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Picard group of an affine curve through its boundary

PDF p. 13: Divisibility of the normalization Picard group

Statement

Let $k$ be algebraically closed and let $C$ be a smooth connected affine curve over $k$ . Then a smooth projective completion $j : C ↪ \overline{C}$ exists and its boundary $B = \overline{C} ∖ C$ is a nonempty finite set of closed points. For any such completion, the sequence $⨁_{b \in B} Z [b] \to Pic (\overline{C}) \to Pic (C) \to 0$ is exact, the first map sending $[b]$ to $[O_{\overline{C}} (b)]$ and the second being restriction. Each $b$ has degree one. Consequently the degree-zero Picard group ${Pic}^{0} (\overline{C}) (k)$ surjects onto $Pic (C)$ .

Here a curve means a nonempty finite-type scheme of pure dimension one; smoothness is over $k$ . A completion means a dense open immersion into a smooth projective integral curve. A line bundle is a sheaf locally isomorphic to the structure sheaf; its isomorphism classes form the Picard group under tensor product. The degree of a closed point is the dimension of its residue field over $k$ . The degree-zero Picard group is the kernel of the degree homomorphism on line-bundle classes, identified with the $k$ -points of the degree-zero Picard variety by the cited geometric input.

Facts & Assumptions

[F1]

A regular integral separated finite-type curve over a perfect field has a smooth projective completion with finite closed-point boundary; if affine, the boundary is nonempty. For smooth integral curves, divisor classes identify naturally with Picard classes and restriction deletes coefficients outside the open. On a smooth projective curve, degree is additive, a point class has its residue-field degree, and the degree-zero kernel is the group of $k$ -points of the degree-zero Picard variety (Completion, divisors and degree on smooth curves).

[F2]

If $d : K \to Z^{(S)}$ is a homomorphism and $B \subseteq S$ , restriction induces an exact sequence $Z^{(B)} \to coker (d) \to coker (r d) \to 0$ (Localization of a presented divisor class group). Exactness transports across compatible isomorphisms (Transporting an exact sequence through isomorphisms).

[F3]

If a surjective homomorphism $q : A \to D$ kills an element of degree one for a homomorphism $\deg : A \to Z$ , then $q (\ker \deg) = D$ (Degree-zero representatives modulo a boundary class).

[F4]

Closed points of a finite-type scheme over an algebraically closed field have residue field $k$ , by the weak Nullstellensatz on affine charts (Over an algebraically closed field, every maximal ideal is an evaluation ideal). Picard groups and restriction use line bundles and tensor product (Picard group of a scheme).

Given: $k$ and $C$ as stated.

Goal: Prove existence and nonemptiness of the boundary, exactness, and the degree-zero surjection.

Proof

1.1F1given

An algebraically closed field is perfect. Smoothness makes $C$ regular. Its regular local rings are domains, so distinct irreducible components cannot meet; finite type gives finitely many components, hence connectedness makes $C$ integral. Since $C$ is affine over $k$ , it is separated over $k$ . Apply [F1] to obtain a smooth projective integral completion with finite boundary. It is nonempty because an affine positive-dimensional curve is not proper. The same argument proves nonemptiness for any completion in the statement.

2.1F1F2step 1.1

Let $S$ be the closed points of $\overline{C}$ and $K = k (\overline{C})^{\times}$ , the multiplicative group of nonzero rational functions. The principal-divisor homomorphism $d : K \to Z^{(S)}$ assigns each function its finite sum of orders at closed points. The open curve has the same function field, and its principal-divisor homomorphism is restriction of $d$ to $S ∖ B$ . Apply [F2], then use the compatible divisor-class and Picard isomorphisms of [F1]. The boundary basis element $[b]$ maps to $[O_{\overline{C}} (b)]$ , giving the exact sequence in the statement.

3.1F1F4step 1.1step 2.1

By [F4], every $b \in B$ has residue field $k$ , hence degree one; [F1] gives $\deg [O_{\overline{C}} (b)] = 1$ . Exactness in step 2.1 shows that this class lies in the kernel of restriction.

4.1F1F3step 2.1step 3.1∎

Restriction is surjective by step 2.1. Applying [F3] with the boundary class from step 3.1 shows that its restriction from the degree-zero kernel is still surjective. The identification in [F1] gives the stated map from ${Pic}^{0} (\overline{C}) (k)$ .

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Divisibility of the Picard group of an affine curve

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