In $\mathbb{C}[a_{11}, \ldots, a_{nn}, b_{11}, \ldots, b_{nn}]$, is $\langle AB - I_n \rangle = \langle BA - I_n \rangle$? By basic linear algebra, we know that for any $A, B \in M_{n\times n}(\mathbb{C})$,  we have $AB = I_n$ if and only if $BA = I_n$.  Therefore, by Hilbert's Nullstellensatz, if we set $\langle AB - I_n \rangle$ to be the ideal of $\mathbb{C}[a_{11}, \ldots, a_{nn}, b_{11}, \ldots, b_{nn}]$ generated by the entries of $AB - I_n$, i.e. $\langle \sum_{k=1}^n a_{ik} b_{kj} - \delta_{ij} \mid i, j = 1, \ldots, n \rangle$, and similarly for $\langle BA - I_n \rangle$, we have that the two ideals have equal radicals.
What I wonder is: are the two ideals actually equal?
(I was able to verify in Mathematica that the case $n=2$ does work.  Try it online!)
 A: Consider the quotient ring $R := \mathbb{C}[\vec a, \vec b] / \langle AB - I_n \rangle$.  Then in $M_{n\times n}(R)$, we have $AB = I_n$, so $(\det A) (\det B) = 1$.  Thus, $\det A$ is a unit of $R$, which implies that $A$ is a unit of $M_{n\times n}(R)$.  From here, it is easy to conclude that $BA = I_n$ also.

In fact, we can also show that $\langle AB - I_n \rangle$ and $\langle BA - I_n \rangle$ are prime ideals (and therefore radical, giving an alternate proof).  To do so, we establish an isomorphism of $R$ with the localization $\mathbb{C}[\vec a][(\det A)^{-1}]$, which is an integral domain.  The inverse maps will be:
$$\mathbb{C}[\vec a, \vec b] / \langle AB - I_n \rangle \to \mathbb{C}[\vec a][(\det A)^{-1}], \\
a_{ij} \mapsto a_{ij}, \\
b_{ij} \mapsto (\det A)^{-1} (-1)^{i+j} \det(A_{ji})
$$
and
$$
\mathbb{C}[\vec a][(\det A)^{-1}] \to \mathbb{C}[\vec a, \vec b] / \langle AB - I_n \rangle, \\
a_{ij} \mapsto a_{ij}, \\
(\det A)^{-1} \mapsto \det B.
$$
The details of showing that both maps are well-defined, and that the compositions are the identity, are straightforward.
