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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!)

Daniel Schepler
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    Use the second identity displayed here. It allows you to write the generators of one in terms of the generators of the other. Note that because the two little matrices on the outsides of the products have determinant $1$ their inverses have entries that are actually polynomials. –  Feb 20 '20 at 19:17

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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.

Daniel Schepler
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  • In the first paragraph, did you prove that the ideals are equal or what? – user26857 Feb 23 '20 at 17:46
  • What I've understood from your question and answer is that the ideals have the same radical by Hilbert's Nullstellensatz, and then you proved that these are prime. – user26857 Feb 23 '20 at 17:49
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    Yes, the first paragraph proves directly that the ideals are equal (without using the Nullstellensatz, so it will even work over $\mathbb{Z}[a_{11}, \ldots, b_{nn}]$). Since $BA - I_n$ in $M_{n\times n}(R)$, that means all the entries of $BA - I_n$ are in $\langle AB - I_n \rangle$. – Daniel Schepler Feb 23 '20 at 20:34
  • Ok. The suggestion you got under the question shows even more: the ideals generated by all $r$-minors of $AB-I_n$, respectively $BA-I_n$ are equal, for every $r\ge 1$. – user26857 Feb 23 '20 at 21:07