I'm trying to solve the following exercise frome Hartshorne's Algebraic Geometry:

Exercise III 3.2. Let $X$ be a reduced noetherian scheme. Show that $X$ is affine if and only if each irreducible component is affine

Clearly, if $X$ is affine, the compoents are closed subschemes so affine as well.

For the converse, I guess one has to use the following Proposition, because the chapter is rather short, but contains this strong criterion to be affine:

Proposition III 3.7. A noetherian scheme $X$ is affine iff $H^1(X, \mathcal{I}) = 0$ for every (coherent) sheaf of ideals $\mathcal{I} \subset \mathcal{O}_X$ iff $H^1(X, \mathcal{F}) = 0$ for all quasi-coherent sheaves $\mathcal{F}$.

But I don't know how to relate sheaf cohomology on $X$ to the cohomology the components. Two thoughts I had:

  1. If $i: Z \hookrightarrow X$ is an irreducible compoent with ideal sheaf $I_Z \subset \mathcal{O}_X$, then the cohomology of the quotient sheaf $i_*\mathcal{O}_Z = \mathcal{O}_X / I_Z$ vanishes, because I can take an injective resolution $\mathcal{O}_Z \to \mathcal{I}^*$ on $Z$, and take the push-forward $i_*\mathcal{I}^*$ of this. Then the sheaves $i_*\mathcal{I}^k$ remain flasque, and I still get a resolution of $i_*\mathcal{O}_Z$, which can be checked on the stalks. So we can use this resolution to commute the cohomology groups of $i_*\mathcal{O}_Z$ on $X$, which will be the same as cohomology of $\mathcal{O}_Z$ on $Z$, i.e. $0$.

Is that reasoning correct?

  1. Do I have to use Exercise 2.3 and 2.4? That seems to give a tool to compute cohomology on $X$, when cohomology on the components can be computed, but I'm not sure what the relation between $H^i_Y(X, \cdot)$ and $H^i(Y, \cdot)$ is. If part 1. here is correct, then I think $$H^i_Y(X, i_*\mathcal{F}) = H^i(Y, \mathcal{F}) = H^i(X, i_*\mathcal{F})$$ is true, because all three groups can by computed by taking the resolution $i_*\mathcal{I}$.

But even if 1. and 2. work out, I still don't know how to show that $H^1(X, I) = 0$ for all ideal sheaves $I$.


1 Answer 1


So I found something here, and polished it a bit. Any feedback would still be appreciated :)

Let $\mathscr{F} \subset \mathcal{O}_X$ be a quasi-coherent sheaf on $X$, and let $\mathscr{I}_1,\dotsc,\mathscr{I}_n$ be the ideal sheaves associated to the irreducible and reduced compontents $X_1,\dots,X_n \subset X$. Consider the filtration $$ \mathcal{F} \supset \mathscr{I}_1 \cdot \mathscr{F} \supset \mathscr{I}_1 \mathscr{I}_2 \cdot \mathscr{F} \supset \dotsb \supset \mathscr{I}_1 \dotsb \mathscr{I}_n \cdot \mathscr{F} = 0. $$ The last equality iholds because $X$ is reduced, so $\mathcal{I}_1 \dots \mathcal{I}_n = 0$.

Let $\mathscr{G}_k$ denote the $k$-th quotient of this filtration. It is a quasi-coherent module on $X_k$, so its cohomology groups (on $X_k$) vanish.

But if $\mathscr{G}$ is any quasi-coherent sheaf on $X_k \xrightarrow{i} X$, then the cohomology groups of $i_* \mathscr{G}$ and $\mathscr{G}$ are the same. This is a special case of Exercise III 8.2, and can also be seen as follows. Take any injective resolution $0 \to \mathscr{G} \to \mathscr{I}^*$. Because $X_k$ is a subspace of $X$, the push-forward is exact in this situation: The stalks at points in $X_k$ are the same, and the stalks outside of $X_k$ are all $0$. So $0 \to i_* \mathscr{G} \to i_* \mathscr{I}^*$ is a resolution, and the $i_* \mathscr{I}^*$ are flasque, so we can use them to compute cohomology. But applying $\Gamma(X, \cdot)$ just reproduces $\Gamma(X_k, \mathscr{I}^i)$, so $H^i(X, i_* \mathscr{G}) = H^i(X_k, \mathscr{G})$.

If we apply the long exact sequence to the sequences $$ 0 \to \mathscr{I}_1 \dotsm \mathscr{I}_{k-1} \cdot \mathscr{F} \to \mathscr{I}_1 \dotsm \mathscr{I}_{k} \cdot \mathscr{F} \to \mathscr{G}_k \to 0, $$ we get a surjection $H^1(X, \mathscr{I}_1 \dotsm \mathscr{I}_{k-1} \cdot \mathscr{F}) \to H^1(X, \mathscr{I}_1 \dotsm \mathscr{I}_{k} \cdot \mathscr{F}) \to 0$. Composing the surjections for all $k$ we get a surjection $0 = H^1(X, \mathscr{I}_1\dotsm\mathscr{I}_n \cdot \mathscr{F}) \to H^1(X, \mathscr{F})$, hence $H^1(X, \mathscr{F}) = 0$.


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