Show that a separated, one-dimensional scheme of finite type over \(k\), none of whose irreducible components is proper over \(k\), is affine.
Hint: Combine (Ex. 1.3) with (III, Ex. 3.1, Ex. 3.2, Ex. 4.2).
Solution
1. It is enough to prove that every irreducible component of\(X_{\mathrm{red}}\), with its reduced induced structure, is affine.
The scheme \(X\) is noetherian because it is of finite type over a field. By Exercise III.3.1, \(X\) is affine if and only if \(X_{\mathrm{red}}\) is affine.
The scheme \(X_{\mathrm{red}}\) is reduced and noetherian. By Exercise III.3.2, it is affine if and only if each of its irreducible components is affine. Thus it suffices to prove the assertion for one reduced irreducible component \[ C\subseteq X_{\mathrm{red}}. \]
2. The component \(C\) is an integral, separated, one-dimensionalscheme of finite type over \(k\), and it is not proper over \(k\).
By construction \(C\) is reduced and irreducible, hence integral. It is a closed subscheme of the separated finite-type \(k\)-scheme \(X\), so it is separated and of finite type over \(k\). The hypothesis says precisely that this irreducible component is not proper.
It remains only to exclude dimension \(0\). If \(\dim C=0\), then \(C\) is the spectrum of its function field, which is a finite extension of \(k\) by Zariski’s lemma. Hence \[ C\longrightarrow\Spec k \] is finite and therefore proper, a contradiction. Thus \[ \dim C=1. \]
3. Let\[ \nu:\widetilde C\longrightarrow C \]be the normalization. Then \(\nu\) is finite and surjective, and\(\widetilde C\) is an integral, separated, regular, one-dimensionalscheme of finite type over \(k\).
Because \(C\) is an integral finite-type \(k\)-scheme, its normalization is finite and birational. In particular \(\nu\) is finite and dominant; since a finite morphism is closed and its image contains the generic point of the irreducible scheme \(C\), it is surjective.
The normalization \(\widetilde C\) is integral and normal. It is finite over the finite-type \(k\)-scheme \(C\), hence itself of finite type over \(k\); it is separated because it is finite over the separated scheme \(C\). A noetherian normal local domain of dimension \(1\) is a discrete valuation ring. Therefore every local ring of the one-dimensional normal scheme \(\widetilde C\) is regular, so \(\widetilde C\) is regular.
4. The normalization \(\widetilde C\) is not proper over \(k\).
3. A surjective image of a universally closed\(k\)-scheme is universally closed over \(k\): after arbitrary basechange, the image of a closed subset of \(C\) equals the image of itsclosed inverse image in \(\widetilde C\).
Thus \(C\to\Spec k\) would be universally closed. Step 2 already shows that \(C\) is separated and of finite type over \(k\). Hence \(C\) would be proper over \(k\), contradicting the hypothesis. Therefore \(\widetilde C\) is not proper.
5. The normalization \(\widetilde C\) is affine.
By steps 3 and 4, \(\widetilde C\) is integral, separated, regular, one-dimensional, of finite type over \(k\), and not proper. Therefore Exercise IV.1.3 applies and gives \[ \widetilde C\ \text{affine}. \]
6. The component \(C\) is affine.
5. Chevalley’s theoremExercise III.4.2therefore implies that \(C\) is affine.
7. The original scheme \(X\) is affine.
Step 6 proves that every irreducible component of \(X_{\mathrm{red}}\) is affine. Exercise III.3.2 then gives that \(X_{\mathrm{red}}\) is affine. Finally Exercise III.3.1 gives that \(X\) itself is affine.
8. Q.E.D.
Step 7 is the required conclusion.