Polynomial identity ring

Remarks

Over a field $k$, every nonzero polynomial is a scalar multiple of a monic one, so a $k$-algebra is PI when it satisfies some nonzero polynomial identity. Monic polynomials are used for rings because, for example, every ring of characteristic $2$ satisfies the identity $2 x_1 = 0$, which says nothing about its multiplication.

Example: a commutative ring satisfies the monic identity $x_1 x_2 - x_2 x_1$.

Example: the ring $M_2(k)$ of $2 \times 2$ matrices over a commutative ring $k$ satisfies the monic identity $[[x_1, x_2]^2, x_3]$, where $[a, b] = ab - ba$. Indeed, $c = [a_1, a_2]$ has trace $0$, so the Cayley–Hamilton theorem gives $c^2 = -\det(c) \, I$, a scalar matrix that commutes with every $a_3$.

Non-example: let $k$ be a field and $A = k\langle x, y \rangle$. If $f(x_1, \ldots, x_n)$ is nonzero, then $f(x, xy, x y^2, \ldots, x y^{n-1}) \ne 0$, because distinct words in the $x y^i$ are distinct words in $x$ and $y$. So the free algebra on two letters satisfies no nonzero identity and is not PI.

Every subring and every homomorphic image of a PI ring is a PI ring. The exterior algebra of an infinite-dimensional vector space over a field of characteristic $0$ is PI, since it satisfies $[[x_1, x_2], x_3]$, but it satisfies no standard identity.