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I want to show that $k[x,y,z,w]/(xy-zw)$, the coordinate ring of $V(xy-zw)\subset\mathbb{A}^4$, is not a unique factorization domain.

Morally, all we need to do is find some nonzero element that can be written in two distinct ways as a product of irreducible elements, but nothing is coming to mind immediately. Thoughts?

user26857
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user96033
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    In this coordinate ring, we have $xy = wz$. – Joel Cohen Sep 21 '13 at 15:48
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    @JoelCohen As one can see from the number of upvotes collected by your comment people like to think that $xy=zw$ solves the problem instantly, but I won't be so sure. Could you post a solution starting from this equation? –  Sep 22 '13 at 10:00

1 Answers1

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Let $S$ be an integral domain. Then $R=S[X,Y,Z,W]/(XY-ZW)$ is not an UFD.

First of all, it isn't hard to prove that $R$ is also an integral domain. (See here.)

In the following we denote by $x,y,z,w$ the residue classes of $X,Y,Z,W$ modulo the ideal $(XY-ZW)$.

  • $x$ is irreducible.

Consider $R$ as a graded ring (with the grading inherited from $S[X,Y,Z,W]$). Assume that $x$ has a factorization into two non-zero non-unit elements. Since $x$ has degree one, the two factors must be a degree one element $ax+by+cz+dw$ with $a,b,c,d\in S$, and a degree zero element $s\in S$. (If $S$ is a field, one can stop here.) This gives $$(1-sa)X+sbY+scZ+sdW\in(XY-ZW),$$ so in $S[X,Y,Z,W]$ a degree one element belongs to the ideal $(XY-ZW)$ whose the non-zero elements are of degree at least two. Consequently, $(1-sa)X+sbY+scZ+sdW=0$ and this implies $sa=1$, so $s$ is a unit, a contradiction.

  • $x$ is not prime.

We have $R/(x)\simeq (S[Z,W]/(ZW))[Y]$ and this is not an integral domain.

Remark. When $S$ is a field one can prove that the divisor class group of $R$ is (isomorphic to) $\mathbb Z$.

user26857
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