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## The Picard groups

Let $R$ be a commutative ring. From $R$ we can construct the category $R\text{-Mod}$ of $R$-modules, which becomes a symmetric monoidal category when equipped with the tensor product of $R$-modules. Now, whenever we have a monoidal operation (for example, the multiplication on a ring), it’s interesting to look at the invertible things with respect to that operation (for example, the group of units of a ring). This suggests the following definition.

Definition: The Picard group $\text{Pic}(R)$ of $R$ is the group of isomorphism classes of $R$-modules which are invertible with respect to the tensor product.

By invertible we mean the following: for $L \in \text{Pic}(R)$ there exists some $L^{-1}$ such that the tensor product $L \otimes_R L^{-1}$ is isomorphic to the identity for the tensor product, namely $R$.

In this post we’ll meander through some facts about this Picard group as well as several variants, all of which capture various notions of line bundle on various kinds of spaces (where the above definition captures the notion of a line bundle on the affine scheme $\text{Spec } R$).

Some propositions

Proposition: An invertible module is finitely presented and projective.

Proof. If $L$ is invertible, then the functor $L \otimes_R (-) : R\text{-Mod} \to R\text{-Mod}$ is an autoequivalence of categories with inverse $L^{-1} \otimes_R (-)$, and consequently it preserves all categorical properties; in addition, it sends $R$ to $L$, so it follows that any categorical property of $R$ is also a categorical property of $L$. In particular, since being projective is a categorical property (namely the property that $\text{Hom}(L, -)$ is exact) and $R$ is projective, $L$ is also projective.

Less obviously, being finitely presented is also a categorical property: an $R$-module $M$ is finitely presented iff $\text{Hom}(M, -)$ preserves filtered colimits (that is, $M$ is a compact object).

We can avoid appealing to this fact with the following more hands-on argument. By assumption, $L \otimes_R L^{-1} \cong R$. Let

$\displaystyle \sum_{i=1}^n a_i \otimes b_i, a_i \in L, b_i \in L^{-1}$

be the element of $L \otimes_R L^{-1}$ representing $1 \in R$. Then the map

$\displaystyle R^n \otimes_R L^{-n} \ni (r_1, \dots r_n) \otimes (\ell_1, \dots \ell_n) \mapsto \sum_{i, j} r_i a_i \otimes \ell_j \in L \otimes_R L^{-1} \cong R$

is surjective, where $L^{-n}$ denotes $(L^{-1})^n$, as can be seen by setting $r_1 = \dots = r_n, \ell_i = b_i$. Surjective morphisms of $R$-modules are precisely the epimorphisms, and this is a categorical property, so tensoring with $L$ we get an epimorphism $R^{n^2} \to L$, from which it follows that $L$ is finitely generated (by the elements $a_i$), and for a projective module this is equivalent to being finitely presented. $\Box$

Proposition: Being invertible is preserved under extension of scalars. More precisely, if $f : R \to S$ is a morphism of commutative rings, then the extension of scalars functor

$\displaystyle f_{\ast} : R\text{-Mod} \ni M \mapsto M \otimes_R S \in S\text{-Mod}$

sends invertible modules to invertible modules, and in fact induces a homomorphism $\text{Pic}(R) \to \text{Pic}(S)$.

Proof. It suffices to show that extension of scalars is a monoidal functor. More or less this boils down to having a natural isomorphism

$\displaystyle (M \otimes_R N) \otimes_R S \cong (M \otimes_R S) \otimes_S (S \otimes_R N)$.

But by the associativity of the tensor product, the RHS is just

$M \otimes_R (S \otimes_S S) \otimes_R N$

and $S \otimes_S S \cong S$, so the conclusion follows using the commutativity of the tensor product of modules over commutative rings. $\Box$

Theorem: The following conditions on an $R$-module $L$ are equivalent.

1. $L$ is invertible.
2. $L$ is locally free of rank $1$: that is, for every prime ideal $P$ of $R$, the localization $L_P \cong L \otimes_R R_P$ is a free $R_P$-module of rank $1$.
3. The natural map $L \otimes_R L^{\ast} \to R$ is an isomorphism, where $L^{\ast} = \text{Hom}_R(L, R)$ is the dual module. In this case $L^{\ast} \cong L^{-1}$.

Proof. 1) $\Rightarrow$ 2): Since localization is a special case of extension by scalars, $L_P$ remains invertible, hence is in particular finitely presented projective. We can give an independent argument that this is true as follows: since localization is exact, it preserves finite presentations, so $L_P$ is finitely presented. Since we have a tensor-hom adjunction

$\text{Hom}_{R_P}(L_P, M) \cong \text{Hom}_R(L, M)$

it follows that if $\text{Hom}_R(L, -)$ is exact then so is $\text{Hom}_{R_P}(L_P, -)$, hence localization preserves projectivity.

But since $R_P$ is a local ring, it follows that $L_P$ must be free, and since rank is multiplicative under tensor product, $L_P$ must be free of rank $1$: that is, we must have $L_P \cong R_P$.

2) $\Rightarrow$ 3): being an isomorphism is a local property, so the natural map $L \otimes_R L^{\ast} \to R$ is an isomorphism iff its localizations $L_P \otimes_{R_P} L^{\ast}_P \to R_P$ are. But if $L$ is locally free of rank $1$ then $L_P \cong L_P^{\ast} \cong R_P$ for all $P$.

3) $\Rightarrow$ 1): by definition. $\Box$

The proposition shows in particular that invertibility is a local condition: $L$ is invertible iff $L_P$ is invertible for all $P$. We still haven’t given any interesting examples of invertible modules, though.

The ideal class group

Proposition: Let $R$ be an integral domain with fraction field $K$. Then every invertible module $L$ over $R$ is isomorphic to a fractional ideal of $R$.

Proof. $L$ is projective, hence in particular torsion-free, so the natural inclusion $L \to L \otimes_R K \cong K$ is an embedding. Since $L$ is finitely generated, we can multiply the image of this embedding by the product of the denominators of the images of its generators in $K$, and we conclude that an $R$-multiple of the image of $L$ lands in $R \subseteq K$ as desired. $\Box$

Proposition: Let $R$ be a Dedekind domain. Then a finitely generated module $M$ over $R$ is projective iff it is torsion-free.

Proof. Projectivity is a local property, so $M$ is projective iff $M_P$ is projective for all $P$. If $M$ is torsion-free, then so is $M_P$. Since the localizations $R_P$ are all DVRs, hence in particular PIDs, it follows by the structure theorem for finitely generated modules over a PID that each $M_P$ is free, hence in particular projective. $\Box$

Proposition: Let $I, J$ be two fractional ideals of a Dedekind domain $R$. Then $IJ \cong I \otimes_R J$.

Proof. WLOG $I, J$ are ideals. We want to show that the natural surjection

$I \otimes_R J \ni i \otimes j \mapsto ij \in IJ \subseteq R$

is an isomorphism. Since $I, J$ are fractional ideals, they are torsion-free, hence projective, hence flat, so $I \otimes_R J$ embeds into $R \otimes_R R$. Any element in the kernel of the natural surjection above must therefore also be in the kernel of the natural map $R \otimes_R R \to R$, but this natural map is an isomorphism. $\Box$

Theorem: The Picard group $\text{Pic}(R)$ of a Dedekind domain $R$ is canonically isomorphic to its ideal class group $\text{Cl}(R)$ of invertible fractional ideals modulo principal invertible fractional ideals. In particular, $\text{Pic}(R)$ is nontrivial iff $R$ is not a UFD.

Proof. By the above proposition, any invertible fractional ideal gives rise to an invertible module, and moreover multiplication of fractional ideals corresponds to tensor product of ideals. The kernel of this map consists of invertible fractional ideals which are isomorphic to the trivial module, which is precisely the principal invertible fractional ideals; hence we get an injection from the ideal class group to $\text{Pic}(R)$. We also showed that every invertible module comes from a fractional ideal, necessarily also invertible, so this injection is a surjection and hence a bijection. $\Box$

Example. Let $R = \mathbb{Z}[\sqrt{-5}]$ be the ring of integers of the number field $K = \mathbb{Q}(\sqrt{-5})$. This is a Dedekind domain which is not a UFD because of the non-unique factorization

$\displaystyle 6 = (1 + \sqrt{-5})(1 - \sqrt{-5}) = 2 \cdot 3$.

Here $1 + \sqrt{-5}, 1 - \sqrt{-5}, 2, 3$ have norms $6, 6, 4, 9$ respectively. An examination of the norm form $N(a + b \sqrt{-5}) = a^2 + 5b^2$ reveals that $\mathbb{Z}[\sqrt{-5}]$ has no elements of norm $2$ or $3$, hence all four of these elements are irreducible. The factorization above refines to unique prime ideal factorizations

$2 = (2, 1 + \sqrt{-5})(2, 1 - \sqrt{-5}) = P^2$

$3 = (3, 1 + \sqrt{-5})(3, 1 - \sqrt{-5}) = Q \overline{Q}$

$1 + \sqrt{-5} = (2, 1 + \sqrt{-5})(3, 1 + \sqrt{-5}) = PQ$

$1 - \sqrt{-5} = (2, 1 - \sqrt{-5})(3, 1 - \sqrt{-5}) = P \overline{Q}$

which gives $[P] = [Q] = [\overline{Q}]$ and $[P]^2 = [1]$ in the ideal class group. Because $P$ has norm $2$ and there exist no elements of $\mathbb{Z}[\sqrt{-5}]$ of norm $2$, we also know that $[P] \neq [1]$ in the ideal class group.

By the Minkowski bound, the ideal class group is generated by ideals of norm at most

$\displaystyle \sqrt{|-20|} \left( \frac{4}{\pi} \right) \frac{2!}{2^2} \approx 2.85$

and since $P$ is the only prime ideal lying over $(2)$, the ideal class group must be generated by $P$. Hence we compute the Picard group to be

$\displaystyle \text{Pic}(\mathbb{Z}[\sqrt{-5}]) \cong \text{Cl}(\mathbb{Z}[\sqrt{-5}]) \cong \mathbb{Z}_2$.

This example turns out to be minimal in the sense that $\mathbb{Q}(\sqrt{-5})$ is the number field of smallest discriminant (in absolute value) whose ideal class group is nontrivial.

Example. Let $R = \mathbb{C}[x, y]/f(x, y)$ be the ring of functions on a smooth affine curve

$C \cong \text{Spec } R \cong \{ (x, y) \in \mathbb{C}^2 \mid f(x, y) = 0 \}$

in the complex plane, and let $\overline{C} \subseteq \mathbb{CP}^2$ denote its projective closure in the complex projective plane. Then ideals of $r$ can be identified with effective divisors $D = \sum n_p p, n_p \ge 0$ on $C$, and principal ideals of $R$ can be identified with effective principal divisors. Since meromorphic functions on $C$ are quotients of functions in $R$, it follows that $\text{Pic}(R)$ is canonically isomorphic to the divisor class group $\text{Cl}(C)$ of $C$, which is closely related to the divisor class group of $\overline{C}$, which is in turn very well-understood and which we will turn to later. The relationship is the following: restriction of divisors gives a surjection

$\displaystyle \text{Cl}(\overline{C}) \to \text{Cl}(C)$

(a priori it only gives a surjection on divisors, but since $C$ and $\overline{C}$ have the same meromorphic functions, the natural map on divisors respects the quotient by principal divisors). The kernel of this map clearly contains the subgroup of $\text{Cl}(\overline{C})$ generated by the points in $\overline{C} \setminus C$, and in fact it must be precisely this subgroup: if $\sum n_p p$ is a divisor in the kernel, then $\sum_{p \not \in C} n_p p$ is the divisor of some function $f$ on $C$ (that is, some element of $R$), but $f$ extends to a meromorphic function on $\overline{C}$ and hence has a principal divisor whose restriction to $C$ is precisely $\sum_{p \not \in C} n_p p$. If there are $k$ points in $\overline{C} \setminus C$, then we have an exact sequence

$\displaystyle \mathbb{Z}^k \to \text{Cl}(\overline{C}) \to \text{Cl}(C) \to 0$.

As we’ll see later, if $\overline{C}$ has genus $g$ then

$\displaystyle \text{Cl}(\overline{C}) \cong \mathbb{Z} \times J(\overline{C}) \cong \mathbb{Z} \times T^{2g}$

where $J(\overline{C})$ denotes the Jacobian and $T^{2g}$ denotes a torus of (real) dimension $2g$. In particular, $\text{Cl}(\overline{C})$ is uncountable as soon as $g \ge 1$, and hence its quotient $\text{Cl}(C)$ by the image of $\mathbb{Z}^k$ is nontrivial as soon as $g \ge 1$.

The topological Picard groups

The characterization of invertible modules as locally free modules of rank $1$ suggests that invertible modules over a commutative ring $R$ should be thought of as (modules of sections of) line bundles on $\text{Spec } R$. This idea is strongly supported by variants of the Serre-Swan theorem, such as the following.

Theorem: Let $X$ be a compact Hausdorff space and let $C(X, \mathbb{R})$ resp. $C(X, \mathbb{C})$ be the ring of continuous real-valued resp. complex-valued functions on $X$. Assigning a real resp. complex vector bundle on $X$ its module of continuous sections gives an equivalence of monoidal categories between real resp. complex vector bundles on $X$ and finitely presented projective modules over $C(X, \mathbb{R})$ resp. $C(X, \mathbb{C})$.

Corollary: $\text{Pic}(C(X, \mathbb{R}))$ resp. $\text{Pic}(C(X, \mathbb{C}))$ is canonically isomorphic to the abelian group of topological real resp. complex line bundles on $X$, which is in turn canonically isomorphic to $H^1(X, \mathbb{Z}_2)$ resp. $H^2(X, \mathbb{Z})$.

This theorem suggests a natural definition of the real resp. complex Picard groups of an arbitrary space $X$, not necessarily compact Hausdorff, namely the group of isomorphism classes of real resp. complex line bundles on $X$.

Example. Let $X = S^1$; this is arguably the simplest example of a space with a nontrivial real line bundle over it. Since $H^1(S^1, \mathbb{Z}_2) \cong \mathbb{Z}_2$, there are exactly two (isomorphism classes of) real line bundles over $S^1$, one trivial and one nontrivial. The nontrivial line bundle is the Möbius bundle. Its $C(S^1, \mathbb{R})$-module of continuous sections can be identified with the function space

$\{ f : \mathbb{R} \to \mathbb{R} \mid f(x + 1) = - f(x) \}$

where $C(S^1, \mathbb{R})$ itself is thought of as the function space

$\displaystyle \{ f : \mathbb{R} \to \mathbb{R} \mid f(x + 1) = f(x) \}$

and the module structure is given by pointwise multiplication. (So here we are thinking of $S^1$ as the quotient $\mathbb{R}/\mathbb{Z}$.)

Example. Let $X = S^2$; this is arguably the simplest space with a nontrivial complex line bundle over it. Since $H^2(S^2, \mathbb{Z}) \cong \mathbb{Z}$, there are countably many (isomorphism classes of) complex line bundles over $S^2$, all of which are powers of a single generator.

Thinking of $S^2$ as the complex projective line $\mathbb{CP}^1$, there are two choices for such a generator, one given by the tautological bundle $\mathcal{O}(-1)$ which assigns to a point in $\mathbb{CP}^1$ the complex line in $\mathbb{C}^2$ it represents, and the other given by its dual $\mathcal{O}(1)$; the other line bundles are given by $\mathcal{O}(n) \cong \mathcal{O}(1)^n$, where if $n$ is negative then $\mathcal{O}(1)^n \cong \mathcal{O}(-1)^{-n}$ as expected.

The bundles $\mathcal{O}(n), n \ge 1$ are important in algebraic geometry because their spaces of algebraic (or equivalently, holomorphic) sections are precisely the homogeneous polynomials of degree $n$. Their $C(S^2, \mathbb{C})$-modules of continuous sections can be identified with the function spaces

$\displaystyle \{ f : \mathbb{C}^2 \setminus \{ (0, 0) \} \to \mathbb{C} \mid f(\lambda x, \lambda y) = \lambda^n f(x, y), \lambda \in \mathbb{C}^{\times} \}$

where $C(S^2, \mathbb{C})$ itself is thought of as the function space

$\displaystyle \{ f : \mathbb{C}^2 \setminus \{ (0, 0) \} \to \mathbb{C} \mid f(\lambda x, \lambda y) = f(x, y), \lambda \in \mathbb{C}^{\times} \}$

and, as above, the module structure is given by pointwise multiplication. (So here we are thinking of $S^2 \cong \mathbb{CP}^1$ as the quotient $(\mathbb{C}^2 \setminus \{ (0, 0) \}) / \mathbb{C}^{\times}$.)

We can make the construction for $S^1$ look more like the construction for $S^2$ by thinking of $S^1$ as the real projective line $\mathbb{RP}^1$ and exhibiting it as the quotient of $\mathbb{R}^2 \setminus \{ (0, 0) \}$ by the action of $\mathbb{R}^{\times}$.

The algebraic and analytic Picard groups

The discussion of $\mathbb{CP}^1$ above blurred the distinction between topological, holomorphic, and algebraic line bundles, so it’s worth making the distinction in general.

To talk about topological vector bundles on a space $X$ only requires that it be equipped with a topology. To talk about holomorphic vector bundles requires that $X$ be equipped with the structure of a complex manifold. Finally, to talk about algebraic vector bundles requires that $X$ be equipped with the structure of a scheme, e.g. $X$ might be a complex variety. We can talk about all three on a smooth complex variety. On $\mathbb{CP}^1$ all three notions coincide, but in general they all differ.

The GAGA principle implies that the classifications of holomorphic and algebraic vector bundles on a smooth projective complex variety coincide; in particular, the classifications of holomorphic and algebraic line bundles on $\mathbb{CP}^1$ coincide. However, it is not true in general that these classifications also coincide with the classification of topological line bundles, although it is true for the complex projective spaces $\mathbb{CP}^n$. In general, if $X$ is a complex manifold, then the exponential sheaf sequence

$\displaystyle 0 \to \mathbb{Z} \to \mathcal{O}_X \xrightarrow{\exp(2\pi i(-))} \mathcal{O}_X^{\times} \to 0$

gives rise to a long exact sequence in sheaf cohomology

$\displaystyle \cdots \to H^1(X, \mathcal{O}_X) \to H^1(X, \mathcal{O}_X^{\times}) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}_X) \to \cdots$

where $H^1(X, \mathcal{O}_X^{\times})$ turns out to be the Picard group $\text{Pic}(X)$ of holomorphic line bundles on $X$ and the connecting homomorphism to $H^2(X, \mathbb{Z})$ sends such a line bundle to its first Chern class $c_1$, which completely determines the underlying topological line bundle but not its holomorphic structure.

The classifications of holomorphic and topological line bundles on $X$ coincide iff this connecting homomorphism $L \mapsto c_1(L)$ is an isomorphism. By exactness, this is guaranteed if $H^1(X, \mathcal{O}_X) \cong H^2(X, \mathcal{O}_X) \cong 0$, which in particular holds if $X$ is a Stein manifold (e.g. a smooth affine variety) by Cartan’s theorem B. More generally, the Oka-Grauert principle asserts that the classifications of holomorphic and topological vector bundles on a Stein manifold coincide.

But $H^1(X, \mathcal{O}_X)$ and $H^2(X, \mathcal{O}_X)$ are both nontrivial in general, so we can’t expect the holomorphic and topological classifications to agree in general. And because the holomorphic and topological classifications agree on smooth affine varieties, a smooth affine variety on which the algebraic and topological classifications disagree also shows that the algebraic and holomorphic classifications disagree in general.

Example. On a smooth projective curve $X = \Sigma_g$ of genus $g \ge 1$, the divisor class group $\text{Cl}(X)$ turns out to be naturally isomorphic to the Picard group, with the first Chern class map $H^1(X, \mathcal{O}_X^{\times}) \to H^2(X, \mathbb{Z})$ corresponding to the degree map

$\displaystyle \deg : \text{Cl}(X) \ni \sum n_p p \mapsto \sum n_p \in \mathbb{Z}$

under the isomorphism $H^2(X, \mathbb{Z}) \cong \mathbb{Z}$ given by pairing with the fundamental class. Hence the degree zero divisor class group $\text{Cl}^0(X)$ is naturally isomorphic to the Picard group $\text{Pic}^0(X)$ of line bundles with vanishing first Chern class. This group measures the difference between the holomorphic / algebraic and the topological classifications of line bundles on $X$.

An inspection of the long exact sequence associated to the exponential sequence shows that this group is in turn isomorphic to the quotient

$\displaystyle J(X) \cong H^1(X, \mathcal{O}_X) / H^1(X, \mathbb{Z})$

which is one description of the Jacobian. As we saw previously, $H^1(X, \mathcal{O}_X) \cong \mathbb{C}^g$, and we know that $H^1(X, \mathbb{Z}) \cong \mathbb{Z}^{2g}$. This exhibits $J(X)$ as a complex torus $\mathbb{C}^g / \mathbb{Z}^{2g}$ (at least provided that we show that the image of $H^1(X, \mathbb{Z})$ in $H^1(X, \mathcal{O}_X)$ is a lattice). In particular, once $g \ge 1$, the holomorphic / algebraic and topological classifications of line bundles on $X$ disagree.

Example. As when we were discussing Dedekind domains, let $X$ be an elliptic curve minus a point. Then $X$ is a smooth affine variety, hence in particular a Stein manifold, so the classifications of holomorphic and topological line bundles agree: both Picard groups are isomorphic to $H^2(X, \mathbb{Z})$, which vanishes since $X$, being topologically a torus minus a point, deformation retracts onto its $1$-skeleton. But the classification of algebraic line bundles is given by the divisor class group, which we saw earlier was uncountable. In particular, the holomorphic and algebraic classifications of line bundles on $X$ disagree.

### 12 Responses

1. In the first proposition you say that a projective finitely generated module is finitely presented. Don’t you need the hypothesis that R is noetherian?

• No, I don’t need that hypothesis. For a projective module, both finitely presented and finitely generated are equivalent to being a retract of a finitely generated free module.

2. Ah yes I see, I was mixing up direct sum and tensor product. Better to think of R^n as nR in this context…

• So this raises another question. What is L generated by? It would seem to require n^2 elements now, instead of n (for instance, the comment about being generated by the a_i seems to need tweaking). Do those elements somehow correspond to a_i x b_j as (i,j) range? That would seem strange since we always paired them as a_i x b_i in the construction.

• Okay sorry for all these posts but I’ve sorted it all out now. L is still generated by just the n elements a_i, but we need a map from R^{n^2} because the map is diagonal (e_i x e_j maps to delta_ij a_i)

To fix the issue in your current definition of the map, just sum over i instead of over i,j.

3. I think there is something funny going on in your proof that invertible modules are finitely presented, when you define the map from R^n x L^-n to R. Specifically, where does v live? It appears like your map is actually from R^n x L^-1 to R. But now I don’t think surjectivity follows immediately since the b_i are different but you use the same v.

Any thoughts?

• Ah, sorry, that was a notational goof. It’s been fixed.

• Okay, I figured that’s what you meant. Then there’s one more slight issue: tensoring now yields a map from R^n to L^n (instead of R^n to L). So you compose with a projection L^n to L and then conclude.

• I think tensoring now yields a map from $R^{n^2}$ to $L$, but after that fix I think everything’s fine.

• Qiaochu, I just realized there’s another issue with your map now; it need not be surjective. The expression $\sum_{i,j}r_i a_i\otimes b_j$ factors into $\left(\sum_i r_i a_i\right)\otimes \left(\sum_i \ell_i\right)$, which is now a “pure” tensor in $L\otimes_R L^{-1}$, and in particular can’t represent 1 (if 1 is chosen to be a mixed tensor).

By the way, do you have a reference on this Picard group material that you could point me towards?

4. Great stuff! Typo in the proof of the first proposition of Ideal Class Group section: “lands in $D\subseteq K$” should read $R\subseteq K$.

• Ah, thanks. I was originally using D but switched to R after revising the post a bit.