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Let $(X,\mathcal{O}_X)$ be a Noetherian integral scheme and let $g$ be a (numerical) additive nonnegative function from coherent $\mathcal{O}_X$-modules to $[0,\infty)$. This question may be well known to the expert but I couldn't find a reference: is $g$ a constant multiple of generic rank? If true, do you know of any reference for this?

Notes:

  1. If $X=\mathrm{Spec}\:R$ is affine with $R$ an integral domain, then a proof can be found in Northcott-Reufel, Theorem 2, p. 303. There are also other proofs.
  2. If $X$ is a projective variety over a field, I think I can prove it, but I don't know any reference for this case.

I have a feeling this question must have been answered in K-theory.

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Basically you ask for homomorphisms $G_0(X) \to \mathbb{R}_+$. –  Martin Brandenburg Dec 13 '11 at 8:48
4  
To Martin: no, he is asking for homomorphisms G0(X)→ℝ that are non-negative on classes of coherent sheaves. –  Angelo Dec 13 '11 at 9:40

1 Answer 1

up vote 10 down vote accepted

I suppose that "additive" means that "additive over short exact sequences". If so, this is does not seem too hard, at least if $X$ is separated.

By noetherian induction, you may assume that for all proper integral subscheme $Y$ of $X$, the restriction of $g$ to $Y$ is given by a multiple of the generic rank at $Y$. But every coherent sheaf with support on $Y$ can be given by a successive extension of coherent sheaves of $\mathcal O_Y$-modules; hence the restriction of $g$ to sheaves supported on $Y$ is given by a multiple of the length of the stalk at the generic point of $Y$. On the other hand, for each $n > 0$ denote by $Y_n$ the subscheme defined by the $n^{\mathrm th}$ power of the sheaf of ideals of $Y$; the length of the stalk of $\mathcal O_{Y_n}$ at the generic point of $Y$ is unbounded, but by the positivity of $g$ the value of $g(\mathcal O_{Y_n})$ is bounded by $g(\mathcal O_X)$. Hence this multiple is 0, and $g$ is 0 on all torsion sheaves. In particular, if $F \to G$ is a generic isomorphism of coherent sheaves, $g(F) = g(G)$.

But on the other hand if $F$ is a coherent sheaf of generic rank $r$, there exists homomorphisms $F \to G$ and $\mathcal O_X^r \to G$ of coherent sheaves that are generic isomorphisms; hence $g(F) = rg(\mathcal O_X)$. The conclusion follows.

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This is great, thank you! –  Mahdi Majidi-Zolbanin Dec 13 '11 at 14:01
2  
Perhaps in line 9, you meant "is bounded by $g(\mathcal O_X)$" instead of $\mathcal O_Y$. –  Hailong Dao Dec 13 '11 at 16:32
    
To Hailong: yes, thank you. I edited the post. –  Angelo Dec 13 '11 at 18:42

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