Is $M_{2}(K)$ a polynomial identity ring? Simply, if there is a polynomial $f$, in noncommuting variables, which vanishes under substitutions from ring $R$, the ring will be called a PI ring (Polynomial Identity ring). For example, commutative rings always satisfy the polynomial $f(x,y) = xy - yx$. 
Is $M_{2}(K)$, the ring of all $2 \times2$ matrices over a field $K$, a PI ring? I have tried to construct a polynomial, assuming that this ring is a PI ring, but couldn't find any. Thanks.
 A: See the Amitsur-Levitzki theorem. 
Any four $2\times 2$ matrices $A_1,A_2,A_3,A_4$ satisfy 
$$\sum_{\sigma\in S_4} \mathrm{sgn}(\sigma)A_{\sigma(1)}A_{\sigma(2)}A_{\sigma(3)}A_{\sigma(4)}=0.$$
Hence $M_2(K)$ satisfies the polynomial identity
$$\sum_{\sigma\in S_4} \mathrm{sgn}(\sigma) x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)}x_{\sigma(4)}$$
of degree $4$ in four variables. This generalizes to $M_{n}(K)$.
A: Besides to first pervasive answer, suppose $A$ and $B$ be such two matrices. The Cayley–Hamilton theorem implies that if $x$= $(AB-BA)$ then $(AB-BA)^2$ commutes with any other $2\times 2$ matrix in $M_{2}(K)$. It leads you to $f(x,y,z)=(xy-yx)^2z-z(xy-yx)^2$.
A: Matrix rings over a given commutative ring (in particular, over any field) are indeed PI, and in some sense are the motivating prototypes.
Since you say this is homework, I am reluctant to give an explicit reference for this fact, at least until you give some more context. Were you given any hints in the course? Is this an optional question for interest, or are you expected to work out the answer using things mentioned in class?
