The Stacks project

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Let $X' \to Y'$ be the base change of $f$. Note that $Y' \to Y$ induces a bijection on points and that $X' \to X$ identifies fibres. Hence we may assume that $Y'$ is reduced, i.e., integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by an affine open. Hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain. Denote $K$ the fraction field of $A$. Pick an affine open $\mathop{\mathrm{Spec}}(B) = U \subset X$ and a section $h_\eta \in \Gamma (U_\eta , \mathcal{O}_{U_\eta }) = B_ K$ which is nonzero and nilpotent. After shrinking $Y$ we may assume that $h$ comes from $h \in \Gamma (U, \mathcal{O}_ U) = B$. After shrinking $Y$ a bit more we may assume that $h$ is nilpotent. Let $I = \{ b \in B \mid hb = 0\} $ be the annihilator of $h$. Then $C = B/I$ is a finite type $A$-algebra whose generic fiber $(B/I)_ K$ is nonzero (as $h_\eta \not= 0$). We apply generic flatness to $A \to C$ and $A \to B/hB$, see Algebra, Lemma 10.118.3, and we obtain a $g \in A$, $g \not= 0$ such that $C_ g$ is free as an $A_ g$-module and $(B/hB)_ g$ is flat as an $A_ g$-module. Replace $Y$ by $D(g) \subset Y$. Now we have the short exact sequence

with $B/hB$ flat over $A$ and with $C$ nonzero free as an $A$-module. It follows that for any homomorphism $A \to \kappa $ to a field the ring $C \otimes _ A \kappa $ is nonzero and the sequence

is exact, see Algebra, Lemma 10.39.12. Note that $B/hB \otimes _ A \kappa = (B \otimes _ A \kappa ) / h(B \otimes _ A \kappa )$ by right exactness of tensor product. Thus we conclude that multiplication by $h$ is not zero on $B \otimes _ A \kappa $. This clearly means that for any point $y \in Y$ the element $h$ restricts to a nonzero element of $U_ y$, whence $X_ y$ is nonreduced. $\square$

Proof. This comes down to the statement that for $y' \in Y'$ with image $y \in Y$ the fibre $X'_{y'} = X_ y \times _ y y'$ is geometrically reduced over $\kappa (y')$ if and only if $X_ y$ is geometrically reduced over $\kappa (y)$. This follows from Varieties, Lemma 33.6.6. $\square$

Proof. Apply Lemma 37.24.7 to get

with all the properties mentioned in that lemma. Let $\eta '$ be the generic point of $Y'$. Consider the morphism $X' \to X_{Y'}$ (which is the reduction morphism) and the resulting morphism of generic fibres $X'_{\eta '} \to X_{\eta '}$. Since $X'_{\eta '}$ is geometrically reduced, and $X_\eta $ is not this cannot be an isomorphism, see Varieties, Lemma 33.6.6. Hence $X_{\eta '}$ is nonreduced. Hence by Lemma 37.26.1 the fibres of $X_{Y'} \to Y'$ are nonreduced at all points $y' \in V'$ of a nonempty open $V' \subset Y'$. Since $g : Y' \to V$ is a homeomorphism Lemma 37.26.2 proves that $g(V')$ is the open we are looking for. $\square$

Proof. Let $Y' \subset Y$ be the reduction of $Y$. Let $X' \to Y'$ be the base change of $f$. Note that $Y' \to Y$ induces a bijection on points and that $X' \to X$ identifies fibres. Hence we may assume that $Y'$ is reduced, i.e., integral, see Properties, Lemma 28.3.4. We may also replace $Y$ by an affine open. Hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain. Denote $K$ the fraction field of $A$. After shrinking $Y$ a bit we may also assume that $X \to Y$ is flat and of finite presentation, see Morphisms, Proposition 29.27.1.

As $X_\eta $ is geometrically reduced there exists an open dense subset $V \subset X_\eta $ such that $V \to \mathop{\mathrm{Spec}}(K)$ is smooth, see Varieties, Lemma 33.25.7. Let $U \subset X$ be the set of points where $f$ is smooth. By Morphisms, Lemma 29.34.15 we see that $V \subset U_\eta $. Thus the generic fibre of $U$ is dense in the generic fibre of $X$. Since $X_\eta $ is reduced, it follows that $U_\eta $ is scheme theoretically dense in $X_\eta $, see Morphisms, Lemma 29.7.8. We note that as $U \to Y$ is smooth all the fibres of $U \to Y$ are geometrically reduced. Thus it suffices to show that, after shrinking $Y$, for all $y \in Y$ the scheme $U_ y$ is scheme theoretically dense in $X_ y$, see Morphisms, Lemma 29.7.9. This follows from Lemma 37.24.4. $\square$

Proof. Let $y \in Y$. We have to show that there exists an open neighbourhood $V$ of $y$ in $Y$ such that $E \cap V$ is constructible in $V$. Thus we may assume that $Y$ is affine. Then $X$ is quasi-compact. Choose a finite affine open covering $X = U_1 \cup \ldots \cup U_ n$. Then the fibres of $U_ i \to Y$ at $y$ form an affine open covering of the fibre of $X \to Y$ at $y$. Hence we may assume $X$ is affine as well. Write $Y = \mathop{\mathrm{Spec}}(A)$. Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as a directed limit of finite type $\mathbf{Z}$-algebras. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $f_ i : X_ i \to \mathop{\mathrm{Spec}}(A_ i)$ of finite presentation whose base change to $Y$ recovers $f$. By Lemma 37.26.2 it suffices to prove the lemma for $f_ i$. Thus we reduce to the case where $Y$ is the spectrum of a Noetherian ring.

We will use the criterion of Topology, Lemma 5.16.3 to prove that $E$ is constructible in case $Y$ is a Noetherian scheme. To see this let $Z \subset Y$ be an irreducible closed subscheme. We have to show that $E \cap Z$ either contains a nonempty open subset or is not dense in $Z$. If $X_\xi $ is geometrically reduced, then Lemma 37.26.4 (applied to the morphism $X_ Z \to Z$) implies that all fibres $X_ y$ are geometrically reduced for a nonempty open $V \subset Z$. If $X_\xi $ is not geometrically reduced, then Lemma 37.26.3 (applied to the morphism $X_ Z \to Z$) implies that all fibres $X_ y$ are geometrically reduced for a nonempty open $V \subset Z$. Thus we win. $\square$

Proof. Assume the special fibre $X_ s$ is reduced. Let $x \in X$ be any point, and let us show that $\mathcal{O}_{X, x}$ is reduced; this will prove that $X$ and $X_\eta $ are reduced. Let $x \leadsto x'$ be a specialization with $x'$ in the special fibre; such a specialization exists as a proper morphism is closed. Consider the local ring $A = \mathcal{O}_{X, x'}$. Then $\mathcal{O}_{X, x}$ is a localization of $A$, so it suffices to show that $A$ is reduced. Let $\pi \in R$ be a uniformizer. If $a \in A$ 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 by flatness of $A$ over $R$. But $a'$ maps to a nonzero element of the reduced ring $A/\pi A = \mathcal{O}_{X_ s, x'}$. This is a contradiction unless $A$ is reduced, which is what we wanted to show. $\square$

Proof. 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, Lemma 32.10.1, 32.8.7, and 32.13.1. Since the formation of the set of points where the fibres are geometrically reduced commutes with base change (Lemma 37.26.2), we may assume the base is Noetherian.

Assume $Y$ is Noetherian. The set is constructible by Lemma 37.26.5. 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 Varieties, Lemma 33.6.4. 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 37.26.6. $\square$