Let \(X\) be a reduced noetherian scheme. Show that \(X\) is affine if and only if each irreducible component is affine.
Solution
Every irreducible component is given its reduced induced closed-subscheme structure. The noetherian scheme has finitely many irreducible components. We use Serre's affineness criterion and vanishing of higher quasi-coherent cohomology on affine schemes [@Har10a, Theorems III.3.5 and III.3.7].
1. If \(X\) is affine, each irreducible component is affine.
For \(X=\Spec A\), an irreducible component has reduced coordinate ring \(A/\mathfrak p\), where \(\mathfrak p\) is a minimal prime of \(A\). It is the affine closed subscheme \(\Spec(A/\mathfrak p)\).
2. Suppose \(X=Y\cup Z\) is a union of two reduced closed subschemes and \(X\) is reduced.If \(Y\) and \(Z\) are affine, then \(H^1(X,F)=0\) for every coherent sheaf \(F\) on \(X\).
Let \(I\) and \(J\) be the coherent ideal sheaves defining \(Y\) and \(Z\). Their intersection is zero. Indeed, on any affine open \(\Spec A\subseteq X\), the ring \(A\) is reduced, the ideals defining \(Y\) and \(Z\) are radical, and their zero sets cover \(\Spec A\). The intersection of those ideals is radical with zero set all of \(\Spec A\), hence is the nilradical \(0\). In particular \(IJ=0\) as ideal sheaves.
For a coherent \(F\), use the exact sequence \[ 0\longrightarrow IF\longrightarrow F\longrightarrow F/IF\longrightarrow0. \] The quotient \(F/IF\) is annihilated by \(I\), so it is the direct image of a coherent sheaf on \(Y\). The subsheaf \(IF\) is annihilated by \(J\), since \(J(IF)=(JI)F=0\), so it is the direct image of a coherent sheaf on \(Z\). These assertions follow locally from the correspondence between modules annihilated by an ideal and modules over the quotient ring; all these sheaves are coherent because \(X\) is noetherian.
Cohomology of a sheaf on a closed subspace agrees with cohomology of its direct image [@Har10a, Lemma III.2.10]. Affineness of \(Y\) and \(Z\) therefore gives \[ H^1(X,IF)=0,\qquad H^1(X,F/IF)=0. \] The long exact sequence of the displayed short exact sequence forces \(H^1(X,F)=0\).
3. If every irreducible component is affine, then \(X\) is affine.
Induct on the number \(r\) of irreducible components. For \(r=0\), the scheme is empty and is \(\Spec(0)\). For \(r=1\), reducedness identifies \(X\) with its sole reduced component, which is affine by hypothesis.
For \(r>1\), take \(Y\) to be one component and let \(Z\) be the union of the other \(r-1\) components, with its reduced induced structure. The scheme \(Z\) is reduced and noetherian, and its irreducible components, with their reduced structures, are precisely the remaining components of \(X\). They are affine by hypothesis, so induction gives affineness of \(Z\). The chosen \(Y\) is affine as well. Step 2 gives \(H^1(X,F)=0\) for every coherent \(F\), in particular for every coherent ideal sheaf. Serre’s criterion therefore makes \(X\) affine.
4. Q.E.D.
Step 1 proves the forward implication and step 3 proves the converse.