Let \(X\) be an integral, separated, regular, one-dimensional scheme of finite type over \(k\), which is not proper over \(k\). Then \(X\) is affine.
Hint: Embed \(X\) in a (proper) curve \(\bar{X}\) over \(k\), and use (Ex. 1.2) to construct a morphism \(f: \bar{X} \to \PP^1\) such that \(f^{-1}(\AA^1)=X\).
Solution
1. There is an open immersion\[ j:X\hookrightarrow\bar X \]into a regular projective integral curve \(\bar X\), and\[ \bar X\setminus X=\{P_1,\ldots,P_r\} \]is a finite nonempty set of closed points.
A regular separated curve of finite type over a field has a regular projective compactification. Equivalently, one may take a projective completion and then normalize it; regularity of \(X\) identifies \(X\) with an open subscheme of the normalization.
Because \(\bar X\) is a noetherian integral curve, every proper closed subset is zero-dimensional and hence finite. Thus the complement of the dense open \(X\) consists of finitely many closed points.
Finally \(r>0\). If the complement were empty, then \[ X=\bar X \] would be projective, hence proper over \(k\), contrary to the hypothesis.
2. There exists a nonconstant rational function\[ f\in K(\bar X) \]whose poles occur precisely at \(P_1,\ldots,P_r\).
Apply Exercise IV.1.2 on the proper regular curve \(\bar X\) to the finite set \[ P_1,\ldots,P_r. \] It gives a rational function \(f\) having a pole of positive order at every \(P_i\) and regular at every point of \[ \bar X\setminus\{P_1,\ldots,P_r\}=X. \] Since \(f\) has at least one pole, it is not constant.
3. The rational function \(f\) determines a morphism\[ \bar f:\bar X\longrightarrow\PP^1 \]such that\[ \bar f^{-1}(\infty)=\{P_1,\ldots,P_r\}. \]
A nonzero rational function on a regular complete curve defines the usual morphism to \(\PP^1\): where \(f\) is regular use the affine coordinate \(z=f\), and near a pole use the affine coordinate \[ w=f^{-1} \] around \(\infty\). Regularity of the local rings of \(\bar X\) makes these two descriptions glue.
A point maps to \(\infty\) exactly when \(f\) has a pole there. By step 2, these are precisely the points \(P_1,\ldots,P_r\). Hence \[ \bar f^{-1}(\AA^1) = \bar X\setminus\{P_1,\ldots,P_r\} = X. \]
4. The morphism\[ \bar f:\bar X\longrightarrow\PP^1 \]is finite.
2. Both source and target areproper integral curves, and \(\bar X\) is regular. Therefore the standardfiniteness theorem for maps of proper curvesfinite morphisms between proper curvesapplies: a nonconstant morphism\[ \bar X\longrightarrow\PP^1 \]is finite.
5. The scheme \(X\) is affine.
By step 3, \[ X=\bar f^{-1}(\AA^1). \] Base change of a finite morphism is finite, so restriction gives a finite morphism \[ f:X\longrightarrow\AA^1. \] Finite morphisms are affine. Since \[ \AA^1=\Spec k[t] \] is affine, its inverse image under the finite morphism \(f\) is affine. Thus \(X\) is affine.
6. Q.E.D.
Step 5 is exactly the required conclusion.