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.
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