76.29 Reduced fibres
This section is the analogue of More on Morphisms, Section 37.26.
Lemma 76.29.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $y \in |Y|$. The following are equivalent
for some morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class of $y$ the algebraic space $X_ k$ is geometrically reduced over $k$,
for every morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class of $y$ the algebraic space $X_ k$ is geometrically reduced over $k$,
for every morphism $\mathop{\mathrm{Spec}}(k) \to Y$ in the equivalence class of $y$ the algebraic space $X_ k$ is reduced.
Proof.
This follows immediately from Spaces over Fields, Lemma 72.11.6 and the definition of the equivalence relation defining $|X|$ given in Properties of Spaces, Section 66.4.
$\square$
Definition 76.29.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $y \in |Y|$. We say the fibre of $f : X \to Y$ at $y$ is geometrically reduced if the equivalent conditions of Lemma 76.29.1 hold.
Here are the obligatory lemmas.
Lemma 76.29.3. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y' \to Y$ be morphisms of algebraic spaces over $S$. Denote $f' : X' \to Y'$ the base change of $f$ by $g$. Then
\begin{align*} \{ y' \in |Y'| : \text{the fibre of }f' : X' \to Y'\text{ at }y' \text{ is geometrically reduced}\} \\ = g^{-1}(\{ y \in |Y| : \text{the fibre of }f : X \to Y\text{ at }y \text{ is geometrically reduced}\} ). \end{align*}
Proof.
For $y' \in |Y'|$ choose a morphism $\mathop{\mathrm{Spec}}(k) \to Y'$ in the equivalence class of $y'$. Then $g(y')$ is represented by the composition $\mathop{\mathrm{Spec}}(k) \to Y' \to Y$. Hence $X' \times _{Y'} \mathop{\mathrm{Spec}}(k) = X \times _ Y \mathop{\mathrm{Spec}}(k)$ and the result follows from the definition.
$\square$
Lemma 76.29.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is quasi-compact and locally of finite presentation. Then the set
\[ E = \{ y \in |Y| : \text{the fibre of }f : X \to Y\text{ at }y \text{ is geometrically reduced}\} \]
is étale locally constructible.
Proof.
Choose an affine scheme $V$ and an étale morphism $V \to Y$. The meaning of the statement is that the inverse image of $E$ in $|V|$ is constructible. By Lemma 76.29.3 we may replace $Y$ by $V$, i.e., we may assume that $Y$ is an affine scheme. Then $X$ is quasi-compact. Choose an affine scheme $U$ and a surjective étale morphism $U \to X$. For a morphism $\mathop{\mathrm{Spec}}(k) \to Y$ the morphism between fibres $U_ k \to X_ k$ is surjective étale. Hence $U_ k$ is geometrically reduced over $k$ if and only if $X_ k$ is geometrically reduced over $k$, see Spaces over Fields, Lemma 72.11.7. Thus the set $E$ for $X \to Y$ is the same as the set $E$ for $U \to Y$. In this way we see that the lemma follows from the case of schemes, see More on Morphisms, Lemma 37.26.5.
$\square$
Lemma 76.29.5. Let $X$ be an algebraic space over a discrete valuation ring $R$ whose structure morphism $X \to \mathop{\mathrm{Spec}}(R)$ is proper and flat. If the special fibre is reduced, then both $X$ and the generic fibre $X_\eta $ are reduced.
Proof.
Choose an étale morphism $U \to X$ where $U$ is an affine scheme. Then $U$ is of finite type over $R$. Let $u \in U$ be in the special fibre. The local ring $A = \mathcal{O}_{U, u}$ is essentially of finite type over $R$, hence Noetherian. Let $\pi \in R$ be a uniformizer. Since $X$ is flat over $R$, we see that $\pi \in \mathfrak m_ A$ is a nonzerodivisor on $A$ and since the special fibre of $X$ is reduced, we have that $A/\pi A$ is reduced. If $a \in A$, $a \not= 0$ then there exists an $n \geq 0$ and an element $a' \in A$ such that $a = \pi ^ n a'$ and $a' \not\in \pi A$. This follows from Krull intersection theorem (Algebra, Lemma 10.51.4). If $a$ is nilpotent, so is $a'$, because $\pi $ is a nonzerodivisor. But $a'$ maps to a nonzero element of the reduced ring $A/\pi A$ so this is impossible. Hence $A$ is reduced. It follows that there exists an open neighbourhood of $u$ in $U$ which is reduced (small detail omitted; use that $U$ is Noetherian). Thus we can find an étale morphism $U \to X$ with $U$ a reduced scheme, such that every point of the special fibre of $X$ is in the image. Since $X$ is proper over $R$ it follows that $U \to X$ is surjective. Hence $X$ is reduced. Since the generic fibre of $U \to \mathop{\mathrm{Spec}}(R)$ is reduced as well (on affine pieces it is computed by taking localizations), we conclude the same thing is true for the generic fibre.
$\square$
Lemma 76.29.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. If $f$ is flat, proper, and of finite presentation, then the set
\[ E = \{ y \in |Y| : \text{the fibre of }f : X \to Y\text{ at }y \text{ is geometrically reduced}\} \]
is open in $|Y|$.
Proof.
By Lemma 76.29.3 formation of $E$ commutes with base change. To check a subset of $|Y|$ is open, we may replace $Y$ by the members of an étale covering. Thus we may assume $Y$ is affine. Then $Y$ is a cofiltered limit of affine schemes of finite type over $\mathbf{Z}$. Hence we can assume $X \to Y$ is the base change of $X_0 \to Y_0$ where $Y_0$ is the spectrum of a finite type $\mathbf{Z}$-algebra and $X_0 \to Y_0$ is flat and proper. See Limits of Spaces, Lemma 70.7.1, 70.6.12, and 70.6.13. Since the formation of $E$ commutes with base change (see above), we may assume the base is Noetherian.
Assume $Y$ is Noetherian. The set is constructible by Lemma 76.29.4. Hence it suffices to show the set is stable under generalization (Topology, Lemma 5.19.10). By Properties, Lemma 28.5.10 we reduce to the case where $Y = \mathop{\mathrm{Spec}}(R)$, $R$ is a discrete valuation ring, and the closed fibre $X_ y$ is geometrically reduced. To show: the generic fibre $X_\eta $ is geometrically reduced.
If not then there exists a finite extension $L$ of the fraction field of $R$ such that $X_ L$ is not reduced, see Spaces over Fields, Lemmas 72.11.4 (characteristic zero) and 72.11.5 (positive characteristic). There exists a discrete valuation ring $R' \subset L$ with fraction field $L$ dominating $R$, see Algebra, Lemma 10.120.18. After replacing $R$ by $R'$ we reduce to Lemma 76.29.5.
$\square$
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