## 64.12 Reduced spaces

We have already defined reduced algebraic spaces in Section 64.7. Here we just prove some simple lemmas regarding reduced algebraic spaces.

Lemma 64.12.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. The following are equivalent

$X$ is reduced,

for every $x \in |X|$ the local ring of $X$ at $x$ is reduced (Remark 64.7.6).

In this case $\Gamma (X, \mathcal{O}_ X)$ is a reduced ring and if $f \in \Gamma (X, \mathcal{O}_ X)$ has $X = V(f)$, then $f = 0$.

**Proof.**
The equivalence of (1) and (2) follows from Properties, Lemma 28.3.2 applied to affine schemes étale over $X$. The final statements follow the cited lemma and fact that $\Gamma (X, \mathcal{O}_ X)$ is a subring of $\Gamma (U, \mathcal{O}_ U)$ for some reduced scheme $U$ étale over $X$.
$\square$

Lemma 64.12.2. Let $S$ be a scheme. Let $Z \to X$ be an immersion of algebraic spaces. Then $|Z| \to |X|$ is a homeomorphism of $|Z|$ onto a locally closed subset of $|X|$.

**Proof.**
Let $U$ be a scheme and $U \to X$ a surjective étale morphism. Then $Z \times _ X U \to U$ is an immersion of schemes, hence gives a homeomorphism of $|Z \times _ X U|$ with a locally closed subset $T'$ of $|U|$. By Lemma 64.4.3 the subset $T'$ is the inverse image of the image $T$ of $|Z| \to |X|$. The map $|Z| \to |X|$ is injective because the transformation of functors $Z \to X$ is injective, see Spaces, Section 63.12. By Topology, Lemma 5.6.4 we see that $T$ is locally closed in $|X|$. Moreover, the continuous map $|Z| \to T$ is a homeomorphism as the map $|Z \times _ X U| \to T'$ is a homeomorphism and $|Z \times _ Y U| \to |Z|$ is submersive.
$\square$

The following lemma will help us construct (locally) closed subspaces.

Lemma 64.12.3. Let $S$ be a scheme. Let $j : R \to U \times _ S U$ be an étale equivalence relation. Let $X = U/R$ be the associated algebraic space (Spaces, Theorem 63.10.5). There is a canonical bijection

\[ R\text{-invariant locally closed subschemes }Z'\text{ of }U \leftrightarrow \text{locally closed subspaces }Z\text{ of }X \]

Moreover, if $Z \to X$ is closed (resp. open) if and only if $Z' \to U$ is closed (resp. open).

**Proof.**
Denote $\varphi : U \to X$ the canonical map. The bijection sends $Z \to X$ to $Z' = Z \times _ X U \to U$. It is immediate from the definition that $Z' \to U$ is an immersion, resp. closed immersion, resp. open immersion if $Z \to X$ is so. It is also clear that $Z'$ is $R$-invariant (see Groupoids, Definition 39.19.1).

Conversely, assume that $Z' \to U$ is an immersion which is $R$-invariant. Let $R'$ be the restriction of $R$ to $Z'$, see Groupoids, Definition 39.18.2. Since $R' = R \times _{s, U} Z' = Z' \times _{U, t} R$ in this case we see that $R'$ is an étale equivalence relation on $Z'$. By Spaces, Theorem 63.10.5 we see $Z = Z'/R'$ is an algebraic space. By construction we have $U \times _ X Z = Z'$, so $U \times _ X Z \to Z$ is an immersion. Note that the property “immersion” is preserved under base change and fppf local on the base (see Spaces, Section 63.4). Moreover, immersions are separated and locally quasi-finite (see Schemes, Lemma 26.23.8 and Morphisms, Lemma 29.19.16). Hence by More on Morphisms, Lemma 37.49.1 immersions satisfy descent for fppf covering. This means all the hypotheses of Spaces, Lemma 63.11.5 are satisfied for $Z \to X$, $\mathcal{P}=$“immersion”, and the étale surjective morphism $U \to X$. We conclude that $Z \to X$ is representable and an immersion, which is the definition of a subspace (see Spaces, Definition 63.12.1).

It is clear that these constructions are inverse to each other and we win.
$\square$

Lemma 64.12.4. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $T \subset |X|$ be a closed subset. There exists a unique closed subspace $Z \subset X$ with the following properties: (a) we have $|Z| = T$, and (b) $Z$ is reduced.

**Proof.**
Let $U \to X$ be a surjective étale morphism, where $U$ is a scheme. Set $R = U \times _ X U$, so that $X = U/R$, see Spaces, Lemma 63.9.1. As usual we denote $s, t : R \to U$ the two projection morphisms. By Lemma 64.4.5 we see that $T$ corresponds to a closed subset $T' \subset |U|$ such that $s^{-1}(T') = t^{-1}(T')$. Let $Z' \subset U$ be the reduced induced scheme structure on $T'$. In this case the fibre products $Z' \times _{U, t} R$ and $Z' \times _{U, s} R$ are closed subschemes of $R$ (Schemes, Lemma 26.18.2) which are étale over $Z'$ (Morphisms, Lemma 29.34.4), and hence reduced (because being reduced is local in the étale topology, see Remark 64.7.3). Since they have the same underlying topological space (see above) we conclude that $Z' \times _{U, t} R = Z' \times _{U, s} R$. Thus we can apply Lemma 64.12.3 to obtain a closed subspace $Z \subset X$ whose pullback to $U$ is $Z'$. By construction $|Z| = T$ and $Z$ is reduced. This proves existence. We omit the proof of uniqueness.
$\square$

Lemma 64.12.5. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $Z \subset X$ be a closed subspace. Assume $Y$ is reduced. A morphism $f : Y \to X$ factors through $Z$ if and only if $f(|Y|) \subset |Z|$.

**Proof.**
Assume $f(|Y|) \subset |Z|$. Choose a diagram

\[ \xymatrix{ V \ar[d]_ b \ar[r]_ h & U \ar[d]^ a \\ Y \ar[r]^ f & X } \]

where $U$, $V$ are schemes, and the vertical arrows are surjective and étale. The scheme $V$ is reduced, see Lemma 64.7.1. Hence $h$ factors through $a^{-1}(Z)$ by Schemes, Lemma 26.12.7. So $a \circ h$ factors through $Z$. As $Z \subset X$ is a subsheaf, and $V \to Y$ is a surjection of sheaves on $(\mathit{Sch}/S)_{fppf}$ we conclude that $X \to Y$ factors through $Z$.
$\square$

Definition 64.12.6. Let $S$ be a scheme, and let $X$ be an algebraic space over $S$. Let $Z \subset |X|$ be a closed subset. An *algebraic space structure on $Z$* is given by a closed subspace $Z'$ of $X$ with $|Z'|$ equal to $Z$. The *reduced induced algebraic space structure* on $Z$ is the one constructed in Lemma 64.12.4. The *reduction $X_{red}$ of $X$* is the reduced induced algebraic space structure on $|X|$.

## Comments (1)

Comment #5097 by Klaus Mattis on