## 70.14 Geometrically integral algebraic spaces

Recall that integral algebraic spaces are by definition decent, see Section 70.4.

Definition 70.14.1. Let $X$ be an algebraic space over the field $k$. We say $X$ is geometrically integral over $k$ if the algebraic space $X_{k'}$ is integral (Definition 70.4.1) for every field extension $k'$ of $k$.

In particular $X$ is a decent algebraic space. We can relate this to being geometrically reduced and geometrically irreducible as follows.

Lemma 70.14.2. Let $k$ be a field. Let $X$ be a decent algebraic space over $k$. Then $X$ is geometrically integral over $k$ if and only if $X$ is both geometrically reduced and geometrically irreducible over $k$.

Proof. This is an immediate consequence of the definitions because our notion of integral (in the presence of decency) is equivalent to reduced and irreducible. $\square$

Lemma 70.14.3. Let $k$ be a field. Let $X$ be a proper algebraic space 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 Spaces, Lemma 67.20.3 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 rings by Algebra, Lemma 10.52.2 and Algebra, Proposition 10.59.6. If $X = Y \amalg Z$ with $Y$ and $Z$ open subspaces of $X$, 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 $|X|$ is a Noetherian topological space (by Morphisms of Spaces, Lemma 65.28.6 and Properties of Spaces, Lemma 64.24.2) 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 (Properties of Spaces, Lemma 64.21.4). Hence the local rings $A_ i = k_ i$ are reduced and therefore fields (for example by Algebra, Lemma 10.24.1). This proves (3).

If $X$ is geometrically reduced, then same thing is true for $A \otimes _ k \overline{k} = H^0(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$ (see Cohomology of Spaces, Lemma 67.11.2 for equality). 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 10.43.1 and 10.43.3. 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). $\square$

Lemma 70.14.4. Let $k$ be a field. Let $X$ be a proper integral algebraic space over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. If $H^0(X, \mathcal{L})$ and $H^0(X, \mathcal{L}^{\otimes - 1})$ are both nonzero, then $\mathcal{L} \cong \mathcal{O}_ X$.

Proof. Let $s \in H^0(X, \mathcal{L})$ and $t \in H^0(X, \mathcal{L}^{\otimes - 1})$ be nonzero sections. Let $x \in |X|$ be a point in the support of $s$. Choose an affine étale neighbourhood $(U, u) \to (X, x)$ such that $\mathcal{L}|_ U \cong \mathcal{O}_ U$. Then $s|_ U$ corresponds to a nonzero regular function on the reduced (because $X$ is reduced) scheme $U$ and hence is nonvanishing in a generic point of an irreducible component of $U$. By Decent Spaces, Lemma 66.20.1 we conclude that the generic point $\eta$ of $|X|$ is in the support of $s$. The same is true for $t$. Then of course $st$ must be nonzero because the local ring of $X$ at $\eta$ is a field (by aforementioned lemma the local ring has dimension zero, as $X$ is reduced the local ring is reduced, and Algebra, Lemma 10.24.1). However, we have seen that $K = H^0(X, \mathcal{O}_ X)$ is a field in Lemma 70.14.3. Thus $st$ is everywhere nonzero and we see that $s : \mathcal{O}_ X \to \mathcal{L}$ is an isomorphism. $\square$

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