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This question is about a technical issue I ran into.

Let $S$ be a connected 1-dimensional Dedekind scheme, and let $X\to S$ be a flat projective integral normal 2-dimensional scheme. (For simplicity, we can also assume that the generic fibre of $X\to S$ is smooth.)

Is every Weil divisor on $X$ a $\mathbf{Q}$-Cartier divisor? That is, does every Weil divisor on $X$ have the property that a certain integer multiple is a Cartier divisor on $X$?

The answer to this question is positive by Lemma 3.3 in Moret-Bailly's Groupes de Picard et problemes de Skolem,I if $S$ is affine, excellent and satisfies Condition (T) on page 162 of Moret-Bailly's article. In particular, if $S$ is the spectrum of the ring of integers of a number field the answer to this question is positive.

I suspect that the answer to the above question is positive for any excellent Dedekind scheme. Is this known?

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Hm, this certainly doesn't hold for general singularities though (especially in the equal characteristic setting). Surface singularities can have non-torsion elements in their divisor class groups. For example, $\mathbb{C}[x,y,z]/\langle x^3 + y^3 + z^3 \rangle$. So somehow the fact that you are over a number field (or satisfy condition (T) more generally) must be forcing enough finiteness to avoid these divisors of infinite order. –  Karl Schwede Nov 11 '11 at 21:37
    
I see. So my intuition was wrong. –  Ari Nov 11 '11 at 22:32
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For a normal local two-dimensional scheme X, the obstruction to a Weil divisor being Cartier comes from the Picard group of the punctured spectrum. Up to finite nuisance, this group is the connected component of the Picard group of the exceptional fibres of a resolution of X (up to some vector spaces over the field of definition of the exceptional fibres). In the number field case, the exceptional fibres are projective varieties over a finite field, so the Pic^0 is finite, as are the vector spaces showing up. In particular, all Weil divisors on X are torsion. –  Bhargav Nov 11 '11 at 23:44
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1 Answer

The property of being a Cartier divisor is a local property, by definition. Similarly, the property of being a Q-Cartier divisor is also local, since one can take the gcd of the relevant $n$s.

Therefore, the statemeent is true for a scheme $S$ if and only if it is true for each affine subscheme. Then, apply the theorem.

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I agree that this proves the statement for excellent Dedekind schemes which fulfill Moret-Bailly's condition (T) locally. But my question is really about excellent Dedekind schemes which don't satisfy Moret-Bailly's condition (T). –  Ari Nov 11 '11 at 19:18
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Ah, I see. I didn't notice that you wanted to drop the condition T. I can't help you with that. –  Will Sawin Nov 12 '11 at 8:08
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