# Prove that R is an integral domain

I'm studying for my qualifying exam and I came across the following question in one of the old question bank.

Consider the affine space given by four $$2\times 2$$ matrices, i.e., $$\mathbb{A}^{16}\cong M(\mathbb{C})_{2\times 2}^4$$. Now, consider the algebraic set $$V$$ given by the vanishing of the relation $$AB-CD=0$$, where the matrices are as follows: $$A=(a_{ij}), B=(b_{ij}), C=(c_{ij})$$ and $$D=(d_{ij})$$. Prove that $$V$$ is irreducible in $$\mathbb{A}^{16}$$.

In other words, I want to prove that the following ring

$$R=\mathbb{C}[a_{11},a_{12},a_{21},a_{22}, b_{11},\dotsc, d_{21},d_{22}]/I$$, where $$I=(a_{11}b_{11}+a_{12}b_{21}−c_{11}d_{11}−c_{12}d_{21},\,a_{11}b_{12}+ a_{12}b_{22}−c_{11}d_{12}−c_{12}d_{22},\,a_{21}b_{11}+a_{22}b_{21}−c_{21}d_{11}−c_{22}d_{21},\,a_{21}b_{12}+a_{22}b_{22}−c_{21}d_{12}−c_{22}d_{22})$$

$$R$$ is an integral domain.

I've been trying to follow the same idea as in this post (https://math.stackexchange.com/a/4303220/884739), but I'm having hard time trying to figure out what the correct change of coordinate should be, so that I can embed this ring $$R$$ inside some field and hence, conclude that $$R$$ is an integral domain.

• I would check if the subset with $\det(ABCD)\ne 0$ is dense (the complex topology makes it easy: if $f$ vanishes on $W=V\cap \det(ABCD)\ne 0$ then it vanishes on the whole $V$), as it is clear that $W$ is irreducible May 7, 2022 at 21:31
• Why is the tag "quiver" applied to this question? May 16, 2022 at 19:48
• @Pierre-GuyPlamondon because the above ring can be seen as the coordinate ring of the representation space of the quiver which looks like a commutative diagram, with dimension vector $(2,2,2,2)$. May 17, 2022 at 3:46

You can consider $$V'$$ the open subset of $$V$$ given by the non-vanishing of the polynomial $$a_{1,1}a_{2,2}-a_{1,2}a_{2,1}$$. $$V'$$ is defined by the relations $$\det A \neq 0,$$ $$B = A^{-1}CD.$$ The second of these expresses the variables $$b_{1,1}, b_{1,2}, b_{2,1}, b_{2,2}$$ in terms of the remaining $$12$$, which implies that the (restriction to $$V'$$ of the) projection onto $$V(a_{1,1}a_{2,2}-a_{1,2}a_{2,1})^c \subseteq \mathbb{A}^{12}$$ is an isomorphism of varieties, the latter being irreducible (it shouldn't be hard to see the polynomial $$(xy-zw)t - 1$$ is irreducible).

To finish, all you have to do is check that $$V'$$ is Zariski-dense in $$V$$, which amounts to showing the canonical map $$\mathbb{C}[a_{1,1}, \dots , d_{2,2}]/I := R \to R[(a_{1,1}a_{2,2}-a_{1,2}a_{2,1})^{-1}]$$is injective (or rather, that its kernel equals the nilradial, which I guess might not be trivial a priori?).

To find this kernel we search for polynomials $$p$$ in $$\mathbb{C}[a_{1,1}, \dots, d_{2,2}]$$ such that $$d^k\cdot p \in I$$ for some $$k\geq 0$$, where $$d := a_{1,1}a_{2,2}-a_{1,2}a_{2,1}$$. We may also suppose $$p$$ to be homogeneous since $$I$$ and $$d$$ both are (whence the considered kernel will also be). I think you can now compare leading terms and work out that $$k=1$$; this shouldn't be a tough computation using that $$p$$ is homogeneous.

I think this last step can be made more direct by arguing that your generating system for $$I$$ is a grobner basis, by computing the $$S$$-polynomials, but I'm not sure you're allowed to use those in your exam :p

Best of luck! I hope this was helpful and I didn't make any silly mistakes :)

P.s. I'd love to see a solution using quiver representations as you remarked!

• The idea is that- you consider the quiver that I mentioned in the comment and then, consider the representation which is given by the $Id_{2\times 2}$ at all the four arrows. Now, you can prove that the representation space is equal to the closure of the orbit of this representation and hence, the ring $R$ is an integral domain. Jun 19, 2022 at 9:39
• Sounds cool! I think this amounts to showing the open subset where $A,B,C$ and $D$ are all invertible is dense in $V$, since this is the orbit you mention if I'm not mistaken. I like this description of the problem much better :) Jun 19, 2022 at 10:12