## 71.4 Integral algebraic spaces

We have not yet defined the notion of an integral algebraic space. The problem is that being integral is not an étale local property of schemes. We could use the property, that $X$ is reduced and $|X|$ is irreducible, given in Properties, Lemma 28.3.4 to define integral algebraic spaces. In this case the algebraic space described in Spaces, Example 64.14.9 would be integral which does not seem right. To avoid this type of pathology we will in addition assume that $X$ is a decent algebraic space, although perhaps a weaker alternative exists.

Definition 71.4.1. Let $S$ be a scheme. We say an algebraic space $X$ over $S$ is integral if it is reduced, decent, and $|X|$ is irreducible.

In this case the irreducible topological space $|X|$ is sober (Decent Spaces, Proposition 67.12.4). Hence it has a unique generic point $x$. In fact, in Decent Spaces, Lemma 67.20.4 we characterized decent algebraic spaces with finitely many irreducible components. Applying that lemma we see that an algebraic space $X$ is integral if it is reduced, has an irreducible dense open subscheme $X'$ with generic point $x'$ and the morphism $x' \to X$ is quasi-compact.

Lemma 71.4.2. Let $S$ be a scheme. Let $X$ be an integral algebraic space over $S$. Let $\eta \in |X|$ be the generic point of $X$. There are canonical identifications

$R(X) = \mathcal{O}_{X, \eta }^ h = \kappa (\eta )$

where $R(X)$ is the ring of rational functions defined in Morphisms of Spaces, Definition 66.47.3, $\kappa (\eta )$ is the residue field defined in Decent Spaces, Definition 67.11.2, and $\mathcal{O}_{X, \eta }^ h$ is the henselian local ring defined in Decent Spaces, Definition 67.11.5. In particular, these rings are fields.

Proof. Since $X$ is a scheme in an open neighbourhood of $\eta$ (see discussion above), this follows immediately from the corresponding result for schemes, see Morphisms, Lemma 29.49.5. We also use: the henselianization of a field is itself and that our definitions of these objects for algebraic spaces are compatible with those for schemes. Details omitted. $\square$

This leads to the following definition.

Definition 71.4.3. Let $S$ be a scheme. Let $X$ be an integral algebraic space over $S$. The function field, or the field of rational functions of $X$ is the field $R(X)$ of Lemma 71.4.2.

We may occasionally indicate this field $k(X)$ instead of $R(X)$.

Lemma 71.4.4. Let $S$ be a scheme. Let $X$ be an integral algebraic space over $S$. Then $\Gamma (X, \mathcal{O}_ X)$ is a domain.

Proof. Set $R = \Gamma (X, \mathcal{O}_ X)$. If $f, g \in R$ are nonzero and $fg = 0$ then $X = V(f) \cup V(g)$ where $V(f)$ denotes the closed subspace of $X$ cut out by $f$. Since $X$ is irreducible, we see that either $V(f) = X$ or $V(g) = X$. Then either $f = 0$ or $g = 0$ by Properties of Spaces, Lemma 65.21.4. $\square$

Here is a lemma about normal integral algebraic spaces.

Lemma 71.4.5. Let $S$ be a scheme. Let $X$ be a normal integral algebraic space over $S$. For every $x \in |X|$ there exists a normal integral affine scheme $U$ and an étale morphism $U \to X$ such that $x$ is in the image.

Proof. Choose an affine scheme $U$ and an étale morphism $U \to X$ such that $x$ is in the image. Let $u_ i$, $i \in I$ be the generic points of irreducible components of $U$. Then each $u_ i$ maps to the generic point of $X$ (Decent Spaces, Lemma 67.20.1). By our definition of a decent space (Decent Spaces, Definition 67.6.1), we see that $I$ is finite. Hence $U = \mathop{\mathrm{Spec}}(A)$ where $A$ is a normal ring with finitely many minimal primes. Thus $A = \prod _{i \in I} A_ i$ is a product of normal domains by Algebra, Lemma 10.37.16. Then $U = \coprod U_ i$ with $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ and $x$ is in the image of $U_ i \to X$ for some $i$. This proves the lemma. $\square$

Lemma 71.4.6. Let $S$ be a scheme. Let $X$ be a normal integral algebraic space over $S$. Then $\Gamma (X, \mathcal{O}_ X)$ is a normal domain.

Proof. Set $R = \Gamma (X, \mathcal{O}_ X)$. Then $R$ is a domain by Lemma 71.4.4. Let $f = a/b$ be an element of the fraction field of $R$ which is integral over $R$. For any $U \to X$ étale with $U$ a scheme there is at most one $f_ U \in \Gamma (U, \mathcal{O}_ U)$ with $b|_ U f_ U = a|_ U$. Namely, $U$ is reduced and the generic points of $U$ map to the generic point of $X$ which implies that $b|_ U$ is a nonzerodivisor. For every $x \in |X|$ we choose $U \to X$ as in Lemma 71.4.5. Then there is a unique $f_ U \in \Gamma (U, \mathcal{O}_ U)$ with $b|_ U f_ U = a|_ U$ because $\Gamma (U, \mathcal{O}_ U)$ is a normal domain by Properties, Lemma 28.7.9. By the uniqueness mentioned above these $f_ U$ glue and define a global section $f$ of the structure sheaf, i.e., of $R$. $\square$

Lemma 71.4.7. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. There are canonical bijections between the following sets:

1. the set of points of $X$, i.e., $|X|$,

2. the set of irreducible closed subsets of $|X|$,

3. the set of integral closed subspaces of $X$.

The bijection from (1) to (2) sends $x$ to $\overline{\{ x\} }$. The bijection from (3) to (2) sends $Z$ to $|Z|$.

Proof. Our map defines a bijection between (1) and (2) as $|X|$ is sober by Decent Spaces, Proposition 67.12.4. Given $T \subset |X|$ closed and irreducible, there is a unique reduced closed subspace $Z \subset X$ such that $|Z| = T$, namely, $Z$ is the reduced induced subspace structure on $T$, see Properties of Spaces, Definition 65.12.5. This is an integral algebraic space because it is decent, reduced, and irreducible. $\square$

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).