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In the definition of smooth morphisms, Hartshorne use the notation $dim_{k(x)}(\Omega_{X/Y}\otimes k(x))$· But $\Omega_{X/Y}\otimes k(x)$ is a sheaf, what is the dimension? Thanks for any intepretation.

edited Feb 11 2012 at 15:07

Mahdi Majidi-Zolbanin
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asked Feb 11 2012 at 7:24

MZWang
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The sheaf in question is a sheaf over the residue field, which means that it can be viewed as vector space over the residue field. – Mike Skirvin Feb 11 2012 at 7:30

Can you explain how to view? – MZWang Feb 11 2012 at 7:46

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Your question is really about notation. Another way to understand this is to first take the stalk at $x$ to get an $\mathcal{O}_x$-module, and then tensor by the residue field. – Donu Arapura Feb 11 2012 at 10:33

Dear MZWang, $k(x)$ is a field. On the other hand, anything tensored with $k(x)$ has an action of $k(x)$ -- multiplication on the right. Thus it's a $k(x)$-vector space. – Karl Schwede Feb 11 2012 at 15:25

Thanks to dear all. I think Donu have resolveed my confusion. – MZWang Feb 12 2012 at 6:54

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Harized · Answer 1 · 2012-02-12 11:16:22Z UTC

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Let $k$ be a field. To give a quasi-coherent sheaf on $\mathrm{Spec} \ k$ is equivalent to giving a $k$-vector space. (To give a coherent sheaf on $\mathrm{Spec} \ k$ is to give a finite-dimensional $k$-vector space.) The dimension of the sheaf is then defined to be the dimension of this $k$-vector space.

edited Feb 12 2012 at 11:16

answered Feb 11 2012 at 9:09

Harized
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I'm not quite sure this true. Surely giving a sheaf on $\textrm{Spec} \ k$ is the same as giving an abelian group? If you want it to be a vector space, don't you need the sheaf to be a sheaf of $\mathcal{O}_{\textrm{Spec} \ k}$-modules? – Daniel Loughran Feb 11 2012 at 16:26

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@Daniel: Strictly speaking, "sheaf" is not a shorthand for "sheaf of abelian groups" either (unless you say so). – a-fortiori Feb 11 2012 at 18:09

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@Harized probably you want to replace the $k$ in "$k$-dimensional" with some other letter. – Keenan Kidwell Feb 11 2012 at 21:25

Mahdi Majidi-Zolbanin · Answer 2 · 2012-02-11 15:25:56Z UTC

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This is a possible answer to the title of your question, it is not intended to interpret $\dim_{k(x)}(\Omega_{X/Y}\otimes k(x))$ (which is already explained in previous responses).

On a Noetherian scheme, dimension of a coherent sheaf could mean dimension of its support. This is in analogy with the affine case. If $R$ is a Noetherian ring and $M$ is a finitely generated $R$-module then $\dim M=\dim\big( R/\mathrm{ann}(M)\big)=\dim\mathrm{Supp}\ M$, since $\mathrm{Supp}\ M=V\big(\mathrm{ann}(M)\big)$.

answered Feb 11 2012 at 15:25

Mahdi Majidi-Zolbanin
1,792●1●2●11

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I'm worried that this post might confuse the OP, as the definition of dimension that you have given does not conincide with the definition that they are interested in. Namely with your definition all locally free sheaves have dimension $\textrm {dim}$ $ X$, whereas $\textrm{dim}_{k(x)}(\Omega_{X/Y} \otimes k(x))$ can certainly vary with $Y$. – Daniel Loughran Feb 11 2012 at 16:21

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Dear Daniel: that is why I began my post by saying: "This is a possible answer to the title of your question, it is not intended to interpret $\dim_{k(x)}(\Omega_{X/Y}\otimes k(x))$". Also, note that the title of this question reads: "*what is the dimension of a sheaf?*" – Mahdi Majidi-Zolbanin Feb 11 2012 at 17:04

What is the dimension of a sheaf?

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