
> 
> Let $A$ be a finitely generated $\mathbb{Z}$-algebra. Is $\operatorname{Pic}(A)$ finitely generated (as an abelian group)?
> 
> 
> 


Thoughts:


1. We may assume that $A$ is reduced since $\operatorname{Pic}(A) = \operatorname{Pic}(A\_{\mathrm{red}})$.
2. If $A$ is reduced, then the group of units $A^{\times}$ is a finitely generated abelian group, see e.g. [1, Appendix 1, no. 3] or [4, Théorème 1] (which I learned about through [this question](https://math.stackexchange.com/q/2445562/)).
3. The case $A$ is normal is proved in [3, Chapter 2, Theorem 7.6].
4. The following argument is from [2, Lemma 9.6]: Let $B$ be the normalization of $A$, set $X := \operatorname{Spec} A$ and $Y := \operatorname{Spec} B$ and let $\pi : Y \to X$ be the normalization morphism. We have the Leray spectral sequence $$ \mathrm{E}\_{2}^{p,q} = \mathrm{H}^{p}(X,\mathbf{R}^{q}\pi\_{\ast}\mathbb{G}\_{m,Y}) \implies \mathrm{H}^{p+q}(Y,\mathbb{G}\_{m,Y}) $$ with differentials $\mathrm{E}\_{2}^{p,q} \to \mathrm{E}\_{2}^{p+2,q-1}$. Since $\pi$ is a finite morphism (e.g. since $\mathbb{Z}$ is Nagata and [5, [030C](https://stacks.math.columbia.edu/tag/030C)]), every invertible sheaf on $Y$ can be trivialized on an open cover obtained as the preimage of an open cover of $X$ (e.g. [5, [0BUT](https://stacks.math.columbia.edu/tag/0BUT)]). Hence $\mathbf{R}^{1}\pi\_{\ast}\mathbb{G}\_{m,Y} = 0$, so we have $\operatorname{Pic}(Y) \simeq \mathrm{H}^{1}(X,\pi\_{\ast}\mathbb{G}\_{m,Y})$ from the Leray spectral sequence. Set $Q := \pi\_{\ast}\mathbb{G}\_{m,Y}/\mathbb{G}\_{m,X}$; then the long exact sequence in cohomology associated to the sequence $1 \to \mathbb{G}\_{m,X} \to \pi\_{\ast}\mathbb{G}\_{m,Y} \to Q \to 1$ gives an exact sequence $$ \Gamma(Y,\mathbb{G}\_{m,Y}) \to \Gamma(X,Q) \stackrel{\partial}{\to} \operatorname{Pic}(X) \to \operatorname{Pic}(Y) $$ where the first and fourth terms are finitely generated. But what can I say about the sheaf $Q$? I know that it is $0$ on a dense open since $\pi$ is an isomorphism on a dense open (e.g. since $A$ is reduced, the regular locus is an open subset containing the generic points [5, [07R5](https://stacks.math.columbia.edu/tag/07R5)]).
5. I should also note that there is a Hartshorne exercise (II, Exercise 6.9) which relates the Picard group of a singular curve (over a field) to that of its normalization.


References:


1. Bass, *Introduction to Introduction to Some Methods of Algebraic K-Theory*, Number 20 in CBMS Regional Conference Series in Mathematics. American Mathematical Society, 1974.
2. Jaffe, "Coherent functors, with application to torsion in the
Picard group", Transactions of the American Mathematical Society, vol. 349, no. 2, 1997, pp. 481–527 [link](http://www.ams.org/journals/tran/1997-349-02/S0002-9947-97-01616-4/)
3. Lang, *Fundamentals of Diophantine Geometry*, Springer-Verlag (1983)
4. Samuel, "A propos du théorème des unités", Bulletin des Sciences
Mathématiques, vol. 90, 1966, pp. 89–96
5. Stacks Project [link](https://stacks.math.columbia.edu/)


Keywords: arithmetic scheme, Picard group, finite type $\mathbb{Z}$-algebra



Picard group of a finite type $\mathbb{Z}$-algebra
This is false. A counterexample is given in [Kahn06, **Rmq. 1 (6)**]. The example uses the cuspidal cubic $B = A[x^2,x^3]$ over a finite type $\mathbb Z$-algebra $A$ that is not a finitely generated $\mathbb Z$-module. For example, take $A$ to be $\mathbb Z[x]$ or $\mathbb F\_p[x]$.


As usual, one shows that (under suitable hypotheses on $A$) there is an isomorphism
$$\operatorname{Pic}(B) \cong \mathbb G\_a(A) = A,$$
which shows that $\operatorname{Pic}(B)$ is not finitely generated. (Details omitted, unfortunately also in the paper.)




---


**References.**


[Kahn06] Kahn, Bruno, [*Sur le groupe des classes d'un schéma arithmétique*](http://dx.doi.org/10.24033/bsmf.2515), Bull. Soc. Math. Fr. 134, No. 3, 395–415 (2006). [ZBL1222.14048](https://zbmath.org/?q=an:1222.14048).


