Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

An effective Cartier divisor on a scheme $X$ (a closed subscheme of $X$ that is locally cut out by one equation that is not a zero-divisor) is always a regular embedding of codimension 1 (in the stalks, it is cut out by one equation that is not a zero-divisor). The converse is true if $X$ is locally Noetherian, using Nakayama's lemma to "lift" from the stalk to an "honest neighborhood".

Is there a (necessarily non-Noetherian) example of a codimension 1 regular embedding that is not an effective Cartier divisor?

(See section 8.4.7 of the March 23, 2013 version of the notes here for some discussion, if you care.)

I am asking this out of idle curiosity rather than any need for it, but I am genuinely curious.

share|improve this question
2  
Stripping away the geometric terminology, it sounds like you seek a commutative ring $A$ and ideal $I$ such $I_{\mathfrak{p}}$ is invertible as an $A_{\mathfrak{p}}$-module for all prime ideals $\mathfrak{p}$ of $A$ but $I$ is not invertible as an $A$-module (equivalently, $I$ is not finitely presented as an $A$-module). Is that correct? –  user28172 May 1 '13 at 10:00
    
@nosr: Yes, that's right! –  Ravi Vakil May 1 '13 at 20:16

1 Answer 1

up vote 11 down vote accepted

Let $(A,m,k)$ be a discrete valuation ring. Consider the subring $R\subset A^\mathbb{N}$ of converging sequences (for the discrete topology on $A$), i.e. ultimately constant sequences. We have a morphism $\varphi:R\to A$ sending any sequence to its limit. Put $\mathfrak{p}=\varphi^{-1}(m)$: this is the maximal ideal of sequences with noninvertible limit.

Now it is easy to check that $\varphi$ induces an isomorphism $R_\mathfrak{p}\cong A$. In particular, $\mathfrak{p}$ becomes principal in $R_\mathfrak{p}$, and of course also in $R_\mathfrak{q}$ for all $\mathfrak{q}\neq\mathfrak{p}$ because then $\mathfrak{p}R_\mathfrak{q}=R_\mathfrak{q}$. So $V(\mathfrak{p})\subset\mathrm{Spec}\,(R)$ is a regular embedding of codimension 1.

On the other hand, $\mathfrak{p}$ is not finitely generated. Indeed, if $J\subset\mathfrak{p}$ is a finitely generated ideal, there is an $N\in\mathbb{N}$ such that for each $u:\mathbb{N}\to A$ in $J$ and each $n>N$ we have $u(n)\in m$. Clearly there is no such $N$ for $\mathfrak{p}$.

share|improve this answer
1  
Beautiful! I'd hoped for not just a counterexample, but also something that gave me some insight, so I am very happy with this. –  Ravi Vakil May 1 '13 at 19:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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