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Let $Y$ be a normal surface and let $X$ be a closed subscheme of codimension 2, i.e., $X$ is a finite set of closed points.

Let $D$ be a Weil divisor on $Y$.

Question. Does there exist a Weil divisor $E$ on $Y$ which is linearly equivalent to $D$ and does not go through $X$? (Edit: I do not assume $E$ to be effective.)

Of course, when $D$ does not go through $X$ the answer is yes.

When I say that $E$ does not go through $X$, I mean that the support of $E$ and $X$ are disjoint.

I'm interested in this question in the most general set-up known. For example, $Y$ is an integral noetherian separated excellent normal 2-dimensional scheme. If you prefer sticking to algebraic surfaces, I would be glad to hear about what's possible in that case too.

The motivation for this question comes from an article by Mumford in which he defines an intersection pairing on a normal surface. In this case $X$ is the singular locus of $Y$.

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Not neccesarily. Take $X$ supported on the exceptional divisor $F$ of the blow-up of $\mathbb{P}^2$ at a point and let $D=F$. Then the only divisors linearly equivalent to $D$ is $D$ itself. If however the linear system $|D|$ is base-point free, then what you want should be true.

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    $\begingroup$ On a smooth surface the answer is certainly yes, since any divisor is lineraly equivalent to the difference of two very ample divisors. Of course, the divisor $E$ lineraly equivalent to $D$ and avoiding $X$ will not be effective in general. But the OP does not require effectiveness. In your example, it is sufficient to take $E=L-F$, where $L$ is the strict transform of a general line and $F$ the strict transform of a line passing to the point you are blowing up. $\endgroup$
    – Francesco Polizzi
    Commented Jul 31, 2011 at 17:16
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For an arbitrary normal surface Weil divisors cannot be moved: If $X$ is the singular locus of $Y$ then if any Weil divisor could be moved off $X$ it would not intersect the singular locus hence would be a Cartier divisor (since linear equivalence preserves Cartier divisors). So any normal surface on which all Weil divisors are not Cartier gives a counterexample.

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  • $\begingroup$ Right. I was just thiking of a line in the quadric cone in $\mathbb{P}^3$. There is no way to move it in its linear equivalence class in order to avoid the vertex $\endgroup$
    – Francesco Polizzi
    Commented Jul 31, 2011 at 17:38
  • $\begingroup$ Does a similar reason hold for $Y$ a normal threefold with isolated singularities ($X$ is again the singular locus) ? $\endgroup$
    – Maharana
    Commented Aug 1, 2011 at 15:59
  • $\begingroup$ Yes. The argument works in any dimension and the singularities need not be isolated. $\endgroup$
    – naf
    Commented Aug 2, 2011 at 8:24
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No. Let $Y$ be the quadric cone $\mathrm{Spec} \ k[x,y,z]/(xz-y^2)$ and let $X$ be the singular point $(x,y,z) = (0,0,0)$. Then $\mathrm{Pic}\ Y \cong \mathbb{Z}/2 \mathbb{Z}$, and any divisor not passing through $X$ is trivial. So any divisor which represents the nontrivial class in $\mathrm{Pic} \ Y$, for example $\{ x=y=0 \}$, will not be equivalent to a divisor avoiding $X$.

Note that this counter-example does not assume divisors are effective, as J.C. Ottem's does.

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  • $\begingroup$ That's beautiful. How do you prove the Picard group of this surface to be Z/2Z? Does it help that it's a cyclic quotient singularity of type 1/2(1,1)? Moreover, based on Fransesco's answer in the other question, I guess this example doesn't give a counter example to the other question, right? $\endgroup$
    – Ariyan Javanpeykar
    Commented Jul 31, 2011 at 19:55

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