1
$\begingroup$

This exercise comes from Ravi Vakil's notes. Suppose that $X$ is an integral Noetherian scheme, and $f \in K(X)^{\times }$ is a nonzero element of its function field. Show that $f$ has a finite number of zeros and poles.

Since $X$ is quasi-compact, we can reduce to the case when $X = \text{Spec } A$ where $A$ is a Noetherian integral domain. If $f = f_1/f_2$ for $f_i \in A$, then it suffices to prove the result for $f_i$.

In the section above the exercise, Vakil states that for any regular codimension 1 point $p$, we can talk about an element of the function field having a zero or a pole at $p$. My idea was to proceed by contradiction, so suppose that there are infinitely many of these regular codimension 1 points $p$. Initially, I thought I could contradict the Noetherianess of the ring $A$, however, I have been unable to do so.

I would appreciate any hints or suggestions on how to proceed by this (or a different proof).

$\endgroup$
4
  • $\begingroup$ Is $X$ normal? It seems hard to talk about zeroes and poles otherwise. Granting this I think the only real ingredient is the finite decomposition of $V(f)$ into irreducible components. $\endgroup$
    – Hoot
    Aug 17 '15 at 23:50
  • 1
    $\begingroup$ Do you really think that $z$ has a finite number of zeros on $\mathbb A^2_\mathbb C=Spec(\mathbb C[z,w])$ ? $\endgroup$ Aug 18 '15 at 0:21
  • $\begingroup$ @Georges That's a good point. Probably what is meant is codimension $1$ zeros/poles. $\endgroup$
    – Hoot
    Aug 18 '15 at 0:28
  • 1
    $\begingroup$ This should just follow, with @Hoot interpretation, from the fact that a noetherian ring has finitely many minimal primes. $\endgroup$ Aug 18 '15 at 3:39
0
$\begingroup$

Let $f\in A$ and $f\neq 0$, where $A$ is a noetherian domain. Let $Y$ be a prime divisor, i.e. an integral closed subset of codimension 1 of $Spec A$. It is $v_Y(f)>0$ if and only if $Y\subset V(f)$. Write $Y=V(\mathfrak{p})$ where $\mathfrak{p}$ is a height 1 prime ideal. So $Y\subset V(f)$ means $f\in \mathfrak{p}$. The primes containing $f$ are the primes of $A/fA$. The height 1 primes are the minimal primes of $A/fA$. But a noetherian ring can only have finitely many minimal primes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.