Proof.
Assume g \geq 2. First we may choose a finite separable extension of degree at most 2g - 2 such that X acquires a rational point, see Lemma 53.13.9. Thus we may assume X has a k-rational point x \in X(k) but now we have to prove the lemma with [k' : k] \leq (n^{2g})!. Let k \subset k^{sep} \subset \overline{k} be a separable algebraic closure inside an algebraic closure. By Lemma 53.17.1 we have
\mathop{\mathrm{Pic}}\nolimits (X_{\overline{k}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}
By Picard Schemes of Curves, Lemma 44.7.2 we conclude that
\mathop{\mathrm{Pic}}\nolimits (X_{k^{sep}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g}
By Picard Schemes of Curves, Lemma 44.7.2 there is a continuous action
\text{Gal}(k^{sep}/k) \longrightarrow \text{Aut}(\mathop{\mathrm{Pic}}\nolimits (X_{k^{sep}})[n]
and the lemma is true for the fixed field k' of the kernel of this map. The kernel is open because the action is continuous which implies that k'/k is finite. By Galois theory \text{Gal}(k'/k) is the image of the displayed arrow. Since the permutation group of a set of cardinality n^{2g} has cardinality (n^{2g})! we conclude by Galois theory that [k' : k] \leq (n^{2g})!. (Of course this proves the lemma with the bound |\text{GL}_{2g}(\mathbf{Z}/n\mathbf{Z})|, but all we want here is that there is some bound.)
If the genus is 1, then there is no upper bound on the degree of a finite separable field extension over which X acquires a rational point (details omitted). Still, there is such an extension for example by Varieties, Lemma 33.25.6. The rest of the proof is the same as in the case of g \geq 2.
\square
Comments (0)