Let $\mathcal{F}$ be a cohorent sheaf on projective scheme $X$. My question is simple... If $\operatorname{dim}\operatorname{Supp}\mathcal{F}$ is zero, then $\mathcal{F}(n) =\mathcal{F}$ for any integer $n$??
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One should read the comments by Matt E before reading the following. I am keeping this just as a record.
Let $X = \hbox{Proj} \, (K [x,y])$ and $Y = \hbox{Proj} \, (K[x,y]/(x^2))$. Write $i: Y \hookrightarrow X$ and $F = i_* \mathcal{O}_Y$. Then $\dim \hbox{Supp} \, F = 0$. One has $\Gamma(X, F) = 1$, but $\Gamma(X, F(i)) = 2$ for $i \ge 1$.
I belive that this has to do with Hilbert-polynomial. Since $\dim F = 0$, the degree of the Hilbert polynomial is $0$, i.e., it is a constant. Since degree of $F$ is $2$, one has $\Gamma(X, F(i)) = 2$ for $i \gg 0$. However, this does not tell us the behavior of the Hilbert-funtion when $i$ is "small".
In the above $\Gamma(X, F) = 1$ is incorrect. The exact sequence of graded $S= K[x,y]$-modules $$ 0 \to (x^2) \to S \to S/(x^2) \to 0 $$ induces an exact sequence of $\mathcal{O}_X$-modules $$ 0 \to \mathcal{O}_X (-2) \to \mathcal{O}_X \to F \to 0. $$ Taking $\Gamma(X,-)$, we obtain an exact sequence $$ H^0(X, \mathcal{O}_X (-2)) \to H^0(X, \mathcal{O}_X) \to H^0(X, F) \to H^1(X, \mathcal{O}_X (-2)) \to H^1(X, \mathcal{O}_X ). $$ Since $H^0(X, \mathcal{O}_X (-2)) = H^1(X, \mathcal{O}_X ) = 0$ and $H^0(X, \mathcal{O}_X) = H^1(X, \mathcal{O}_X (-2)) = 1$, we see that $H^0(X, F) = \Gamma(X,F) = 2$.
Youngsu
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Dear Youngsu, I don't understand your answer. Is $F$ a sheaf or a scheme? You seem to have it meaning both. Regards, – Matt E Dec 18 '13 at 21:19
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P.S. If by $\Gamma(X,F(i))$ you mean $\Gamma(X,\mathcal O_F(i))$, then in fact the dimension is $2$ for any $i \geq 0$. – Matt E Dec 18 '13 at 21:23
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P.P.S. The point is that the map from $\Gamma(X, \mathcal O_X)$ (which is just $K$) to $\Gamma(X,\mathcal O_F)$ (which is two-dimensional) is not surjective, and so the graded ring $K[x,y]/(x^2)$ doesn't see all the elements of $\Gamma(X,\mathcal O_F)$. – Matt E Dec 18 '13 at 21:26
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@MattE: Thanks for the comments. They are very helpful. Is there a way to see that $\Gamma(X,F) = 2$ directly? – Youngsu Dec 19 '13 at 02:54
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Dear Youngsu, Sure. Pretty obviously, $F =$ Spec $k[x]/x^2$, and so $\Gamma(X,\mathcal O_F) = \Gamma(F,\mathcal O_F) = k[x]/x^2$, which is $2$-dimesional. Regards, – Matt E Dec 19 '13 at 04:05
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Let $Z$ be the support of $F$. Then $\mathcal F(n) := \mathcal F \otimes \mathcal O(n) = \mathcal F \otimes (\mathcal O(n)_{| Z})$ (i.e. we can restrict $\mathcal O(n)$ to $Z$ before computing the tensor product).
Because $Z$ is zero-dimensional, it is a union of Specs of Artinian local rings, and so (exercise) any invertible sheaf on $Z$ is trivial. Thus $\mathcal O(n)_{|Z} \cong \mathcal O_Z$, and so $\mathcal F(n) \cong \mathcal F$, as required.
Matt E
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Thanks Matt. But I can't prove your exercise and I wonder why $Z$ is a union of specs of Artinian "local"(??) rings... – Sang Cheol Lee Dec 19 '13 at 03:13
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Dear Sang Cheol Lee, Regarding the exercise, any locally free sheaf over a local ring is free. (This is almost by definition.) As for the assertion about $Z$, any zero dimensional finite type $k$-algebra is Artinian, and so is a product of Artinian local rings. Intuitively, just think about what a zero-dimensional scheme (of finite type) over $k$ has to look like: can you see that is will just have to be a bunch of (possibly thickened) points? Incidentally, I don't understand why you have "local" in quotes. Regards, – Matt E Dec 19 '13 at 04:10
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