86.8 The reduction

All formal algebraic spaces have an underlying reduced algebraic space as the following lemma demonstrates.

Lemma 86.8.1. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. There exists a reduced algebraic space $X_{red}$ and a representable morphism $X_{red} \to X$ which is a thickening. A morphism $U \to X$ with $U$ a reduced algebraic space factors uniquely through $X_{red}$.

Proof. First assume that $X$ is an affine formal algebraic space. Say $X = \mathop{\mathrm{colim}}\nolimits X_\lambda$ as in Definition 86.5.1. Since the transition morphisms are thickenings, the affine schemes $X_\lambda$ all have isomorphic reductions $X_{red}$. The morphism $X_{red} \to X$ is representable and a thickening by Lemma 86.5.3 and the fact that compositions of thickenings are thickenings. We omit the verification of the universal property (use Schemes, Definition 26.12.5, Schemes, Lemma 26.12.7, Properties of Spaces, Definition 65.12.5, and Properties of Spaces, Lemma 65.12.4).

Let $X$ and $\{ X_ i \to X\} _{i \in I}$ be as in Definition 86.7.1. For each $i$ let $X_{i, red} \to X_ i$ be the reduction as constructed above. For $i, j \in I$ the projection $X_{i, red} \times _ X X_ j \to X_{i, red}$ is an étale (by assumption) morphism of schemes (by Lemma 86.5.11). Hence $X_{i, red} \times _ X X_ j$ is reduced (see Descent, Lemma 35.17.1). Thus the projection $X_{i, red} \times _ X X_ j \to X_ j$ factors through $X_{j, red}$ by the universal property. We conclude that

$R_{ij} = X_{i, red} \times _ X X_ j = X_{i, red} \times _ X X_{j, red} = X_ i \times _ X X_{j, red}$

because the morphisms $X_{i, red} \to X_ i$ are injections of sheaves. Set $U = \coprod X_{i, red}$, set $R = \coprod R_{ij}$, and denote $s, t : R \to U$ the two projections. As a sheaf $R = U \times _ X U$ and $s$ and $t$ are étale. Then $(t, s) : R \to U$ defines an étale equivalence relation by our observations above. Thus $X_{red} = U/R$ is an algebraic space by Spaces, Theorem 64.10.5. By construction the diagram

$\xymatrix{ \coprod X_{i, red} \ar[r] \ar[d] & \coprod X_ i \ar[d] \\ X_{red} \ar[r] & X }$

is cartesian. Since the right vertical arrow is étale surjective and the top horizontal arrow is representable and a thickening we conclude that $X_{red} \to X$ is representable by Bootstrap, Lemma 79.5.2 (to verify the assumptions of the lemma use that a surjective étale morphism is surjective, flat, and locally of finite presentation and use that thickenings are separated and locally quasi-finite). Then we can use Spaces, Lemma 64.5.6 to conclude that $X_{red} \to X$ is a thickening (use that being a thickening is equivalent to being a surjective closed immersion).

Finally, suppose that $U \to X$ is a morphism with $U$ a reduced algebraic space over $S$. Then each $X_ i \times _ X U$ is étale over $U$ and therefore reduced (by our definition of reduced algebraic spaces in Properties of Spaces, Section 65.7). Then $X_ i \times _ X U \to X_ i$ factors through $X_{i, red}$. Hence $U \to X$ factors through $X_{red}$ because $\{ X_ i \times _ X U \to U\}$ is an étale covering. $\square$

Example 86.8.2. Let $A$ be a weakly admissible topological ring. In this case we have

$\text{Spf}(A)_{red} = \mathop{\mathrm{Spec}}(A/\mathfrak a)$

where $\mathfrak a \subset A$ is the ideal of topologically nilpotent elements. Namely, $\mathfrak a$ is a radical ideal (Lemma 86.4.10) which is open because $A$ is weakly admissible.

Lemma 86.8.3. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$ which is representable by algebraic spaces and smooth (for example étale). Then $X_{red} = X \times _ Y Y_{red}$.

Proof. (The étale case follows directly from the construction of the underlying reduced algebraic space in the proof of Lemma 86.8.1.) Assume $f$ is smooth. Observe that $X \times _ Y Y_{red} \to Y_{red}$ is a smooth morphism of algebraic spaces. Hence $X \times _ Y Y_{red}$ is a reduced algebraic space by Descent on Spaces, Lemma 73.8.5. Then the univeral property of reduction shows that the canonical morphism $X_{red} \to X \times _ Y Y_{red}$ is an isomorphism. $\square$

Lemma 86.8.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$ which is representable by algebraic spaces. Then $f$ is surjective in the sense of Bootstrap, Definition 79.4.1 if and only if $f_{red} : X_{red} \to Y_{red}$ is a surjective morphism of algebraic spaces.

Proof. Omitted. $\square$

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