# Picard Group of the spectrum of a Noetherian UFD.

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.

• 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. Dec 6, 2021 at 0:20
• @KReiser Yeah I'm actually interested in UFDs. Is the Picard group trivial because of its correspondence to the class group? Dec 6, 2021 at 22:50
• Yes. I've posted an answer with the relevant material. Dec 6, 2021 at 23:24

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 $$\operatorname{Cl} X\cong \operatorname{Pic} X$$.