If $A$ is an abelian variety over a finite field $\mathbf{F}_q$, then $A(\mathbf{F}_q)$ (resp. $A(\overline{\mathbf{F}}_q)$) is a finite (resp. infinite torsion) group, but $A(\mathbf{F}_q(t))$ is a finitely generated and possibly infinite abelian group.

This question concerns a similar situation for étale group schemes over a field, and it came up while thinking about Serre's method of specialization for Mordell-Weil groups via Hilbert irreducibility.

Suppose $G$ is a commutative étale group scheme over $\mathbf{F}_q$ (I am not assuming $G$ is of finite type. Only locally of finite type).

Is there an example of such $G$ such that $G(\mathbf{F}_q)$ is finite, or at least torsion, and $G(\mathbf{F}_q(t))$ is finitely generated and with a point of infinite order?

Suppose now that $f: G_1\to G_2$ is a surjective map of commutative étale group schemes over $\overline{\mathbf{F}}_q$ (not necessarily of finite type). Assume that $G_2(\overline{\mathbf{F}}_q)$ is a finitely generated free abelian group.

Is there a homomorphism of commutative étale group schemes over $\overline{\mathbf{F}}_q$, $g : G_2\to G_1$, such that $f\circ g$ is the identity on $G_2$?

In other words, a section to $f$ that is a homomorphism.


1 Answer 1


If $k \to \ell$ is any regular field extension (i.e. $k$ is algebraically closed in $\ell$), then $X(k) \to X(\ell)$ is a bijection when $X \to \operatorname{Spec} k$ is étale. Indeed, it suffices to show this when $X = \operatorname{Spec} K$ for some finite separable field extension $k \to K$, and then the statement is that the natural map $\operatorname{Hom}_k(K,k) \to \operatorname{Hom}_k(K,\ell)$ is a bijection. If $k \to K$ is an isomorphism, then both sides are a singleton, and otherwise both sides are empty by assumption.

Note that $k \to \ell$ is regular if and only if $\ell$ is a geometrically integral $k$-algebra [Tags 037Q and 030W]. In particular, this holds when $\ell = k(X)$ is the function field of a geometrically integral $k$-variety $X$. Notably, $k \to k(t)$ (in this case regularity is easily checked by hand).

(A previous version of this answer contained a more geometric argument when $\ell = k(C)$ for a smooth geometrically integral curve with a rational point, but neither dimension $1$ nor the rational point is needed, and the current proof is actually easier.)

For the second question, recall that the category of finite étale $k$-schemes is equivalent to the category of finite sets with a continuous $\Gamma=\operatorname{Gal}(\bar k/k)$-action. Taking disjoint unions gives an equivalence between étale $k$-schemes and discrete sets with a continuous $\Gamma$-action. The group objects are therefore discrete groups with a continuous $\Gamma$-action, i.e. groups $A$ with a group homomorphism $\phi \colon \Gamma \to \operatorname{Aut}(A)$ such that for every $a \in A$, the stabiliser $\Gamma_a$ is open (equivalently, closed of finite index).

This gives the required counterexample: let $U \subseteq \Gamma$ be a subgroup of index $2$ (when $k = \mathbf F_q$ there is a unique such subgroup), and write $H$ for the quotient $\Gamma/U$, with nontrivial element $\sigma$. Now consider the augmentation $\mathbf Z[H] \twoheadrightarrow \mathbf Z$ of $H$-modules. The quotient is the constant étale scheme $\mathbf Z$ corresponding to the trivial $\Gamma$-action. But the surjection is not split as the only invariant elements in $\mathbf Z[H]$ are multiples of $\sigma + 1$, which maps to $2$ under the augmentation.


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