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Let $X$ be a projective scheme and $Y \subset X$ be a projective subscheme. Assume $X, Y$ are connected. Denote by $i:Y \hookrightarrow X$ the closed immersion. Is it always true that the induced morphism $H^0(\mathcal{O}_X) \to H^0(i_*\mathcal{O}_Y)$ surjective?

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migrated from mathoverflow.net Feb 5 '14 at 0:20

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I suggest learning a very little bit of algebraic geometry before asking this kind of question -- and please use mathstackexchange. –  abx Feb 4 '14 at 19:30

3 Answers 3

No. Let $X = P^1$ and $Y$ be a double point (locally given by the ideal $(x^2)$). Then $H^0(X,O_X) = k$ while $H^0(i_*O_Y) = k^2$.

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If you are looking at varieties over an alg. closed field, the answer will be yes, since both $H^0$'s will just equal $k$. But in general the answer is no. Sasha gives a non-reduced example. A reduced example is given by considering $Y = V(X_0^2 + X_1^2) \hookrightarrow \mathbb P^1$, over $\mathbb Q$ (or any field that doesn't contain $\sqrt{-1}$).

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Suppose $\mathcal{I}$ is the ideal sheaf of $\mathcal{O}_X$ that defines $Y$.That is, $i_*\mathcal{O}_Y=\mathcal{O}_X/\mathcal{I}$. Then we have a short exact sequence of sheaves $$0\rightarrow\mathcal{I}\rightarrow\mathcal{O}_X\rightarrow i_*\mathcal{O}_Y\rightarrow0.$$ This gives us a long exact sequence of cohomology $$0\rightarrow H^0(X,\mathcal{I})\rightarrow H^0(X,\mathcal{O}_X)\rightarrow H^0(X,i_*\mathcal{O}_Y)\rightarrow H^1(X,\mathcal{I})\rightarrow H^1(X,\mathcal{O}_X)\rightarrow\ldots.$$ Clearly, what you want is equivalent to the map $H^1(X,\mathcal{I})\rightarrow H^1(X,\mathcal{O}_X)$ being injective (thanks MB!), which is not necessarily true under the conditions you specified.

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$H^1(X,I)=0$ is sufficient, but not necessary. An equivalent condition is that $H^1(X,I) \to H^1(X,\mathcal{O}_X)$ is injective. –  Martin Brandenburg Feb 5 '14 at 2:24
    
Good point! I edited it. –  Must Feb 5 '14 at 2:28

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