The Stacks project

Proof. We observe that the statement makes sense because $k'$ is a finite separable algebra over $k$ (Varieties, Lemma 33.9.3) and hence $\mathop{\mathrm{Spec}}(k')$ is smooth and projective over $k$. The compatibility of $H^*$ with direct sums (Lemmas 45.9.9 and 45.10.1) shows that it suffices to prove the lemma when $X$ is connected. Hence we may assume $X$ is irreducible and we have to show the equivalence of (1), (2), and (3). Set $d = \dim (X)$. This implies that $k'$ is a field finite separable over $k$ and that $X$ is geometrically irreducible over $k'$, see Varieties, Lemmas 33.9.3 and 33.9.4.

By Lemma 45.8.1 we see that the closed points in (3) may be assumed to have separable residue fields over $k$. By axioms (A)(a) and (A)(b) we see that conditions (1) and (3) are equivalent.

If (2) holds, then pick any closed point $x \in X$ whose residue field is finite separable over $k'$. Then $H^0(\mathop{\mathrm{Spec}}(k')) = H^0(X) \to H^0(x)$ is injective for example by Lemma 45.10.8.

Assume the equivalent conditions (1) and (3) hold. Choose $x_1, \ldots , x_ r \in X$ as in (1). Choose a finite separable extension $k''/k'$. By Lemma 45.10.9 we have

Thus in order to show that $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X)$ is an isomorphism we may replace $k'$ by $k''$. Thus we may assume $x_1, \ldots , x_ r$ are $k'$-rational points (this replaces each $x_ i$ with multiple points, so $r$ is increased in this step). By Lemma 45.10.7 $\gamma (x_1) = \gamma (x_2) = \ldots = \gamma (x_ r)$. By axiom (A)(b) all the maps $H^0(X) \to H^0(x_ i)$ are the same. This means (2) holds.

Finally, Lemma 45.10.2 implies $H^0(X)$ is a separable $F$-algebra if (1) holds. $\square$