Lemma 50.20.2. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$. The map

$\text{d} : H^0(X, \mathcal{O}_ X) \to H^0(X, \Omega ^1_{X/k})$

is zero.

Proof. Since $X$ is smooth over $k$ it is geometrically reduced over $k$, see Varieties, Lemma 33.25.4. Hence $H^0(X, \mathcal{O}_ X) = \prod k_ i$ is a finite product of finite separable field extensions $k_ i/k$, see Varieties, Lemma 33.9.3. It follows that $\Omega _{H^0(X, \mathcal{O}_ X)/k} = \prod \Omega _{k_ i/k} = 0$ (see for example Algebra, Lemma 10.158.1). Since the map of the lemma factors as

$H^0(X, \mathcal{O}_ X) \to \Omega _{H^0(X, \mathcal{O}_ X)/k} \to H^0(X, \Omega _{X/k})$

by functoriality of the de Rham complex (see Section 50.2), we conclude. $\square$

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