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I had a homework problem (II.6.5 in Hartshorne) to compute the (Weil divisor) class group of $X=\operatorname{Spec} k[x,y,z,w]/(xy-zw)$. I have accomplished this; however, I used some results I don't fully understand. I was left with the conviction that there is a simpler way to do it using tools I do understand. I am hoping you can help me find it.

The part I do understand:

Let $Z$ be the prime divisor associated with the ideal $(y,z)$. I have an exact sequence

$$\mathbb{Z}\rightarrow \operatorname{Cl} X \rightarrow \operatorname{Cl} \left(X\setminus Z\right) \rightarrow 0$$

where the first map sends $1\mapsto [Z]$ and the second map sends a divisor to its intersection with $X\setminus Z$. (This is Hartshorne proposition II.6.5.) Now $$X\setminus Z = \operatorname{Spec} k[x,y,y^{-1},z,z^{-1},w]/(xy-zw)$$ but this ring is isomorphic to $k[x,y,y^{-1},z,z^{-1}]$ because $w=xyz^{-1}$. This is a unique factorization domain, so $\operatorname{Cl} \left(X\setminus Z\right)=0$. Thus $\operatorname{Cl} X$ is a cyclic $\mathbb{Z}$-module, generated by $[Z]$.

The question is whether $[Z]$ is torsion or not. I've obtained that $[Z]$ is non-torsion (thus $\operatorname{Cl}X \cong \mathbb{Z}$) by showing $\operatorname{Cl}X$ is infinite, in turn by relating $\operatorname{Cl}X$ to the class group of the projective quadric surface $Q$ of which it is the affine cone. I used the result of an exercise in Hartshorne (II.6.3) relating a projective variety's class group to that of its affine cone. I feel confident that the argument works, but I am personally unsatisfied both because I'm not fully comfortable with the reasoning in exercise II.6.3, and because the whole argument is rather indirect.

What I'm looking for:

What I would like is a direct way of seeing that $[Z]$ is non-torsion, i.e. that $n[Z]$ is not principal for $n\in \mathbb{Z}$.

Can you offer a direct argument that $[Z]$ is non-torsion in $\operatorname{Cl}X$?

I have given this some thought (below), but I am stuck.

My work so far:

Let $A=k[x,y,z,w]/(xy-zw)$ and let $\mathfrak{p}=(y,z)$. Suppose there is some $f\in K(X)=\operatorname{Frac}A$ such that $div(f)=n[Z]$ for some $n\in\mathbb{Z}\setminus\{0\}$. By replacing $f$ with $f^{-1}$ if necessary, we can assume that $n>0$. I think that in this case $f\in A$ and $\mathfrak{p}^{(n)}$, the $n$th symbolic power of $\mathfrak{p}$, is principal and generated by $f$. I think this because since $A$ is an integrally closed noetherian domain, it is the intersection of its localizations at the height 1 primes, which are precisely the DVRs associated to the valuations induced by the prime divisors. $div(f)$ is effective, so this means $f$ is in the intersection of these DVRs, thus $\in A$. Furthermore, for any element $g$ of $A$ whose $Z$-valuation is $\geq n$, we must have $g/f\in A$ for the same reason. (Hartshorne uses reasoning like this several times in section II.6.) It follows that $f$ generates the ideal $\{g\in A\mid v_Z(g)\geq n\}$. But this ideal is the contraction in $A$ of $\mathfrak{p}_\mathfrak{p}^nA_\mathfrak{p}$; this is $\mathfrak{p}^{(n)}$.

Therefore, proving that $[Z]$ is non-torsion in $\operatorname{Cl} X$ is equivalent to proving that $\mathfrak{p}^{(n)}\triangleleft A$ is never principal. I had the thought to mimic Hartshorne's reasoning (in example II.6.5.2) and show that $\mathfrak{p}^{(n)}$'s image in some appropriate vector space, e.g. let $(x,y,z,w)=\mathfrak{m}$ and find the right $\mathfrak{m}^n/\mathfrak{m}^{n+1}$, is never one-dimensional. However, in general, $\mathfrak{p}^{(n)}$ needn't be contained in $\mathfrak{m}^n$, so I wasn't sure how to get this going.

Thanks in advance for your thoughts.

NB: Because this question arose out of a homework problem for me, I'm using the homework tag to be safe, but I intend to turn in the proof I described above, so this question is not intended to help me with my assignment, just my understanding.

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    $\begingroup$ Suppose $n[Z]$ is a principal divisor defined by a rational function $f$ on $X$. Then $f$ is regular and invertible on $U:=X\setminus Z$. Now compute $O_X(U)^{\star}$ (invertible elements, you should find $k^*y^{\mathbb Z}z^{\mathbb Z}$) and observe that the support of $\mathrm{div}(f)$ is then always reducible, hence different from $Z$. $\endgroup$
    – user18119
    May 2, 2013 at 15:03
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    $\begingroup$ It's interesting to note that if you work over $\mathbb{C}$, then $\mathbb{C}[x,y,z,w]/(xy - zw)$ is a toric variety, and calculating class groups of toric varieties is very, very, very easy, as it amounts to finding a Smith normal form of some matrix. $\endgroup$
    – xyzzyz
    May 2, 2013 at 16:08
  • $\begingroup$ @xyzzyz:Toric varieties are defined for any field and what you said is valid over any field. $\endgroup$
    – user18119
    May 3, 2013 at 0:41
  • $\begingroup$ @QiL'8, that's true, but I only worked with them over $\mathbb{C}$, and I didn't know how many of the foundational results transfer to other fields, or characteristic p. The proof of orbit-cone correspondence I've seen, for one thing, relied pretty heavily on the classical topology of varieties. That's why I specifically mentioned $\mathbb{C}$ to make sure that if I don't say the whole truth, at least I say the half-truth instead of half-lie. $\endgroup$
    – xyzzyz
    May 3, 2013 at 1:14
  • $\begingroup$ @xyzzyz: No problem, I just wanted to give the information for arbitrary fields. It is true that all books on toric varieties I know are written for $\mathbb C$. Now can you find out the rational convex polyhedral defining the variety we are interesting here and give its class group by this way ? $\endgroup$
    – user18119
    May 3, 2013 at 2:17

2 Answers 2

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Let me first recap the setup and known parts. $A = k[x,y,z,w]/(xy-zw)$ and $p = (y,z)A$. In order to show that $Cl(A)\cong \mathbb{Z}$, it suffices to show that for any $n\ge 1$, $p^{(n)}$ is not principal.

Suppose that for some $n \ge 1$, $p^{(n)}$ is principal. Write $S = k[x,y,z,w]$, and let $f \in S$ whose image in $A$ generates $p^{(n)}$. Since $p^n \subset p^{(n)}$ in $A$, $(y,z)^n \subset (f, xy-zw)$ in $S$. We are going to show that this cannot happen.

The containment $(y,z)^n \subset (f, xy-zw)$ in $S$ implies $$ (y^n, z^n) \subset (f, xy-zw) \subset (f, x,w). $$ In $S/ (x,w) \cong k[y,z]$, the ideal $(y^n,z^n) S/(x)$ is contained in the ideal $(f,x,w) S/(x,w) = f S/(x,w)$ whose height (codimension) is at most 1 (Krull's height theorem). This is a contradiction because the height of $(y^n,z^n)S/(x,w) = (y^n,z^n)k[y,z]$ is 2.

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  • $\begingroup$ +1 Thanks for this. (a) In the last paragraph, do you mean "the ideal $(y^n,z^n)S/(x,w)$"? (b) Forgive if this is a stupid question, but at the moment I'm unclear on why to believe that the preimage of $fA_p \cap A$ in $S$ is $(f,xy-zw)$. It is clear to me that this is the preimage of $fA$, but not why $fA_p\cap A$ couldn't be bigger. It must have to do with the integral closure. Could you elucidate? $\endgroup$ Jul 28, 2016 at 12:27
  • $\begingroup$ @BenBlum-Smith (a) You are right. (b) So, $p^{(n)} = p^n A_p \cap A$ is the n-th symbolic power. If $p^{(n)}$ is principal, then we may write it as $p^{(n)} = f'A$. Since $S \to A$ is surjective, one may choose $f$ in $S$ whose image is $f'$. You are right in general, for an ideal $I$, we have $I \subset IA_p \cap A$, but if $I$ is p-primary, then they are equal. Here $fA = p^{(n)}$ is $p$-primary. However, I don't think one needs this in the proof. $\endgroup$
    – Youngsu
    Jul 28, 2016 at 23:35
  • $\begingroup$ Thanks, that's what I needed. $\endgroup$ Jul 30, 2016 at 1:46
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Let $A=k[X,Y,Z,W]/(XY-ZW)$, and $\mathfrak p=(y,z)$. As you noticed the divisor class group of $A$ is cyclic generated by $[\mathfrak p]$.

We want to prove that $n[\mathfrak p]=0$ for $n\ge0$ implies $n=0$.

(This leads us to the conclusion that $\operatorname{Cl}(A)\simeq\mathbb Z$.)

Suppose $n\ge1$. Since $n[\mathfrak p]=0$ there is $f\in A$, $f\ne0$ such that $\mathfrak p^{(n)}=fA$. Now extend these ideals to $A[y^{-1}]$ and observe that $f$ is invertible in $A[y^{-1}]$. Since $A[y^{-1}]$ is the localization of the polynomial ring $k[y,z,w]$ at $y$, that is, $A[y^{-1}]=k[y,z,w][y^{-1}]$, it follows that $f=ay^m$ with $a\in k^\times$ and $m\in\mathbb Z$. Thus we have $\mathfrak p^{(n)}=y^mA$. It's easily seen that $m\ge 1$. Since $\mathfrak p^n\subseteq\mathfrak p^{(n)}$ we get $z^n\in y^mA$, that is, $Z^n\in(Y^m,XY-ZW)$. In particular, by sending $Y$ to $0$, we get $Z^n\in(ZW)$, a contradiction.

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  • $\begingroup$ How do you conclude $f$ is invertible in $A[y^{-1}]$? $\endgroup$ Nov 22, 2020 at 16:00
  • $\begingroup$ $\mathfrak p^{(n)}A[y^{-1}]=A[y^{-1}]$ since $\mathfrak p^{(n)}$ contains $y^n$, which is invertible in $A[y^{-1}]$. $\endgroup$
    – user26857
    Nov 22, 2020 at 20:27

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