# $p$-divisibility of Picard groups

Let $$p$$ be a prime number and let $$k$$ be a field with $$char(k)\neq p$$ such that all finite extensions have degree coprime to $$p$$. (For example, we can take $$k=\mathbb{R}$$ and $$p\neq 2$$ or let $$k$$ the union of $$\mathbb{F}_{l^{(p^n)}}$$ for all $$n\geq 1$$.)

Let $$C$$ be a regular affine curve over $$k$$ with (regular) projective compactification $$\widetilde{C}$$.

Then my question is as follows:

Is $$\operatorname{Pic}(C)$$ a $$p$$-divisible group? In other words, is every divisor on $$C$$ linearly equivalent to $$pD$$ for a divisor $$D$$ on $$C$$?

In a result I have proven related to strong approximation over function fields, the $$p$$-divisibilty is one of the necessary and sufficient conditions, so a positive answer here would allow me to simplify the statement.

One of the reasons why I suspect the above statement could be true is that the statement is true if $$k$$ is algebraically closed and the assumption on $$k$$ implies that $$l^\times$$ is a $$p$$-divisible group for all finite extensions $$l/k$$. If the statement is false, then I would be very interested in a counterexample and under what additional assumptions the statement is true.

My attempt thus far: If $$\widetilde{C}\setminus C$$ contains a rational point, or more generally, there exists a divisor of degree $$1$$ supported in $$\widetilde{C}\setminus C$$, then the restriction map $$\operatorname{Pic}^0(\widetilde{C})\rightarrow \operatorname{Pic}(C)$$ is surjective so under this assumption we can reduce the question to

Is $$\operatorname{Pic}^0(\widetilde{C})$$ a $$p$$-divisible group? In other words, is every divisor on $$C$$ of degree $$0$$ linearly equivalent to $$pD$$ for a divisor $$D$$ on $$\widetilde{C}$$?

If $$\widetilde{C}$$ is smooth (automatic if $$k$$ is perfect) and has a rational point, then $$\operatorname{Pic}^0(\widetilde{C})$$ are the $$k$$-points of the Jacobian of $$\widetilde{C}$$. This may allow us to use tools from the theory of abelian varieties. Maybe $$A(k)$$ is $$p$$-divisible for any (principally polarized) abelian variety $$A$$?

• Yes, it’s true that $A(k)$ is always divisible. What worries me is the situation when you’re removing a point of degree $p$ from the compact curve. Commented Aug 17, 2023 at 13:34
• Could you explain why $A(k)$ is $p$-divisible? I know that it is true for separably closed fields, but it is not clear to me why it should hold for the fields I described. Commented Aug 17, 2023 at 15:04
• I thought you meant $k$ is algebraically closed in the last sentence, sorry. In fact I misunderstood your question and thought $k$ has characteristic $p$, and anyway your assumption on $k$ disallows the removal of a point of degree $p$ from $\tilde C$. Commented Aug 17, 2023 at 17:39
• If $\widetilde{C}$ is smooth, it suffices to know that $\widetilde{C} \setminus C$ supports a divisor of degree prime to $p$, which always holds by your assumption on $k$. This implies that $\mathrm{Jac}(\widetilde{C})$ is $p$-divisible (see the answer by R. van Dobben de Bruyn below). It follows that the Picard group is also $p$-divisible, since if $L$ is a line bundle whose class in the Jacobian is divisible by $p$, then the obstruction to finding a $p$-th root in the Picard group is an element of $\mathrm{Br}(k)[p]$, but $\mathrm{Br}(k)[p]=0$ (by the same cohomological argument).
– naf
Commented Aug 18, 2023 at 5:23

$$\newcommand{\wt}{\widetilde}$$ $$\newcommand{\mr}{\mathrm}$$

The question has a positive answer, in fact, regularity of $$C$$ is not needed. The proof as written below works under the assumption that $$C$$ is (geometrically) irreducible and reduced, but both these conditions could be relaxed to some extent. (Note that $$\mr{Pic}(C)$$ depends only on the scheme $$C$$, not the base field.)

Let $$\wt{C}$$ be a projective compactification of $$C$$ such that $$\wt{C}$$ is regular at all points of $$\wt{C} \setminus C$$; this is possible since we assume $$C$$ is reduced.

Let $$J = \mr{Jac}(\wt{C})$$, i.e., the identity component of the Picard scheme of $$\wt{C}$$. Since $$p \neq \mr{char}(k)$$, the multiplication by $$p$$ map $$[p]:J \to J$$ is finite and etale (consider the induced map on the Lie algebra of $$J$$). The other assumption on $$k$$ implies that $$J(k)$$ is $$p$$-divisible: This follows from the fact that for any finite $$G$$-module $$M$$ of order a power of $$p$$, where $$G$$ is the absolute Galois group of $$k$$, $$H^1(G, M) = 0$$. (This is well-known and follows easily from the inflation-restriction sequence since any finite quotient of $$G$$ has order prime to $$p$$.) Applying this to the $$G$$-module $$J(k^{\mr{sep}})[p]$$ one sees that $$H^1(G, J(k^{\mr{sep}})[p]) = 0$$. The short exact sequence of $$G$$-modules $$0 \to J(k^{\mr{sep}})[p] \to J(k^{\mr{sep}}) \stackrel{[p]}{\to} J(k^{\mr{sep}}) \to 0$$ (for surjectivity we use that $$[p]$$ is finite etale) gives rise to a long exact sequence of Galois cohomology from which the $$p$$-divisibility of $$J(k)$$ follows by the vanishing of $$H^1(G, J(k^{\mr{sep}})[p])$$.

Now I claim that $$\mr{Pic}^0(\wt{C})$$ is also $$p$$-divisible: If $$L$$ is a line bundle of degree $$0$$ on $$\wt{C}$$ then it follows from the $$p$$-divisibility of $$J(k)$$ that the obstruction to existence of a $$p$$-th root of $$L$$ is an element of $$\mr{Br}(k)$$ of order $$p$$, but $$\mr{Br}(k)[p] = 0$$ from the Kummer sequence and the assumption on $$k$$.

Since $$\wt{C}$$ is regular at all points of $$\wt{C} \setminus C$$, the restriction map $$\mr{Pic}(\wt{C}) \to \mr{Pic}(C)$$ is a surjection. Since all the points in $$\wt{C} \setminus C$$ are of degree prime to $$p$$ (by the assumption on $$k$$), it follows that there is an exact sequence $$\mr{Pic}^0(\wt{C}) \to \mr{Pic}(C) \to Q \to 0$$ where $$Q$$ is a cyclic group of order prime to $$p$$. Then $$p$$-divisibility for $$\mr{Pic}^0(\wt{C})$$ and $$Q$$ implies it for $$\mr{Pic}(C)$$.

Note: If we drop the assumption that $$p \neq \mr{char}(k)$$, then the above proof still works if $$\wt{C}$$ is smooth. (In fact, the assumption on $$k$$ would imply that it is perfect, so regularity implies smoothness.)

• Thank you for your answer! For the counterexample, you say that a form of the additive group is killed by $p$, but I do not see why this is the case, as I assume $p$ is different from the characteristic. Am I misunderstanding something here? Commented Aug 21, 2023 at 6:37
• @BoazMoerman: You are right, probably the proof works in general. I will edit the answer soon.
– naf
Commented Aug 21, 2023 at 15:44
• Thank you for the update! Your answer is exactly what I was looking for and is a nice proof of the result. Commented Aug 22, 2023 at 8:41
• $\newcommand{\mr}{\mathrm}$ It seems the only place where geometric integrality is used (which is an assumption I prefer to avoid, though integrality is fine), is potentially in the exact sequence $$0\rightarrow \mr{Pic}^0(C)\rightarrow J(k)\rightarrow \mr{Br}(k)$$ (which I assumed you were referring to with the reference to the Brauer group). Is this the right sequence, and do you know a good reference for it? It is mentioned by Milne in jmilne.org/math/xnotes/JVs.pdf, but he does not provide a reference. Commented Aug 22, 2023 at 9:42
• One does not need geometric integrality, or even geometric irreducibility (as I had written). The sequence is very general, see Prop 4 on p. 204 of the book "Neron models" by Bosch, Raynaud and Lutkebohmert. If the curve is not geometrically integral then the (geometric) components of the Picard scheme would not be parametrized by $\mathbb{Z}$, so a little more care needs to be taken to get the last equation. Also, note that one can replace the base field $k$ by the field of constants of $\widetilde{C}$ to simplify this problem and also for the Brauer group issue.
– naf
Commented Aug 22, 2023 at 10:58

Inspired by the $$\mathbf G_m$$ case (Kummer theory), here is a positive result if you assume that $$E(k)$$ contains full $$\ell$$-torsion:

Lemma. Let $$k$$ be a field and $$\ell$$ a prime invertible in $$k$$, such that all finite separable extensions of $$k$$ have degree prime to $$\ell$$. Let $$A$$ be an abelian variety over $$k$$ such that $$A(k)$$ has full $$\ell$$-torsion, and let $$P \in A(k)$$. Then all points $$Q \in A(k^{\text{sep}})$$ with $$[\ell] Q = P$$ are defined over $$k$$.

Proof. The scheme-theoretic preimage of $$P$$ under $$[\ell] \colon A \to A$$ is naturally an étale $$A[\ell]$$-torsor over $$k = \kappa(P)$$, and the assumption on $$A$$ implies that the étale group scheme $$A[\ell] \to \operatorname{Spec} k$$ is constant. Then $$H^1(k,A[\ell]) = \operatorname{Hom}^{\text{cts}}(\operatorname{Gal}(k^{\text{sep}}/k),A[\ell])$$ is trivial by the assumption on $$k$$, so the torsor is trivial. $$\square$$.

For $$\mathbf G_m$$, the only alternative is that $$k^\times$$ has no $$\ell$$-torsion, in which case it is trivially $$\ell$$-divisible (as $$[\ell]$$ is invertible on $$A(k)$$). But on abelian varieties, there are intermediate cases with some but not all $$\ell$$-torsion, on which I have little to say at the moment.

• The assumption on $k$ implies that the image of $\mathrm{Gal}(k^{\mathrm{sep}}/k)$ in the automorphisms of $A[\ell]$ has order prime to $\ell$. One can then use the Hoschschild--Serre spectral sequence to show that $H^1(k,A[\ell])$ vanishes.
– naf
Commented Aug 18, 2023 at 4:28
• @naf you could turn that into an answer! Commented Aug 18, 2023 at 10:53
• Thank you for your answer! I had thought that using something using étale group schemes could be useful, but I had missed this argument. Commented Aug 18, 2023 at 11:34
• @naf Could you turn this into an answer (together with your other comment)? I am not too familiar with spectral sequences, so I have to look into it more, but your argument looks good and solves the important case when $C$ is smooth. Do you also happen to know a way to handle regular curves? Commented Aug 18, 2023 at 11:39