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

Thoughts:


*

*We may assume that $A$ is reduced since $\operatorname{Pic}(A) = \operatorname{Pic}(A_{\mathrm{red}})$.

*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).

*The case $A$ is normal is proved in [3, Chapter 2, Theorem 7.6].

*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]), 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]). 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]).

*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:


*

*Bass, Introduction to Introduction to Some Methods of Algebraic K-Theory, Number 20 in CBMS Regional Conference Series in Mathematics. American Mathematical Society, 1974.

*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

*Lang, Fundamentals of Diophantine Geometry, Springer-Verlag (1983)

*Samuel, "A propos du théorème des unités", Bulletin des Sciences
Mathématiques, vol. 90, 1966, pp. 89–96

*Stacks Project link
Keywords: arithmetic scheme, Picard group, finite type $\mathbb{Z}$-algebra
 A: Here I write out some details in R. van Dobben de Bruyn's answer. This argument is from p. 5 of these notes by Konrad Voelkel.

Let $R$ be a Noetherian normal domain with $\mathrm{Pic}(R) = 0$. Set \begin{align*} A := R[x,y]/(x^{3} = y^{2}) \end{align*} which is the cuspidal cubic over $R$. Then \begin{align*} \operatorname{Pic}(A) \simeq R \end{align*} as abelian groups.

The normalization of $A$ is \begin{align*} B := R[t] \end{align*} via the $R$-algebra map $A \to B$ sending $(x,y) \mapsto (t^{2},t^{3})$ which identifies $A$ with the subring $R[t^{2},t^{3}]$ of $B$. Let \begin{align*} I := \{a \in A \;:\; aB \subset A\} \end{align*} be the conductor ideal of the inclusion $A \subseteq B$; it is the largest ideal of $B$ contained in $A$. We have \begin{align*} I = \langle x,y \rangle A \end{align*} (for this, use that $A,B$ are $\mathbb{Z}_{\ge 0}$-graded rings and that $A \subset B$ is a graded ring map, hence $I$ is also a graded ideal of $A$; certainly $\langle x,y \rangle A \subset I$ and if $a \in A$ is nonzero then $a \not\in I$ since otherwise $at \in A$). We have a Milnor square  $\require{AMScd}$
\begin{CD}
A @>>> B \\
@VVV @VVV\\
A/I @>>> B/I
\end{CD} where $A/I \simeq R$ and $B/I \simeq R[t]/(t^{2})$. By [Wei13, I, Theorem 3.10] we have an exact sequence \begin{align*} (A/I)^{\times} \oplus B^{\times} \stackrel{\alpha}{\to} (B/I)^{\times} \to \operatorname{Pic}(A) \to \operatorname{Pic}(A/I) \oplus \operatorname{Pic}(B) \end{align*} of abelian groups (the Units-Pic sequence). Here $\operatorname{Pic}(A/I) = \operatorname{Pic}(R) = 0$ and $\operatorname{Pic}(B) \simeq \operatorname{Pic}(R) = 0$ where the first isomorphism follows from e.g. Traverso's theorem [Wei13, I, Theorem 3.11]. We have $(A/I)^{\times} = R^{\times}$ and $B^{\times} = R^{\times}$ (since $R$ is reduced) and $(B/I)^{\times} = (R[t]/(t^{2}))^{\times} \simeq R \oplus R^{\times}$ (where the "$R$" in "$R \oplus R^{\times}$" is viewed as an abelian group under addition). Under these identifications, the map $\alpha$ sends $(u_{1},u_{2}) \mapsto (0,u_{1}u_{2}^{-1})$ so $\operatorname{coker} \alpha \simeq R$.
References:
[Wei13] Weibel, The K-book: An Introduction to Algebraic K-theory, volume 145 of Graduate Studies in Mathematics. American Mathematical Society (2013)
A: 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, Bull. Soc. Math. Fr. 134, No. 3, 395–415 (2006). ZBL1222.14048.
