$A$ one
considers a maximal commutative separable subalgebra~$H$ of~$A$ and a
generic element~$x\in A$.  Such data determine geometrically $4$~lines
in projective $3$-space:  The subalgebra~$H$ determines
geometrically~$4$ points and for each of these points there is the
line through it and its image under~$x$.  The $2$~lines which meet all
those lines determine a quadratic subalgebra~$L$ of~$A$.

It seems to be rather complicated to give explicit generators for~$L$
in terms of $H$ and~$x$, in particular in characteristic~$2$
(cf.~Section~\ref{sec:more-quadr-elem}).

However there is a way to construct quadratic elements (described in
Sections~\ref{sec:preliminaries}-\ref{sec:constr-quadr-elem}) which is
algebraically simpler.  This construction is symmetric with respect to
reversing the product in $A$ (i.~e., replacing $A$ with~$A^\op$).  It
works smoothly for central simple algebras~$A$ of degree~$4$ with
involution of second kind and the base ring can be any local ring.  A
geometric interpretation is given in Section~\ref{sec:geom-interpr}.

Once a quadratic subalgebra~$L$ is at hand, it easy to extend it to a
biquadratic subalgebra:  Let $y$ be a generic element in the
orthogonal complement of the centralizer of~$L$.  Then $L$ and $y^2$
generate a biquadratic subalgebra.  See
Section~\ref{sec:exist-biqu-subext}.

In the Appendix we have added (mainly for fun) some formulas for the
coefficients of the characteristic polynomial in a central simple
algebra of degree~$4$.  They occurred when trying to establish
Lemma~\ref{lem:s-wedge}.

\section{Preliminaries}
\label{sec:preliminaries}

Let $F$ be a field and let $A$ be a central simple algebra of
degree~$4$ over~$F$.  For the characteristic polynomial of $x\in A$ we
use the notation
\begin{displaymath}
  \Nrd(t-x)=t^4-\Trd(x)t^3+Q(x)t^2-S(x)t+\Nrd(x)
\end{displaymath}
where $N\colon A\to F$ is the reduced norm of~$A$.

For $x\in A$ we consider the linear form
\begin{gather*}
  \Phi(x)\colon \Lambda^3A\to F
  \\
  \Phi(x)( y_1\wedge y_2\wedge y_3)=\Trd( xy_3x y_2x y_1-xy_1x y_2x
  y_3)
\end{gather*}

\begin{lemma}
  \label{lem:s-wedge}
  For $x$, $y\in A$ one has
  \begin{equation*}
    S(xy-yx)=\Phi(x)(1\wedge y\wedge y^2)
  \end{equation*}
\end{lemma}

\begin{proof}
  This amounts to
  \begin{equation*}
%\label{eq:cubic-identity}
    S(xy-yx)=\Trd(x^2y^2xy-x^2yxy^2)
  \end{equation*}
%%   This follows easily from the formula for~$S$ in
%%   Lemma~\ref{lem:Sx+y}.  Alternatively, one may argue as follows.  It
%%   suffices to consider the case $A=M_4(\LQ)$.  One has
  To prove this identity it suffices to consider the case
  $A=M_4(\LQ)$.  One has
  \begin{displaymath}
    6S(u)=\Trd(u)^3-3\Trd(u^2)\Trd(u)+2\Trd(u^3)
  \end{displaymath}
  as can be easily seen from the case of diagonal matrices.  Thus, if
  $\Trd(u)=0$ one has
  \begin{displaymath}
    3S(u)=\Trd(u^3)
  \end{displaymath}
%%  and \eqref{eq:cubic-identity} follows from this with $u=xy-yx$.
  and the claim follows from this with $u=xy-yx$ and from
  $\Trd(vw)=\Trd(wv)$.
\end{proof}

\begin{remark}
  The formula for~$S$ in Lemma~\ref{lem:monsters} in the Appendix
  yields a ``denominator free'' proof of Lemma~\ref{lem:s-wedge}.
\end{remark}

\section{Construction of quadratic elements}
\label{sec:constr-quadr-elem}

Let $H$ be a commutative subalgebra of~$A$ of dimension~$4$ and let
$x\in A$.

The form~$\Phi(x)$ restricts to a linear form $\Lambda^3H\to F$.  We
consider the associated dual map
\begin{displaymath}
  \alpha(x)\colon \Lambda^4H\to H
\end{displaymath}
defined by
\begin{equation}
  \label{eq:define-dual}
  \mu\wedge\alpha(x)(\omega) = \Phi(x)( \mu)\omega
\end{equation}
for $\mu\in \Lambda^3H$ and $\omega\in \Lambda^4H$.

We fix a basis element~$\omega\in\Lambda^4H$ and write
$\alpha=\alpha(x)(\omega)$.

\begin{corollary}
  \label{cor:rho}
  Let
  \begin{displaymath}
    \rho=\rho(x)=x\alpha-\alpha x
  \end{displaymath}
  Then
  \begin{displaymath}
    \Trd(\rho)=S(\rho)=0
  \end{displaymath}
\end{corollary}
\begin{proof}
  Clearly $\Trd(\rho)=0$ since $\rho$ is a commutator.  From
  Lemma~\ref{lem:s-wedge} and~\eqref{eq:define-dual} we get
  \begin{displaymath}
    S(\rho)\omega=1\wedge \alpha\wedge \alpha^2\wedge\alpha=0
  \end{displaymath}
\end{proof}

Let us write $$\Delta(x)=Q(\rho)^2-4\Nrd(\rho)$$

\begin{corollary}
  \label{cor:main}
  Let $$\theta=\theta(x)=\rho^2$$ One has
  \begin{displaymath}
    \theta^2+Q(\rho)\theta+\Nrd(\rho)=0
  \end{displaymath}
  The characteristic polynomial of \/$\theta$ is
  \begin{displaymath}
    \bigl(t^2+Q(\rho)t+\Nrd(\rho)\bigr)^2
  \end{displaymath}
  Let $x\in A$ with $\Delta(x)\neq 0$.  Then $\theta(x)$ generates a
  uniform quadratic subalgebra of~$A$.  (A quadratic subalgebra $L$
  of~$A$ is called uniform if it is separable and if $A$ is as
  $L$-module of constant rank~$8$.)
\end{corollary}
\begin{proof}
  The first claim is clear from Corollary~\ref{cor:rho}.  The second
  claim follows from
  \begin{displaymath}
    \Nrd(t^2-\theta)=\Nrd(t-\rho)\Nrd(t+\rho)=
    \bigl(t^4+Q(\rho)t^2+\Nrd(\rho)\bigr)^2
  \end{displaymath}
  The final claim is then clear as well.
\end{proof}

Suppose $(A,H)=\bigl(M_4(F),D_4(F)\bigr)$ where $D_4(F)$ denotes the
$4\times4$ diagonal matrices.  Write $H=e_1F\oplus e_2F\oplus
e_3F\oplus e_4F$ with $e_i$ the standard idempotents.  Then, with
$\omega=e_1\wedge e_2\wedge e_3\wedge e_4$,
\begin{equation}
  \label{eq:alpha-concrete}
  \alpha(x)(\omega)=\diag
  \begin{pmatrix}
    x_{23}x_{34}x_{42}-x_{24}x_{43}x_{32}
    \\
    x_{31}x_{14}x_{43}-x_{34}x_{41}x_{13}
    \\
    x_{41}x_{12}x_{24}-x_{42}x_{21}x_{14}
    \\
    x_{13}x_{32}x_{21}-x_{12}x_{23}x_{31}
  \end{pmatrix}
\end{equation}
For
\begin{equation}
  \label{eq:explicit}
  x=
  \begin{pmatrix}
    0&amp;1&amp;0&amp;0
    \\
    1&amp;0&amp;0&amp;1
    \\
    0&amp;1&amp;0&amp;0
    \\
    0&amp;0&amp;1&amp;0
  \end{pmatrix}
\end{equation}
one finds $\Delta(x)=1$.
%%   Also, let $u$, $\bar u\in F$ and let
%% \begin{equation}
%%   \label{eq:explicit-bar}
%%   x=
%%   \begin{pmatrix}
%%     0&amp;1&amp;0&amp;0
%%     \\
%%     1&amp;0&amp;\bar u&amp;1
%%     \\
%%     0&amp;u&amp;0&amp;1
%%     \\
%%     0&amp;1&amp;1&amp;0
%%   \end{pmatrix}
%% \end{equation}
%% One finds $\Delta(x)=(u-\bar u)^4$.

\begin{corollary}[Albert]
  \label{cor:albert2}
  There exists a uniform quadratic subalgebra in~$A$.
\end{corollary}

\begin{proof}
  Choose some~$H$, $\omega$ as above with~$H$ separable.  By
  Corollary~\ref{cor:main} it suffices to find $x\in A$ with
  $\Delta(x)\neq 0$.

  As the computation for \eqref{eq:explicit} shows, the
  polynomial~$\Delta(x)$ is nonzero.  Hence if $F$ is infinite, there
  exists $x$ with $\Delta(x)\neq 0$.

  If $F$ is finite, then $A=M_4(F)$ and the claim is clear anyway.
\end{proof}

\begin{remark}
  Note that we have made no assumption on the characteristic on~$F$.
  In fact, everything works over an arbitrary local ring.
\end{remark}

\begin{remark}
  Everything extends easily to the case of a central simple algebra of
  degree~$4$ with involution of second kind.
%%   (Example~\eqref{eq:explicit-bar} might be useful here, at least if
%%   one works over local rings).
\end{remark}

\begin{remark}
  One has $\alpha(x+h)=\alpha(x)$ for $h\in H$.  Thus $x\mapsto
  \alpha(x)$ is a cubic form on~$A/H$ with values
  in~$\Hom(\Lambda^4H,H)$.
\end{remark}

%% \begin{remark}
%%   Note that for the definition of~$\alpha$ we have not assumed
%%   that~$H$ is separable.  More generally, the tensor~$\alpha$ is
%%   defined for any Azumaya algebra~$A$ and a homomorphism $H\to A$ from
%%   a flat algebra~$H$ of rank~$4$.
%% \end{remark}

\section{Existence of biquadratic subextensions}
\label{sec:exist-biqu-subext}

Let $L$ be a uniform quadratic subalgebra in~$A$ and let~$\theta\in L$
be a generator.
\begin{lemma}
  \label{lem:biquadr}
  For $y\in A$ let
  \begin{displaymath}
    \sigma=\sigma(y)=y\theta-\theta y
  \end{displaymath}
  Then
  \begin{displaymath}
    \Trd(\sigma)=S(\sigma)=0
  \end{displaymath}

  Let $$\Delta'(y)=Q(\sigma)^2-4\Nrd(\sigma)$$ Let $y\in A$ with
  $\Delta'(y)\neq 0$.  Then $\theta$ and~$\sigma(y)^2$ generate a
  biquadratic subalgebra of~$A$.

  Suppose $A=M_4(F)$,
  \begin{equation*}
%    \label{eq:theta-special}
    \theta=
    \begin{pmatrix}
      1&amp;0&amp;0&amp;0
      \\
      0&amp;1&amp;0&amp;0
      \\
      0&amp;0&amp;0&amp;0
      \\
      0&amp;0&amp;0&amp;0
    \end{pmatrix}
  \end{equation*}
  and
  \begin{equation*}
%  \label{eq:y-explicit}
    y=
    \begin{pmatrix}
      0&amp;0&amp;1&amp;0
      \\
      0&amp;0&amp;0&amp;0
      \\
      1&amp;0&amp;0&amp;0
      \\
      0&amp;0&amp;0&amp;0
    \end{pmatrix}
  \end{equation*}
  Then $\Delta'(y)=1$.
\end{lemma}

\begin{proof}
%%   This follows along similar lines as for Corollary~\ref{cor:main},
%%   but somewhat simpler.  Details are left to the reader.  (One uses
%%   \eqref{eq:cubic-identity} and $\sigma\theta=\bar\theta\sigma$ where
%%   $\bar\theta$ is the conjugate of~$\theta$ in~$L$.)
  This can be shown by inspection in the split case.  Suppose
  $A=M_4(F)$ and that~$L$ is the subalgebra of matrices of the form
  \begin{displaymath}
    \begin{pmatrix}
      a&amp;0&amp;0&amp;0
      \\
      0&amp;a&amp;0&amp;0
      \\
      0&amp;0&amp;b&amp;0
      \\
      0&amp;0&amp;0&amp;b
    \end{pmatrix}
  \end{displaymath}
  Then a commutator $\sigma=y\theta-\theta y$ is of the form
  \begin{displaymath}
    \sigma=
    \begin{pmatrix}
      0&amp;U
      \\
      V&amp;0
    \end{pmatrix}
  \end{displaymath}
  with $U$, $V\in M_2(F)$.  The characteristic polynomial of~$\sigma$
  is
  \begin{displaymath}
    t^4-t^2\trace(UV)+\det(UV)
  \end{displaymath}
  and one has
  \begin{displaymath}
    \sigma^2=
    \begin{pmatrix}
      UV&amp;0
      \\
      0&amp;VU
    \end{pmatrix}
  \end{displaymath}
  The element~$\sigma^2$ commutes with the elements of~$L$ and
  generates together with~$L$ a biquadratic subalgebra as long as
  $\bigl(\trace(UV)\bigr)^2\neq 4\det(UV)$.  The last claim is easy to
  check.
\end{proof}

\begin{corollary}[Albert]
  \label{cor:albert2x2}
  There exists a biquadratic subalgebra in~$A$.
\end{corollary}

\begin{proof}
%Similar as for Corollary~\ref{cor:albert2}.
  If $F$ is finite, then $A=M_4(F)$ and the claim is clear.  Otherwise
  one chooses $L$ and $\theta$ as above and picks~$y$ with
  $\Delta'(y)\neq0$ (cf.~Corollary~\ref{cor:albert2} and its proof).
\end{proof}

\section{More quadratic elements}
\label{sec:more-quadr-elem}

We have seen that for generic~$x$ the element~$\theta(x)$ generates a
uniform quadratic subalgebra~$L$.  The centralizer~$Q$ of~$L$ is a
quaternion algebra over~$L$.

In the following we give a description of generators for~$Q$.  This
appears to be somewhat complicated.

We assume $\car F\neq 2$.

%% {\tt Caution:  The might be sign errors in the following.}

For $x\in A$ consider the linear form
\begin{gather*}
  \varphi(x) \colon \Lambda^3H\to F
  \\
  \varphi(x)(\mu)=\sum_{s\in \mathcal S_3}\sgn(s) \Trd( x^2h_{s(3)}x
  h_{s(2)}x h_{s(1)}) -2\Trd(x) \Phi(x)(\mu)
\end{gather*}
with $\mu=h_1\wedge h_2\wedge h_3$, $h_i\in H$, and let
\begin{displaymath}
  \beta(x) \colon \Lambda^4H\to H
\end{displaymath}
be the linear map with
\begin{equation*}
  \mu \wedge \beta(x)(\omega) = \varphi(x)( \mu)\omega
\end{equation*}
for $\mu\in \Lambda^3H$ and $\omega\in \Lambda^4H$.

Let further
\begin{gather*}
  \psi(x),\,\psi'(x)\colon \Lambda^4H\to A
  \\
  \psi(x)=x\alpha(x)+\frac12\beta(x)
  \\
  \psi'(x)=\alpha(x)x+\frac12\beta(x)
\end{gather*}

We fix a basis element~$\omega\in\Lambda^4H$ and write
$\psi=\psi(x)(\omega)$, $\psi'=\psi'(x)(\omega)$.

Let $Q$ be the subalgebra of~$A$ generated by $\psi$ and~$\psi'$.
Note that $\rho=\psi-\psi'\in Q$.  Let $L\subset Q$ be the subalgebra
generated by $\theta=\rho^2$.

\begin{lemma}
  One has $\psi^2={\psi'}^2\in F$.  The algebra~$L$ is in the center
  of~$Q$.

  Suppose~$H$ is separable.  Then for generic~$x$, $L$ is a uniform
  quadratic subalgebra of~$A$ and $Q$ is the centralizer of~$L$
  in~$A$.
\end{lemma}

\begin{proof}
  I know only a tiresome check in the split case
  $(A,H)=\bigl(M_4(F),D_4(F)\bigr)$.
\end{proof}

\section{Geometric interpretation}
\label{sec:geom-interpr}

There is an interpretation of our construction in terms of a classical
theorem in projective geometry:  For $4$ generic lines in $\LP^3$
there exist exactly $2$ lines meeting them all.

Let $V$ be a $4$-dimensional vector space, let $A=\End(V)$, let
$V=V_1\oplus V_2\oplus V_3\oplus V_4 $ be a decomposition into
$1$-dimensional subspaces and $H$ be the subalgebra of elements $h\in
A$ with $h(V_i)\subset V_i$, $i=1$, $2$, $3$,~$4$.  Let further $x\in
A$ be a generic element.

In the projective space $\LP(V)$ let $L_i$ be the line through the
points $[V_i]$, $[xV_i]$ and let $K_1$, $K_2$ be the lines meeting
every~$L_i$.

The lines $K_1$, $K_2$ are in fact given by the $2$ $2$-dimensional
eigenspaces of the element~$\psi(x)$ above.

Let $V^*=V^*_1\oplus V^*_2\oplus V^*_3\oplus V^*_4 $ be the dual
decomposition of the dual space (i.e., $V^*_i(V_j)=\delta_{ij}F$).

In the projective space $\LP(V^*)$ let $L_i'$ be the line through the
points $[V_i^*]$, $[V_i^*x]$ and let $K_1'$, $K_2'$ be the lines
meeting every~$L_i'$.

The lines $K_1'$, $K_2'$ are in fact given by the $2$ $2$-dimensional
eigenspaces of the element~$\psi'(x)$ above.

Every line $L$ in $\LP(V^*)$ defines a line $L^*$ in $\LP(V)$.  Let
$S_1$, $S_2$ be the two lines which meet all of $K_1$, $K_2$,
$(K_1')^*$, $(K_2')^*$.

The lines $S_1$, $S_2$ are in fact given by the $2$ $2$-dimensional
eigenspaces of the element~$\theta(x)$ above.

There is also an interpretation of the element~$\rho=\rho(x)$.  Since
$\rho^2$ is the identity on~$S_i$, the element~$\rho$ is an involution
on each~$S_i$.  An involution on a line is given by a pair of points.
How to get a pair of points on~$S_i$?  We have two pairs of points
already:  the intersections of $S_i$ with $K_1$, $K_2$ and with
$(K_1')^*$, $(K_2')^*$, respectively.  Now a pair of points on a line
gives a point in the $2$-fold symmetric product of the line,
a~$\LP^2$.  The diagonal embedding of the line to that~$\LP^2$
identifies it with a conic.  Thus the two pairs of points determine a
line in~$\LP^2$, which intersects the conic in a pair of points.  This
yields certainly the pair given by~$\rho$ (I haven't really checked
this).

\section*{Appendix}

Lemma~\ref{lem:s-wedge} follows also from the additivity rule for~$S$
in the following Lemma.

%% \begin{lemma}
%%   For $x$, $y\in A$ one has
%%   \begin{align*}
%%     2Q(u)&amp;=\Trd(u)^2-\Trd(u^2)
%%     \\
%%     6S(u)&amp;=\Trd(u)^3-3\Trd(u^2)\Trd(u)+2\Trd(u)
%%   \end{align*}
%% \end{lemma}
\begin{lemma}
  \label{lem:monsters}
  For $x$, $y\in A$ one has
  \begin{align*}
    T(x+y)&amp;=\Trd(x)+\Trd(y)
    \\
    Q(x+y)&amp;=Q(x)+\Trd(x)\Trd(y)-\Trd(xy)+Q(y)
    \\
    S(x+y)&amp;=S(x) +Q(x)\Trd(y)-\Trd(x)\Trd(xy)+\Trd(x^2y)
    \\
    &amp;\qquad +\Trd(x)Q(y)-\Trd(xy)\Trd(y)+\Trd(xy^2) + S(y)
    \\
    N(x+y)&amp;=N(x) + S(x)T(y) - Q(x)T(xy) +T(x)T(x^2y)-T(x^3y)
    \\
    &amp;\qquad +Q(x)Q(y) -T(x)T(xy)T(y) +T(x)T(xy^2)+T(x^2y)T(y)
    \\
    &amp;\qquad\qquad +Q(xy)-T(x^2y^2)
    \\
    &amp;\qquad + T(x)S(y) - T(xy)Q(y) +T(xy^2)T(y)-T(xy^3) + N(y)
  \end{align*}
\end{lemma}

\begin{proof}
  In the power series ring~$A\pseriesr t$ one has
%%   \begin{gather*}
%%     (1+tx)\inv\bigl(1+t(x+y)\bigr)(1+ty)\inv
%%     = \\ 1 -t^2 xy +t^3 (x^2y+xy^2)
%%     -t^4 (x^3y+x^2y^2+xy^3) + \text{higher terms}
%%   \end{gather*}
%%
%%     (1+tx)\inv\bigl(1+t(x+y)\bigr)(1+ty)\inv
%%     &amp;= \bigl(1+t^2(1+tx)\inv xy(1+ty)\inv\bigr)
%%     \\ &amp;= 1 +t^2 xy + t^3 (x^2y+xy^2) + \cdots
%%     \frac 1{1+tx}\bigl(1+t(x+y)\bigr)\frac 1{1+ty} &amp;= 1-t^2\frac
%%     x{1+tx}\frac y{1+ty}
%%     \\ &amp;= 1 -t^2 xy + t^3 x(x+y)y
%%     \\ &amp;\qquad- t^4x(x^2+xy+y^2)y +\cdots
  \begin{displaymath}
    1+t(x+y)= (1+tx) \bigl[1-t^2\frac x{1+tx}\frac y{1+ty}\bigr](1+ty)
  \end{displaymath}
  The middle term expands as follows:
  \begin{displaymath}
    1-t^2\frac x{1+tx}\frac y{1+ty} = 1 -t^2 xy + t^3 x(x+y)y -
    t^4x(x^2+xy+y^2)y +\cdots
  \end{displaymath}
  Taking norms gives in $F\pseriesr t/(t^5)$
  \begin{displaymath}
    \begin{split}
      N\bigl(1+t(x+y)\bigr)=N(1+tx)N(1+ty)\bigl[1-t^2T(xy)
      +t^3T(x^2y+xy^2) \\ {}+ t^4
      \bigl(Q(xy)-T(x^3y+x^2y^2+xy^3)\bigr)\bigr]
    \end{split}
  \end{displaymath}
  Multiplying out yields the claims.
\end{proof}

\begin{remark}
  This argument works for central simple algebras of any degree.  By
  further expansion one gets expressions for the coefficients of the
  characteristic polynomial of~$x+y$ as integral polynomials in the
  coefficients of the characteristic polynomials of noncommutative
  monomials in~$x$,~$y$.
%%  Can one make the resulting formulae more precise?
\end{remark}

%% We summarize some basic rules.

%% \begin{lemma}
%%   \label{lem:Sx+y}
%%   For $x$, $y\in A$ one has
%% %%   $T(xy)=T(yx)$, $Q(xy)=Q(yx)$, $S(xy)=S(yx)$.
%%   \begin{displaymath}
%%     T(xy)=T(yx), \quad Q(xy)=Q(yx), \quad S(xy)=S(yx)
%%   \end{displaymath}
%%   Moreover
%%   \begin{align*}
%%     T(x+y)&amp;=\Trd(x)+\Trd(y)
%%     \\
%%     Q(x+y)&amp;=Q(x)+\Trd(x)\Trd(y)-\Trd(xy)+Q(y)
%%     \\
%%     S(x+y)&amp;=S(x) +Q(x)\Trd(y)-\Trd(x)\Trd(xy)+\Trd(x^2y)
%%     \\
%%     &amp;\qquad +\Trd(x)Q(y)-\Trd(xy)\Trd(y)+\Trd(xy^2) + S(y)
%%   \end{align*}
%% \end{lemma}

%% \begin{proof}
%%   The first claim follows from the identity of polynomials
%% %% There are better reasons.
%% %% How to define the reduced norm in the beginning?
%%   \begin{displaymath}
%%     \Nrd(t-xy)N(x)=\Nrd(tx-xyx)=N(x)\Nrd(t-yx)
%%   \end{displaymath}
%%   Further, in the power series ring~$A\pseriesr t$ one has
%%   \begin{align*}
%% %%     (1+tx)\inv\bigl(1+t(x+y)\bigr)(1+ty)\inv
%% %%     &amp;= \bigl(1+t^2(1+tx)\inv xy(1+ty)\inv\bigr)
%% %%     \\ &amp;= 1 +t^2 xy + t^3 (x^2y+xy^2) + \cdots
%%     \frac 1{1+tx}\bigl(1+t(x+y)\bigr)\frac 1{1+ty} &amp;= 1-t^2\frac
%%     x{1+tx}\frac y{1+ty}
%%     \\ &amp;= 1 -t^2 xy + t^3 x(x+y)y - \cdots
%%   \end{align*}
%%   Taking norms gives in $F\pseriesr t/(t^4)$
%%   \begin{displaymath}
%%     \begin{split}
%%       N\bigl(1+t(x+y)\bigr)=N(1+tx)N(1+ty)\bigl[1-t^2T(xy)
%%       +t^3T(x^2y+xy^2)\bigr]
%%     \end{split}
%%   \end{displaymath}
%%   Multiplying out yields the claims.
%% \end{proof}

%% \begin{remark}
%%   Here is another way to deduce identities
%%   like~\eqref{eq:cubic-identity}.  Let $u$, $v\in A$.  Then in the
%%   power series ring~$A\pseriesr t$ one has
%%   \begin{gather*}
%%     (1+tu)\inv\bigl(1+t(u+v)\bigr)(1+tv)\inv
%%     = \\ 1 -t^2 uv +t^3 (u^2v+uv^2)
%%     -t^4 (u^3v+u^2v^2+uv^3) + \text{higher terms}
%%   \end{gather*}
%%   Taking (reduced) norms gives in $F\pseriesr t/(t^5)$
%%   \begin{displaymath}
%%     \begin{split}
%%       N\bigl(1+t(u+v)\bigr)=N(1+tu)N(1+tv)\bigl[1-t^2T(uv)
%%       +t^3T(u^2v+uv^2) \\ {}+ t^4
%%       \bigl(Q(uv)-T(u^3v+u^2v^2+uv^3)\bigr)\bigr]
%%     \end{split}
%%   \end{displaymath}
%% %%   \begin{gather*}
%% %%     N\bigl(1+t(u+v)\bigr)=N(1+tu)N(1+tv)\cdot {}\qquad\qquad
%% %%     \\
%% %%     \bigl[1-t^2T(uv)
%% %%     +t^3T(u^2v+uv^2) + t^4
%% %%     \bigl(Q(uv)-T(u^3v+u^2v^2+uv^3)\bigr)\bigr]
%% %%   \end{gather*}
%% %%
%% %%   \begin{align*}
%% %%     N\bigl(1+t(u+v)\bigr)&amp;=N(1+tu)N(1+tv)\cdot {}
%% %%     \\
%% %%     &amp;\qquad \bigl[1-t^2T(uv)
%% %%     +t^3T(u^2v+uv^2)  \\ &amp;\quad\qquad +t^4
%% %%     \bigl(Q(uv)-T(u^3v+u^2v^2+uv^3)\bigr)\bigr]
%% %%   \end{align*}
%%   This readily expresses the characteristic polynomial of $u+v$ in
%%   terms of the coefficients of the characteristic polynomials of
%%   certain monomials in $u$,~$v$.

%%   In order to get~\eqref{eq:cubic-identity} let $u=xy$ and $v=-yx$.
%%   Then
%%   \begin{displaymath}
%%     N(1+tv)=N(1-tu)
%%   \end{displaymath}
%%   Calculating $\amod t^4$ one finds
%%   \begin{displaymath}
%%     N(1+tu)N(1+tv)=N(1-t^2u^2)=1-t^2T(u^2)%+t^4Q(u^2)
%%   \end{displaymath}
%%   and
%%   \begin{align*}
%%     N\bigl(1+t(xy-yx)\bigr)=1+t^2T(x^2y^2-(xy)^2)
%%     +t^3T(x^2y^2xy-x^2yxy^2)
%%   \end{align*}
%%   which yields in particular~\eqref{eq:cubic-identity}.
%% \end{remark}

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\begin{thebibliography}{1}

  \bibitem{Albert:39}
  A.~Albert, \emph{Structure of algebras}, American Mathematical
  Society Colloquium Publications, vol.~24, American Mathematical
  Society, Providence, RI, 1961, Revised printing of the 1939 edition.

  \bibitem{Haile:89}
  D.~Haile, \emph{A useful proposition for division algebras of small
  degree}, Proc. Amer. Math. Soc. \textbf{106} (1989), no.~2,
  317--319.

\end{thebibliography}

\end{document}
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