The Stacks project

Sketch of proof. Unwinding definitions, it is enough to check that $H^1(X, A_0) \to \mathop{\mathrm{lim}}\nolimits H^1(X_ n, A_0)$ is not surjective. As $X$ is regular and projective, by Lemmas 110.72.2 and 110.72.1 each $A_0$-torsor over $X$ is torsion. In particular, the group $H^1(X, A_0)$ is torsion. It is thus enough to show: (a) the group $H^1(X_0, A_0)$ is non-torsion, and (b) the maps $H^1(X_{n + 1}, A_0) \to H^1(X_ n, A_0)$ are surjective for all $n$.

Ad (a). One constructs a nontorsion $A_0$-torsor $P_0$ on the nodal curve $X_0$ by glueing trivial $A_0$-torsors on each component of $X_0$ using non-torsion points on $A_0$ as the isomorphisms over the nodes. More precisely, let $x \in X_0$ be a node which occurs in a loop consisting of rational curves. Let $X'_0 \to X_0$ be the normalization of $X_0$ in $X_0 \setminus \{ x\} $. Let $x', x'' \in X'_0$ be the two points mapping to $x_0$. Then we take $A_0 \times _{\mathop{\mathrm{Spec}}(k)} X'_0$ and we identify $A_0 \times {x'}$ with $A_0 \times \{ x''\} $ using translation $A_0 \to A_0$ by a nontorsion point $a_0 \in A_0(k)$ (there is such a nontorsion point as $k$ is algebraically closed and not the algebraic closure of a finite field – this is actually not trivial to prove). One can show that the glueing is an algebraic space (in fact one can show it is a scheme) and that it is an nontorsion $A_0$-torsor over $X_0$. The reason that it is nontorsion is that if $[n](P_0)$ has a section, then that section produces a morphism $s : X'_0 \to A_0$ such that $[n](a_0) = s(x') - s(x'')$ in the group law on $A_0(k)$. However, since the irreducible components of the loop are rational to section $s$ is constant on them ( More on Groupoids in Spaces, Lemma 79.11.3). Hence $s(x') = s(x'')$ and we obtain a contradiction.

Ad (b). Deformation theory shows that the obstruction to deforming an $A_0$-torsor $P_ n \to X_ n$ to an $A_0$-torsor $P_{n + 1} \to X_{n + 1}$ lies in $H^2(X_0, \omega )$ for a suitable vector bundle $\omega $ on $X_0$. The latter vanishes as $X_0$ is a curve, proving the claim. $\square$