Let $X$ be a projective scheme of pure dimension $1$. Let $U$ be a open subscheme and $j:U \to X$ the open immersion. Let $\mathcal{F}$ be a coherent sheaf on $U$. Denote by $j_!(\mathcal{F})$ the extension of $\mathcal{F}$ by zero (as mentioned in Hartshorne, Algebraic geometry Ex. II.$1.19$). Is it true that $H^1(U,\mathcal{F})=H^1(X,j_!(\mathcal{F}))$?
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2 Answers
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Actually, this is almost never true.
Claim Let $X$ be a separated one-dimensional scheme of finite type over an algebraically closed field. Let $U\subseteq X$ be a dense open set that does not contain any irreducible component of $X$. Further let $\mathscr F$ be a coherent sheaf on $X$. Then
- $H^1(U,\mathscr F|_U)=0$, and
- $H^1(X,j_!(\mathscr F|_U))\twoheadrightarrow H^1(X,\mathscr F)$ is surjective.
In particular, if $H^1(X,\mathscr F)\neq 0$, then $H^1(U,\mathscr F|_U)\neq H^1(X,j_!(\mathscr F|_U))$
Proof The assumptions imply that
- $U$ is affine, and
- $Z=X\setminus U$ is zero-dimensional.
The above 1. implies 1. in the Claim and via the short exact sequence $$ 0\to j_!(\mathscr F|_U) \to \mathscr F \to j_*(\mathscr F|_Z) \to 0 $$
the above 2. implies 2. in the Claim. $\square$
answered Jan 29, 2014 at 5:28
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$\begingroup$ do you perhaps want Z to also intersect every irreducible component of X? (Otherwise it seems you could have U = X.) $\endgroup$ Commented Jan 29, 2014 at 14:37
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$\begingroup$ Roy, you're right, thanks. I meant that, but neglected to write it there. Without that $U$ is not necessarily affine. $\endgroup$ Commented Jan 31, 2014 at 0:11
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No, this is not true. This works for extensions from closed subsets. Take for $X$ an irrational curve, so that $H^1(X;\mathcal{O}_X)\ne0$, and let $U=X\setminus\text{point}$ and $\mathcal{F}=\mathcal{O}_U$, which is the restriction of $\mathcal{O}_X$. You immediately get $H^1(X,j_!\mathcal{F})\twoheadrightarrow H^1(X;\mathcal{O}_X)$, whereas $H^1(U,\mathcal{F})=0$.
answered Jan 28, 2014 at 20:00