Lemma 33.9.3. Let $k$ be a field. Let $X$ be a proper scheme over $k$.

1. $A = H^0(X, \mathcal{O}_ X)$ is a finite dimensional $k$-algebra,

2. $A = \prod _{i = 1, \ldots , n} A_ i$ is a product of Artinian local $k$-algebras, one factor for each connected component of $X$,

3. if $X$ is reduced, then $A = \prod _{i = 1, \ldots , n} k_ i$ is a product of fields, each a finite extension of $k$,

4. if $X$ is geometrically reduced, then $k_ i$ is finite separable over $k$,

5. if $X$ is geometrically connected, then $A$ is geometrically irreducible over $k$,

6. if $X$ is geometrically irreducible, then $A$ is geometrically irreducible over $k$,

7. if $X$ is geometrically reduced and connected, then $A = k$, and

8. if $X$ is geometrically integral, then $A = k$.

Proof. By Cohomology of Schemes, Lemma 30.19.2 we see that $A = H^0(X, \mathcal{O}_ X)$ is a finite dimensional $k$-algebra. This proves (1).

Then $A$ is a product of local Artinian $k$-algebras by Algebra, Lemma 10.53.2 and Proposition 10.60.7. If $X = Y \amalg Z$ with $Y$ and $Z$ open in $X$, then we obtain an idempotent $e \in A$ by taking the section of $\mathcal{O}_ X$ which is $1$ on $Y$ and $0$ on $Z$. Conversely, if $e \in A$ is an idempotent, then we get a corresponding decomposition of $X$. Finally, as $X$ has a Noetherian underlying topological space its connected components are open. Hence the connected components of $X$ correspond $1$-to-$1$ with primitive idempotents of $A$. This proves (2).

If $X$ is reduced, then $A$ is reduced. Hence the local rings $A_ i = k_ i$ are reduced and therefore fields (for example by Algebra, Lemma 10.25.1). This proves (3).

If $X$ is geometrically reduced, then $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ (equality by Cohomology of Schemes, Lemma 30.5.2) is reduced. This implies that $k_ i \otimes _ k \overline{k}$ is a product of fields and hence $k_ i/k$ is separable for example by Algebra, Lemmas 10.44.1 and 10.44.3. This proves (4).

If $X$ is geometrically connected, then $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ is a zero dimensional local ring by part (2) and hence its spectrum has one point, in particular it is irreducible. Thus $A$ is geometrically irreducible. This proves (5). Of course (5) implies (6).

If $X$ is geometrically reduced and connected, then $A = k_1$ is a field and the extension $k_1/k$ is finite separable and geometrically irreducible. However, then $k_1 \otimes _ k \overline{k}$ is a product of $[k_1 : k]$ copies of $\overline{k}$ and we conclude that $k_1 = k$. This proves (7). Of course (7) implies (8). $\square$

Comment #7106 by WhatJiaranEatsTonight on

In (5), $A$ should be geometrically connected but not geometrically irreducible.

Comment #7269 by on

Well, the proof shows that $A$ is geometrically irreducible in case (5). Also, recall that $A$ is a finite dimensional $k$-algebra here so being geometrically connected is the same thing as being geometrically irreducible.

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