Amitsur-Levitsky Theorem

In a previous post, we claimed that $M_n(R)$ satisfies a polynomial identity of degree $2n$ for any commutative $F$-algebra. In fact this is true for commutative rings, as well.

For each $k$, define the standard polynomial $\displaystyle S_k=\sum_{\sigma\in\mathcal S_k}(-1)^{|\sigma|}x_{\sigma(1)}\cdots x_{\sigma(k)}$ in the noncommuting variables $x_1,\ldots, x_k$.

Theorem: (Amitsur-Levitsky 1950) $M_n(R)$ satisfies the identity $S_{2n}=0$.

We will outline an elegant short proof due to Shmuel Rosset (1976). It is clearly sufficient to prove the theorem for $M_n(\mathbb Z)$, and then since $\mathbb Z\subset \mathbb Q$, for $M_n(\mathbb Q)$. The point being that we now need only consider coefficients in a field of characteristic $0$.

Consider the alternating (Grassman) algebra $G$ with $2n$ anticommuting generators $\xi_1,\ldots,\xi_{2n}$. $G$ is a superalgebra: $G=G_0\oplus G_1$, where $G_0$ is spanned by products of an even number of generators and $G_1$ is spanned by products of odd numbers of generators. In particular we will use that $G_0$ is a commutative algebra.

Now let $A_1,\ldots, A_{2n}\in M_n(\mathbb Q)$. Consider
$$X=A_1\otimes \xi_1+\cdots+A_{2n}\otimes \xi_{2n}.$$
Note that $$X^{2n}=S_{2n}(A_1,\ldots,A_{2n})\otimes \xi_1\cdots \xi_{2n}.$$ Thus, to show that $S_{2n}(A_1,\ldots, A_{2n})=0$, it is sufficient to show that $X^{2n}=0$. Note that $X^2\in M_n(G_0)$ is a matrix taking values in the commutative algebra $G_0$.

Lemma: Let $B\in M_n(R)$ for a commutative algebra $R$. Then if $\mathrm{tr}(B)=\mathrm{tr}(B^2)=\cdots=\mathrm{tr}(B^{n})=0$, then $B^n=0$.
Proof of lemma:
The matrix $B$ satisfies its characteristic polynomial, and the coefficients of the characteristic polynomial are polynomials in traces of powers of $B$. For example, in degree $2$, the characteristic polynomial can be written
$$\lambda^2-\mathrm{tr}(B)\lambda+\frac{1}{2}(\mathrm{tr}(B)^2-\mathrm{tr}(B^2)).$$
Plugging in $B$, we get   $B^2-\mathrm{tr}(B)B+\frac{1}{2}(\mathrm{tr}(B)^2-\mathrm{tr}(B^2))=0$, so if the traces are trivial, $B^2=0$ as well. One can prove that the characteristic polynomial of a $2\times 2$ matrix can be written this way by first embedding our field in its algebraic closure and considering the Jordan normal form $\begin{bmatrix}\lambda _1&*\\0&\lambda_2\end{bmatrix} $. Then the characteristic polynomial is $$(x-\lambda_1)(x-\lambda_2)=x^2-(\lambda_1+\lambda_2)+(\lambda_1\lambda_2).$$
On the other hand $\mathrm{tr}(B)=\lambda_1+\lambda_2$ and $\mathrm{tr}(B^2)=\lambda_1^2+\lambda_2^2$. In general, the coefficients of the characteristic polynomial for a matrix of arbitrary size will be symmetric polynomials in the eigenvalues and are expressible as polynomials in $\sum \lambda_i,\sum \lambda_i^2,\ldots$. $\Box$

So to finish the proof, we need to show that $\mathrm{tr}(X^{2k})=0$ for all $k\leq n$. Now
$$X^{2k}=\sum A_{i_1}\cdots A_{i_{2k}}\otimes \xi_{i_1}\cdots \xi_{i_{2k}},$$
and the trace will cancel in pairs since $\mathrm{tr}(A_{i_1}\cdots A_{i_{2k}})=\mathrm{tr}(A_{i_2}\cdots A_{i_{2k}}A_{i_1})$ and these two terms appear with opposite sign.