In the last post, we saw that an upper triangular $n\times n$ matrix ring $T_n$ over a commutative ring $R$ for $n \geq 2$ is not a separable $R$-algebra. We did this by invoking the commutator theorem: if $A$ is a central separable algebra and $B$ is a separable subalgebra then $C = A^B$ is also separable and $A^C = B$. The notation $A^B$ means $\{ a\in A : ab = ba~\forall b\in B\}$.

For $M_n(R)$, we have $M_n(R)^{T_n} = R$, whereas $M_n(R)^R = M_n(R)$. Therefore, $T_n$ cannot be separable.

Instead of using the commutator theorem, we could use Azumaya's theorem:

**Theorem (Azumaya).**Let $A$ be a finitely generated faithful $R$-algebra with generating set $a_1,\dots,a_n$. Then $A$ is a central separable free $R$-algebra with basis $a_1,\dots,a_n$ if and only if the matrix $(a_{ij}) = (a_ia_j)$ is invertible in $M_n(A)$.

For example, take the upper triangular matrices $T_n$. For expository purposes, let's suppose $n = 2$, but the general case is only more difficult to write down. So $T_2$ is certainly a free module, with basis $e_{1,1},e_{1,2},e_{2,2}$ where $e_{ij}$ is the matrix whose only nonzero entry is $1$ in the $i,j$-spot. So $T_n$ is a free $R$-module. Since the center of $T_n$ is $R$, Azumaya's theorem tells us that $T_n$ is separable if and only if the following matrix is invertible in $M_3(T_n)$:

So, we need to check whether this matrix is invertible in $M_3(T_2)$. Now matrix multiplication here is the same thing as if were were just multiplying in $M_9(R)$, so if the determinant in $M_9(R)$ is not a unit, then it certainly cannot be invertible in $M_3(T_2)$. And by inspection, the determinant of this matrix, considered as a matrix in $M_9(R)$, is zero. Therefore, it is not invertible in $M_3(T_2)$ and $T_2$ as a result is not separable.