Lemma 45.11.2. Assume given (D0), (D1), (D2), and (D3) satisfying (A), (B), and (C). Let $X$ be a smooth projective scheme over $k$. Set $k' = \Gamma (X, \mathcal{O}_ X)$. The following are equivalent

1. there exist finitely many closed points $x_1, \ldots , x_ r \in X$ whose residue fields are separable over $k$ such that $H^0(X) \to H^0(x_1) \oplus \ldots \oplus H^0(x_ r)$ is injective,

2. the map $H^0(\mathop{\mathrm{Spec}}(k')) \to H^0(X)$ is an isomorphism.

If this is true, then $H^0(X)$ is a finite separable algebra over $F$. If $X$ is equidimensional of dimension $d$, then (1) and (2) are also equivalent to

1. the classes of closed points generate $H^{2d}(X)(d)$ as a module over $H^0(X)$.

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

$H^0(X) \otimes _{H^0(\mathop{\mathrm{Spec}}(k'))} H^0(\mathop{\mathrm{Spec}}(k'')) = H^0(X \times _{\mathop{\mathrm{Spec}}(k')} \mathop{\mathrm{Spec}}(k''))$

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. $\square$

Comment #7511 by Hao Peng on

maybe the last assertion shoule be placed prior to condition 3 to avoid confusion that X is equidimebsional or not.

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