Let $A$ be a Noetherian UFD, how can I compute $\text{Pic}(A):=\text{Pic}(\text{Spec}(A))?$

The Picard group of a scheme $X$ is defined as the group of isomorphism classes of invertible $\mathcal{O}_X$ modules, where an invertible $\mathcal{O}_X$ module is a quasi-coherent module $\mathcal{L}$ for which there's another quasi-coherent $\mathcal{O}_X$ module $\mathcal{N}$ such that $\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{N} \cong \mathcal{O}_X. $

Now if we assume that $X$ is spectral, ie $X=\text{Spec}(A)$, then quasi-coherent $\mathcal{O}_X$ correspond to $A-$modules, thus invertible, quasi-coherent $\mathcal{O}_X$-modules correspond to invertible $A$-modules.

Hence $\text{Pic}(X)=\text{Pic}(A)=\{\left[M \right] \ | \ M \text{ is an invertible } A \text{-module} \}.$

But now I don't know how to compute this group.

  • 3
    $\begingroup$ Are you sure you want to ask about a UFD? The Picard group of a UFD is trivial, and this is a classical result - most algebraic geometry texts will prove it in their section on class/picard groups. $\endgroup$
    – KReiser
    Dec 6, 2021 at 0:20
  • $\begingroup$ @KReiser Yeah I'm actually interested in UFDs. Is the Picard group trivial because of its correspondence to the class group? $\endgroup$
    – Alessandro
    Dec 6, 2021 at 22:50
  • $\begingroup$ Yes. I've posted an answer with the relevant material. $\endgroup$
    – KReiser
    Dec 6, 2021 at 23:24

1 Answer 1


The Picard group of a noetherian UFD is trivial. This follows from the following two statements:

Proposition (Hartshorne II.6.2 or Bourbaki's Commutative Algebra Ch. 1, S3): Let $A$ be a noetherian domain. Then $A$ is a unique factorization domain if and only if $X=\operatorname{Spec} A$ is normal and $\operatorname{Cl} X=0$.

The direction you care about is easy here: if $A$ is a UFD, then every height-one prime is principal, so the class group vanishes.

Proposition (Hartshorne Cor. II.6.16 with more assupmtions, or Vakil 14.2.10): Let $X$ be a noetherian locally factorial scheme. Then $\operatorname{Cl} X\cong \operatorname{Pic} X$.

I've previously written a little about the second claim here.


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