$p$-divisibility of Picard groups

Question

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 $A$?

Answer 1

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

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.)

Answer 2

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.