# Serres Vanishing Theorem II

I am having problem with the proof in Serre's Vanishing Theorem. If we were to translation line 9 of the proof of Lemma 29.3.1 in generality it seems to say that:

Let $$X$$ be scheme, $$I$$ a sheaf of ideals supported in $$U \subseteq X$$. If $$I(U)=J$$, then $$I(X)=J$$.

Is this true?

## 1 Answer

(To start, note that $$U$$ should be an affine open.)

No, this is nowhere close to being true. Consider $$X=\Bbb P^1_k$$, $$U=\Bbb A^1_k\subset \Bbb P^1_k$$, and $$I$$ the sheaf of ideals of a closed point in $$\Bbb A^1_k$$, which WLOG we can take to be $$0$$. $$I(U)=(x)$$, which is infinite-dimensional as a $$k$$-vector space. On the other hand $$0\to I\to \mathcal{O}_{\Bbb P^1_k} \to \mathcal{O}_{\{0\}} \to 0$$ is exact, which means that $$0\to I(X) \to \mathcal{O}_{\Bbb P^1_k}(X) \to \mathcal{O}_{\{0\}}(X)$$ is exact so $$I(X)$$ injects in to $$k$$, and is in particular finite dimensional as a $$k$$-vector space, contradicting $$I(U)=I(X)$$. In fact, for any quasicompact non-affine scheme, one can find counterexamples like this - pick some $$I$$ with nonvanishing $$H^1$$ (which exists as otherwise the scheme would be affine by the proposition) and play the same game. The real hero of the story here is $$H^1(I)=0$$: this says that you can solve any extension problem you want - I'd recommend re-reading the proof and carefully examining how it's used.

• Thanks, I looked at it again, but am still fundamentally confused with another question (in new post). – CL. May 21 at 7:42