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

is zero.

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. $\square$

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: