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


  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).
  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]), 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]).
  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.


  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

  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

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


2 Answers 2


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.)


[Kahn06] Kahn, Bruno, Sur le groupe des classes d'un schéma arithmétique, Bull. Soc. Math. Fr. 134, No. 3, 395–415 (2006). ZBL1222.14048.

  • $\begingroup$ There are details in Hartshorne either for this or for the nodal cubic, I forget which (the other is an exercise) $\endgroup$ Feb 28, 2018 at 19:35
  • $\begingroup$ @user45150: only over a(n algebraically closed) field. It's not entirely straightforward to treat the more general case. $\endgroup$ Feb 28, 2018 at 21:57

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 $\operatorname{coker} \alpha \simeq R$.


[Wei13] Weibel, The K-book: An Introduction to Algebraic K-theory, volume 145 of Graduate Studies in Mathematics. American Mathematical Society (2013)


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