Is there an abelian variety $A$ over a field $k$, such that $A(k^{\rm sep})$ is not a divisible group?
The motivation of my question is the following : if $L$ is any algebraically closed field, then $A(L)$ is a divisible group, that is the multiplication map $ [n] : A(L) \to A(L)$ is a surjective group morphism, for every $n \neq 0$ (see corollary 5.10). But if $L$ is only separably closed (e.g. $L = \Bbb F_p(T)^{\rm sep}$), it might not be true anymore ; however I have no counterexample. According to the cited reference, "if the ground field $k'$ is only assumed to be separably closed then it is not true in general that $X(k')$ is a divisible group", but no example is given.
If one can find a one-dimensional abelian variety $A$ such that $A(k^{\rm sep})$ is not divisible (in particular, $A$ is an elliptic curve), then I don't understand how one gets the surjective map $[n] : A(k^{\rm sep}) \to A(k^{\rm sep})$ at the beginning of the section 2 here).
[My ultimate goal is to get a long exact sequence in Galois cohomology for abelian varities over $\Bbb F_p(T)$.]
$\newcommand{\F}{\Bbb F} $ After thinking again to this question, I finally came up with a counter-example myself.
Let me first notice that whenever $n$ is an integer coprime to the characteristic of a field $K$, then the multiplication-by-2 map $$ [n] : E(K^s) \to E(K^s)$$ is surjective on the points of an elliptic curve $E$ with coordinates in the separable closure $K^s$ of $K$. This is because under the coprimality assumption, $[n] : E \to E$ is an étale morphism (then apply Surjective étale morphisms on points.).
However, this fails if $n$ is no longer coprime to $\mathrm{char}(K)$. Namely, let $E$ be the elliptic curve given by $y^2 + xy = x^3 + t^4$ over $K = \F_2(t)$. We prove that the multiplication-by-2 map $$ [2] : E(K^s) \to E(K^s)$$ is not surjective on the separable closure $K^s$ of $K$.
Let $Q = (t, t^2) \in E(K) \subset E(K^s)$. Let $P = (x, y) \in E(\overline K)$ be such that $Q = 2P$. Then we get (see Silverman's book AEC)
$$t = x(2P) = \dfrac{x^4 - t^4}{x^2},$$ which implies $x^4 - t x^2 - t^4 = 0$. If we prove that the polynomial $$f(z) := z^4 + t z^2 + t^4 \in \F_2(t)[z] = K[z]$$ is irreducible, then from $f(x) = 0$ and $f' = 0$ we deduce that $f$ is the minimal polynomial of $x$ over $K$ and is not separable. Hence $x \not \in K^s$ and $P \not \in E(K^s)$, which shows that $ [2] : E(K^s) \to E(K^s)$ is not surjective.
It remains to prove that $f(z) := z^4 + t z^2 + t^4 \in \F_2(t)[z] = K[z]$ is irreducible. First, by Gauss' lemma, $f$ is irreducible over $K$ iff it is irreducible over $\F_2[t]$. Then, observe that $f$ has no roots : if $z(t) \in \F_2[t]$ is a root of $f$, then plugging $t=1$ yields $z(1)^4 + z(1)^2 + 1 = 0$ but this is impossible since $z(1) \in \F_2$.
Then assume $z^4 + t z^2 + t^4 = (az^2 + bz + c)(\alpha z^2 + \beta z + \gamma)$. Then $a, \alpha \in \F_2[t]^{\times} = \{1\}$ and $$z^4 + z^3(b + \beta) + z^2(\gamma + c + b \beta) + z(b \gamma + c \beta) + c \gamma \;=\; z^4 + t z^2 + t^4.$$
This gives $\beta = -b$ and so $b(\gamma - c) = 0$. The case $b=0$ would imply $(t+c)c = t^4$ but again evaluating at $t=1$ makes it impossible to happen. So $\gamma = c$ and we are left with $b \beta = t = -b^2$, and for degree reasons, this cannot happen either.
All in all, we proved that the above factorization is impossible. Therefore $z^4 + t z^2 + t^4$ is indeed irreducible over $\F_2[t]$, and this concludes the proof. $\hspace{4cm} \square$