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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 $D=(d_{ij})$.

In other words, I have the following ring $$R=\mathbb{C}[a_{11},a_{12},a_{21},a_{22}, b_{11},\dotsc, d_{21},d_{22}]/I,$$ where $$I=(a_{11}b_{11}+a_{12}b_{21}−c_{11}d_{11}−c_{12}d_{21},\,a_{11}b_{12}+ a_{12}b_{22}−c_{11}d_{12}−c_{12}d_{22},\,a_{21}b_{11}+a_{22}b_{21}−c_{21}d_{11}−c_{22}d_{21},\,a_{21}b_{12}+a_{22}b_{22}−c_{21}d_{12}−c_{22}d_{22}).$$

The question is:

Prove that $\overline{a}_{11}$ is a prime element in $R$.

I don't know if there are any nice trick/strategy using commutative algebra (maybe using some change of coordinate or defining some norm of the ideal). I'm thinking in terms of quiver representation. The above setup can be interpreted as: $$\require{AMScd}\begin{CD} \mathbb{C}^2 @>B>> \mathbb{C}^2\\ @VDVV @VVAV \\ \mathbb{C}^2 @>>C> \mathbb{C}^2. \end{CD}$$

The ring $R$ is the coordinate ring of the representation space of the above quiver with relation, where the dimension vector is $(2,2,2,2).$

Any idea/suggestion will be apreciated.

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    $\begingroup$ What background are you supposed to know? That ring is a complete intersection ring whose singular locus appears to be of codimension $4$. Thus, by Grothendieck's proof of the Samuel conjecture, the ring is factorial. Thus, the given element is prime if and only if it is irreducible. $\endgroup$
    – Jason Starr
    Commented May 10, 2022 at 20:39
  • $\begingroup$ Dear Sir @JasonStarr , I don't have advanced knowledge in algebraic geometry. But it'll be nice if you can elaborate your approach little bit. I would like to know. $\endgroup$
    – It'sMe
    Commented May 10, 2022 at 20:47
  • $\begingroup$ Dear Sir @JasonStarr, My main aim is to prove $R$ is a UFD. So, I was trying to use Nagata's criterion for UFD (first prove that $R$ is an integral domain and then $\overline{a}_{11}$ and $\overline{c}_{22}$ are prime elements and then, the localization of $R$ by multiplicative set generated by these two prime elements is a UFD and hence, $R$ is UFD). But, it seems you have better approach. $\endgroup$
    – It'sMe
    Commented May 10, 2022 at 20:49
  • $\begingroup$ Dear sir @JasonStarr isn't the Samuel's conjecture and Grothendieck's proof of Samuel's conjecture involve Local rings? The ring $R$ in my case is not a local ring(unless I'm missing something) $\endgroup$
    – It'sMe
    Commented May 11, 2022 at 6:06
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    $\begingroup$ @FreePawn. The ring in your case is a $\mathbb{Z}_{\geq 0}$-graded ring, and the element $a_{1,1}$ is a homogeneous element. Thus, the element $a_{1,1}$ is prime if and only if the image in the associated local ring is prime (localization at the maximal ideal generated by all positive degree homogeneous elements). $\endgroup$
    – Jason Starr
    Commented May 11, 2022 at 10:33

1 Answer 1

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Now that I have thought about this further, I realize that you need much less than the Samuel Conjecture to solve this problem. By a dimension count, the ring $R$ is a complete intersection ring, hence Cohen-Macaulay. Thus, by the Unmixedness Theorem, to prove that the ideal $I=\langle a_{1,1} \rangle$ is prime, it suffices to prove that the corresponding closed subscheme of $\text{Spec}(R)$ is irreducible and generically reduced. This can be proved using Bertini's Theorem and a homogeneity argument.

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