The Stacks project

Proof. The meaning of the phrase “has a $k$-rational point” is exactly that the structure morphism $f : X \to \mathop{\mathrm{Spec}}(k)$ has a section, which verifies the first condition. By Varieties, Lemma 33.26.2 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a field extension of $k$. Since $X$ has a $k$-rational point there is a $k$-algebra homomorphism $k' \to k$ and we conclude $k' = k$. Since $k$ is a field, any morphism $T \to \mathop{\mathrm{Spec}}(k)$ is flat. Hence we see by cohomology and base change (Cohomology of Schemes, Lemma 30.5.2) that $\mathcal{O}_ T \to f_{T, *}\mathcal{O}_{X_ T}$ is an isomorphism. This finishes the proof. $\square$