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Given a proper, integral scheme $X$ over an arbitrary field and an invertible sheaf $L$ on $X$, does there exist any integer $k \gt 0$ such that the maps

$H^0(X,L^{\otimes k})^{\otimes t} \to H^0(X,L^{\otimes kt})$

become surjective for all integers $t \ge 0$? Or are there any counterexamples?

Georges Elencwajg
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Rührei
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2 Answers2

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No, such a $k\gt 0$ needn't exist.

Take for $X$ a smooth projective curve of genus $g\gt 0$ and for $L$ take $\mathcal O(p)$, where $p$ is some rational point of $X$.
Then $ \operatorname \:{dim } (H^0(X,L))=1$

On the other hand, Riemann-Roch implies that for $N=\operatorname {deg}(L^{\otimes N})\geq g+1$ we have $ \operatorname \:{dim }H^0(X,L^{\otimes N})\geq 2$.
Since of course $ \operatorname \:{dim } (H^0(X,L)^{\otimes N})=1$, the map $H^0(X,L)^{\otimes N} \to H^0(X,L^{\otimes N})$ cannot be surjective and a fortiori the $k\geq 0$ you are asking about doesn't exist.

Edit
The question was edited after I answered it. The original question was about surjectivity of $H^0(X,L)^{\otimes kt} \to H^0(X,L^{\otimes kt})$. So I stand by my answer with the caveat that the non-existence of $k$ refers to the $k $ of the original question, about surjectivity of $H^0(X,L)^{\otimes kt} \to H^0(X,L^{\otimes kt})$. Thanks a lot to QiL'8 for drawing my attention to the modification of the question.

Georges Elencwajg
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    Dear Georges, isn't it true that if $k$ is big enough, then $H^0(X, O(kp))^{\otimes t} \to H^0(X, O(ktp))$ is surjective for all $t\ge 0$ ? –  Jul 27 '13 at 22:02
  • Dear @QiL'8: the question has been edited without mentioning the modification. The original question asked about surjectivity of $H^0(X,L)^{\otimes kt} \to H^0(X,L^{\otimes kt})$, and this is what I answered. It is not the first time that a surreptitious edit of the question has made my answer look completely ridiculous, so thanks a thousand times for your alert. I'll write an edit to explain what went on. – Georges Elencwajg Jul 28 '13 at 07:52
  • Ah, and now I see Marci's answer which also alludes to the original version of the question. – Georges Elencwajg Jul 28 '13 at 08:12
  • Thanks Georges ! I should have noticed the modifications of the question myself. –  Jul 28 '13 at 13:02
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I think you meant the surjectivity of $H^0(X,L^k)^t\rightarrow H^0(X,L^{kt})$. In any case your statement is true for ample linebundles, see the Castelnuovo-Mumford regularity section of Lazarsfeld's Positivity.

Marci
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