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 $k$.

## 70.14 Geometrically integral algebraic spaces

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

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 $k$.

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

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

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

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

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

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

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

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.
$\square$

## Post a comment

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)