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Divisibility of the Picard group of an affine curve

PDF p. 13: Ful98 Example 1.6.6

Statement

Assume the Axiom of Choice. The Picard group of a smooth affine curve over an algebraically closed field is divisible.

A curve is a finite-type scheme of pure dimension one. A line bundle is a locally free module of rank one, and the Picard group consists of its isomorphism classes under tensor product. Divisible means that for each nonzero integer $n$ and each Picard class $L$ there is a class $M$ with $n M = L$ ; in multiplicative notation this means that the $n$ -fold tensor power of $M$ is $L$ for positive $n$ .

Facts & Assumptions

[F1]

A smooth connected affine curve $C$ over an algebraically closed field has a smooth projective integral completion $X$ , and restriction maps the degree-zero Picard group ${Pic}^{0} (X)$ surjectively onto $Pic (C)$ (Picard group of an affine curve through its boundary).

[F2]

For such $X$ , ${Pic}^{0} (X)$ is isomorphic to $J (k)$ for an abelian variety $J$ over $k$ ; multiplication by every positive integer is surjective on $J (k)$ , without a characteristic restriction (Degree-zero Picard group as points of an abelian variety, Divisibility of rational points of an abelian variety).

[F3]

Roots descend along surjective homomorphisms, and finite products retain surjectivity of a fixed positive multiplication map (Roots descend through a surjective homomorphism, Roots in a finite product of groups).

[F4]

Under the Axiom of Choice, a smooth affine curve is a finite disjoint union of open and closed connected smooth affine curves, and its Picard group is the product of their Picard groups (Connected components of a smooth affine curve, Picard groups of finite disjoint unions).

[F5]

Surjectivity of multiplication by every positive integer is equivalent to divisibility of an abelian group (Positive roots suffice for divisibility).

[A1]

Assume the Axiom of Choice, used in the component suppliers (The Axiom of Choice).

Given: A smooth affine curve $C$ over an algebraically closed field $k$ , and the Axiom of Choice.

Goal: Prove $Pic (C)$ divisible.

Proof

1.1F1F2A1givenchoose

First suppose $C$ is connected and fix a positive integer $n$ . Choose $X$ and the surjective restriction map ${Pic}^{0} (X) \to Pic (C)$ from [F1]. By [F2], transport multiplication by $n$ across the group isomorphism ${Pic}^{0} (X) ≅ J (k)$ ; it is surjective on ${Pic}^{0} (X)$ , including if the characteristic divides $n$ .

2.1F3step 1.1

Apply the surjective-image result in [F3] to restriction. Multiplication by $n$ is therefore surjective on $Pic (C)$ . Since $n$ was arbitrary, this holds for every positive integer.

3.1F3F4A1step 2.1

In general, decompose $C$ into finitely many connected smooth affine components as in [F4]. Step 2.1 applies to each component. By [F3], multiplication by each positive $n$ is surjective on the finite product of their Picard groups, and [F4] transports that conclusion to $Pic (C)$ . The empty product case is included.

4.1F5step 3.1∎

By [F5], $Pic (C)$ is divisible, as required.

Depends on

Used by · 1

The divisor presentation and actual Picard surjection

Dependency tree · 3 levels

Arrows run from prerequisites to the items that use them. External source nodes link directly to the cited text. The view shows up to three dependency steps; open a prerequisite to continue.

External dependencies

Jean Fasel and Maria Yakerson, Vector bundles on singular surfaces and threefolds · PDF page 13, paragraph citing Fulton Example 1.6.6

The Picard group of a smooth affine curve over an algebraically closed field is divisible.

Hypotheses: The curve is smooth and affine over an algebraically closed field.

Reading limits: Read the article paragraph in the supplied extracted text. Fulton Example 1.6.6 was not separately read; the proof below instead uses the explicitly identified Jacobian inputs.

Sources

Jean Fasel and Maria Yakerson, Vector bundles on singular surfaces and threefolds · PDF page 13, paragraph citing Fulton Example 1.6.6