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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.

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    $\begingroup$ 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 $\endgroup$
    – reuns
    May 7, 2022 at 21:31
  • $\begingroup$ Why is the tag "quiver" applied to this question? $\endgroup$ May 16, 2022 at 19:48
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    $\begingroup$ @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 $(2,2,2,2)$. $\endgroup$
    – It'sMe
    May 17, 2022 at 3:46

1 Answer 1

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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!

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    $\begingroup$ 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. $\endgroup$
    – It'sMe
    Jun 19, 2022 at 9:39
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    $\begingroup$ 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 :) $\endgroup$ Jun 19, 2022 at 10:12

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