Generic matrix algebra

Remarks

The algebra $k\{X_1, \ldots, X_m\}$ is an associative algebra over $k$, and it is a subring of $M_n(P)$, so it satisfies every polynomial identity of $M_n(P)$. Let $\pi \colon k\langle x_1, \ldots, x_m \rangle \to k\{X_1, \ldots, X_m\}$ be the homomorphism from the free algebra with $\pi(x_l) = X_l$. If $k$ is infinite, the kernel of $\pi$ is the set of polynomial identities of $M_n(k)$ in $m$ variables. Indeed, $f(X_1, \ldots, X_m) = 0$ means that each entry of the matrix $f(X_1, \ldots, X_m)$ is the zero polynomial. Over an infinite field, a polynomial is zero exactly when it vanishes at all points, that is, when $f(a_1, \ldots, a_m) = 0$ for all $a_l \in M_n(k)$.

Example: for $n = 1$ the generic matrices are the variables $x^{(l)}_{11}$, and $k\{X_1, \ldots, X_m\}$ is the polynomial ring in $m$ variables.

Non-example: for $n \ge 2$, the scalar matrix $x^{(1)}_{11} I$ lies in $M_n(P)$ but not in $k\{X_1, \ldots, X_m\}$. The entries of $f(X_1, \ldots, X_m)$ have degree-$1$ part equal to the corresponding entry of $\sum_l c_l X_l$, where $c_l$ is the coefficient of $x_l$ in $f$. If $f(X_1, \ldots, X_m) = x^{(1)}_{11} I$, comparing the $(1, 1)$ entries gives $c_1 = 1$, and comparing the $(2, 2)$ entries gives $\sum_l c_l x^{(l)}_{22} = x^{(1)}_{11}$, which is impossible. So the subalgebra generated by the $X_l$ over the commutative ring $P$ is strictly larger than the generic matrix algebra.

For $n \ge 2$ and $m \ge 2$, the generic matrix algebra is a noncommutative domain.