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\title{\textbf{On a Polynomial Identity for $n \times n$ Matrices}\thanks{This research is supported by NSERC of Canada and MPI of Italy.}}

\author{A. Giambruno\\
\textit{ Dipartimento di Matematica, Universit\`a di Palermo,} \\ \textit{90123 Palermo, Italy}
%\and
\\[8pt]
S. K. Sehgal\\
\textit{ Department of Mathematics, University of Alberta,} \\ \textit{Edmonton, Alberta, Canada T6G 2G1}}

\date{Received December 1987\\[4pt]
Published in: \textit{Journal of Algebra} \textbf{126} (1989), pp.~451--453.}

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\maketitle

\noindent
Digitized with GPT-5.5 based on \href{https://doi.org/10.1016/0021-8693(89)90312-8}{a scan} from \href{https://www.elsevier.com/about/policies-and-standards/open-access-licenses/elsevier-user}{Elsevier Open Archive} \\(original scan: \href{https://doi.org/10.1016/0021-8693(89)90312-8}{doi:10.1016/0021-8693(89)90312-8}). \\
This is an unofficial re-edition of an article that appeared in an Elsevier publication. Elsevier has not endorsed this re-edition.\\
Proofread by Darij Grinberg 27 May 2026.

\begin{abstract}
We prove that the polynomial
\[
 h_k(x_1,\ldots,x_k,y_1,\ldots,y_k)
 = \sum_{\sigma,\tau\in S_k} (\sgn \sigma\tau)
 x_{\sigma(1)} y_{\tau(1)} \cdots x_{\sigma(k)} y_{\tau(k)}
\]
vanishes on $n\times n$ matrices over a commutative ring for $k=2n$ and for no smaller value of $k$.
\end{abstract}

\bigskip

\noindent
Let $C$ be a commutative ring with 1 and $M_n(C)$ the ring of $n\times n$ matrices over $C$.
If $\{x_1,\ldots,x_k,\ldots\}$ and $\{y_1,\ldots,y_k,\ldots\}$ are two distinct sets of non-commuting variables, for each $k\geq 1$ we define the polynomial
\[
 h_k(x_1,\ldots,x_k,y_1,\ldots,y_k)
 = \sum_{\sigma,\tau\in S_k} (\sgn \sigma\tau)
 x_{\sigma(1)} y_{\tau(1)} \cdots x_{\sigma(k)} y_{\tau(k)},
\]
where $S_k$ is the symmetric group of degree $k$.

It is clear that, for some $k$, $h_k$ is a polynomial identity for\footnote{\textit{Comment by DG:} ``$M_n(C)$'' was ``$M_k(C)$'' in the original, but the intended meaning was likely ``$M_n(C)$''.} $M_n(C)$; in fact, since $h_k$ is alternating in the $x_i$ (and also in the $y_i$), $h_{n^2+1}$ is an identity\footnote{\textit{Comment by DG:} Corrected ``$h_{n^2}$'' to ``$h_{n^2+1}$''.} for $M_n(C)$.

The purpose of this note is to prove that $2n$ is the smallest value of $k$ for which $h_k$ is a polynomial identity for $M_n(C)$. This answers a question of Formanek.

We have the following:

\begin{theorem*}
$h_{2n}$ is a polynomial identity for $M_n(C)$. Moreover if $h_k$ is a polynomial identity for $M_n(C)$, then $k\geq 2n$.
\end{theorem*}

Our approach will be based on a proof of the Amitsur--Levitzki theorem given by Rosset in \cite{Rosset}.

Before proceeding to the proof of this theorem we need some preliminaries.

Let $E$ be the exterior algebra on a $4n$-dimensional vector space $V$ over the field of rational numbers $\QQ$ and let $\{v_1,\ldots,v_{4n}\}$ be a basis of $V$ over $\QQ$. $E$ may be viewed as the free algebra on $V$ modulo the relations $v_i v_j=-v_jv_i$. We write $E=E_0+E_1$, where $E_0$ is the subalgebra generated by $\QQ$ and the monomials in the $v_i$ of even degree and $E_1$ is the space generated by the monomials of odd degree.

In the following proposition we study some properties of the algebra
\[
 M_k(E)\simeq M_k(\QQ)\otimes_{\QQ} E.
\]

\begin{proposition*}
\begin{enumerate}[label=\textup{(\roman*)}, leftmargin=2.2em]
\item If $U\in M_k(E_0)$ is such that
$\tr(U)=\tr(U^2)=\cdots=\tr(U^k)=0$, then $U^k=0$.
\item If $U,T\in M_k(E_1)$, then $\tr(UT)=-\tr(TU)$.
\end{enumerate}
\end{proposition*}

\begin{proof}
Since $E_0$ is a commutative algebra over $\QQ$, (i) follows from Newton's formulas for symmetric functions (see \cite{Rosset}\footnote{\textit{Comment by DG:} See Corollary 4.1 \textbf{(b)} in \href{https://arxiv.org/abs/2510.20689v2}{Darij Grinberg, \textit{The trace Cayley-Hamilton theorem}, arXiv:2510.20689v2} for the precise fact being used.}). To prove (ii), write $U=\sum A_i w_i$, $T=\sum B_i w_i$, where $A_i,B_i\in M_k(\QQ)$ and the $w_i$ are monomials in $E_1$. Recalling that $\tr$ is a symmetric bilinear form on $M_k(\QQ)$ and $w_iw_j=-w_jw_i$, we have\footnote{\textit{Comment by DG:} Corrected ``$\tr(AB)$'' and ``$\tr(BA)$'' to ``$\tr(UT)$'' and ``$\tr(TU)$''.}
\[
 \tr(UT)=\sum \tr(A_iB_j) w_iw_j
 = -\sum \tr(B_jA_i) w_jw_i = -\tr(TU).
\]
\end{proof}

\begin{proof}[Proof of the Theorem]
Since, for $k>1$,
\begin{align*}
& h_k(x_1,\ldots,x_k,y_1,\ldots,y_k) \\
&\qquad = \sum_{i,j=1}^k (-1)^{i+j}x_i y_j\,
 h_{k-1}(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_k,
 y_1,\ldots,y_{j-1},y_{j+1},\ldots,y_k),
\end{align*}
to prove the second part of the theorem it is enough to check that $h_{2n-1}$ is not an identity for $M_n(C)$. To this end, consider the substitution (double staircase)
\[
 x_1=e_{11},\quad y_1=e_{12},\quad x_2=e_{22},\quad y_2=e_{23},\quad \ldots,\quad x_n=e_{nn},
\]
\[
 y_n=e_{nn},\quad x_{n+1}=e_{n,n-1},\quad y_{n+1}=e_{n-1,n-1},\quad \ldots,\quad
 x_{2n-1}=e_{21},\quad y_{2n-1}=e_{11}.
\]
Then
\[
 h_{2n-1}(e_{11},\ldots,e_{21},e_{12},\ldots,e_{11})= e_{11} + 2\sum_{i>1}e_{ii}\ne 0
\]
\footnote{\textit{Comment by DG:} Corrected ``$\sum e_{ii}$'' to ``$e_{11} + 2\sum_{i>1}e_{ii}$''.}
and $h_{2n-1}$ is not an identity for $M_n(C)$.

For the first part of the proof notice that, since $h_{2n}$ is multilinear and each monomial has coefficient $\pm 1$, it is enough to prove that $h_{2n}$ vanishes on $M_n(\QQ)$ (see \cite{Rowen}).

Let $A_1,\ldots,A_{2n}, B_1,\ldots,B_{2n}\in M_n(\QQ)$ and let
\[
 A=\sum_{i=1}^{2n} A_i v_i, \qquad
 B=\sum_{i=1}^{2n} B_i v_{2n+i}.
\]
Then $A,B\in M_n(E_1)$ and
\[
 (AB)^{2n}=h_{2n}(A_1,\ldots,A_{2n},B_1,\ldots,B_{2n})
 v_1v_{2n+1}v_2v_{2n+2}\cdots v_{2n}v_{4n}.
\]
This last equality can be verified by noticing that $v_k^2=0$ ($k=1,\ldots,4n$) and for $\sigma,\tau\in S_{2n}$,
\[
 v_{\sigma(1)}v_{2n+\tau(1)}v_{\sigma(2)}v_{2n+\tau(2)}\cdots
 v_{\sigma(2n)}v_{2n+\tau(2n)}
\]
\[
 = (\sgn \sigma\tau)
 v_1v_{2n+1}v_2v_{2n+2}\cdots v_{2n}v_{4n}.
\]
Take now the matrix
\[
 D=\begin{pmatrix}
 AB & 0\\
 0 & BA
 \end{pmatrix}\in M_{2n}(E_0).
\]
Since for $i\geq 1$, we have\footnote{\textit{Comment by DG:} Corrected ``$(BA)^i B$'' to ``$(BA)^{i-1} B$'', and added ``we have''.} $(BA)^{i-1} B$, $A\in M_n(E_1)$, by Proposition (ii),
$\tr((AB)^i)=-\tr((BA)^i)$\ \ \ \ \footnote{\textit{Comment by DG:} This follows by applying Proposition (ii) to $U = A$ and $T = (BA)^{i-1} B$, and observing that $TU = (BA)^{i-1} BA = (BA)^i$ and $UT = A (BA)^{i-1} B = (AB)^i$.}; thus
\[
 \tr(D^i)=\tr((AB)^i)+\tr((BA)^i)=0.
\]
But then Proposition (i) forces $D^{2n}=0$ and so $(AB)^{2n}=0$. This last equality is equivalent to
$h_{2n}(A_1,\ldots,A_{2n},B_1,\ldots,B_{2n})=0$. The proof is now complete.
\end{proof}

It has come to our attention that a different proof of the above theorem has been announced by Qing Chang.\footnote{\textit{Comment by DG:} This is likely a reference to: \href{https://doi.org/10.1090/S0002-9939-1988-0964846-8}{Qing Chang, \textit{Some consequences of the standard polynomial}, Proc. Amer. Math. Soc. \textbf{104} (1988), pp. 707--710}.}

\begin{thebibliography}{9}

\bibitem{Rosset}
\textsc{S. Rosset}, A new proof of the Amitsur--Levitzki identity, \emph{Israel J. Math.} \textbf{23} (1976), 187--188.

\bibitem{Rowen}
\textsc{L. H. Rowen}, ``\emph{Polynomial Identities in Ring Theory}'', Academic Press, New York, 1980.

\end{thebibliography}

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