The Stacks project

Proof. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(\varphi )$. Our goal is to prove that $\mathcal{K} = 0$. In order to do this it suffices to prove that $\text{WeakAss}_ X(\mathcal{K}) = \emptyset $, see Divisors, Lemma 31.5.5. We have $\text{WeakAss}_ X(\mathcal{K}) \subset \text{WeakAss}_ X(\mathcal{F})$, see Divisors, Lemma 31.5.4. As $\mathcal{F}$ is flat we see from Lemma 38.13.5 that $\text{WeakAss}_ X(\mathcal{F}) \subset \text{Ass}_{X/S}(\mathcal{F})$. By purity any point $x'$ of $\text{Ass}_{X/S}(\mathcal{F})$ is a generalization of a point of $X_ s$, and hence is the specialization of a point $x \in \text{Ass}_{X_ s}(\mathcal{F}_ s)$, by Lemma 38.18.1. Hence the injectivity of $\varphi _ x$ implies the injectivity of $\varphi _{x'}$, whence $\mathcal{K}_{x'} = 0$. $\square$

Proof. The implication (1) $\Rightarrow $ (2) is Lemma 38.17.4. Assume $\mathcal{F}$ is pure relative to $S$. Note that by Lemma 38.18.3 this implies $\mathcal{F}$ remains pure after any base change. By Descent, Lemma 35.7.7 it suffices to prove $f_*\mathcal{F}$ is fpqc locally projective on $S$. Pick $s \in S$. We will prove that the restriction of $f_*\mathcal{F}$ to an étale neighbourhood of $s$ is locally projective. Namely, by Lemma 38.12.5, after replacing $S$ by an affine elementary étale neighbourhood of $s$, we may assume there exists a diagram

of schemes affine and of finite presentation over $S$, where $g$ is étale, $X_ s \subset g(X')$, and with $\Gamma (X', g^*\mathcal{F})$ a projective $\Gamma (S, \mathcal{O}_ S)$-module. Note that in this case $g^*\mathcal{F}$ is universally pure over $S$, see Lemma 38.17.4. Hence by Lemma 38.18.2 we see that the open $g(X')$ contains the points of $\text{Ass}_{X/S}(\mathcal{F})$ lying over $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$. Set

By More on Morphisms, Lemma 37.25.5 $E$ is a constructible subset of $S$. We have seen that $\mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) \subset E$. By Morphisms, Lemma 29.22.4 we see that $E$ contains an open neighbourhood of $s$. Hence after replacing $S$ by an affine neighbourhood of $s$ we may assume that $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$. By Lemma 38.7.4 this means that

is $\Gamma (S, \mathcal{O}_ S)$-universally injective. By Algebra, Lemma 10.89.7 we conclude that $\Gamma (X, \mathcal{F})$ is Mittag-Leffler as an $\Gamma (S, \mathcal{O}_ S)$-module. Since $\Gamma (X, \mathcal{F})$ is countably generated and flat as a $\Gamma (S, \mathcal{O}_ S)$-module, we conclude it is projective by Algebra, Lemma 10.93.1. $\square$

Proof. During the proof we may replace $X$ by an open neighbourhood of $x$ and we may replace $S$ by an elementary étale neighbourhood of $s$. Hence, by openness of flatness (see More on Morphisms, Theorem 37.15.1) we may assume that $\mathcal{F}$ is flat over $S$. We may assume $S$ and $X$ are affine. After shrinking $X$ some more we may assume that any point of $\text{Ass}_{X_ s}(\mathcal{F}_ s)$ is a generalization of $x$. This property is preserved on replacing $(S, s)$ by an elementary étale neighbourhood. Hence we may apply Lemma 38.12.5 to arrive at the situation where there exists a diagram

of schemes affine and of finite presentation over $S$, where $g$ is étale, $X_ s \subset g(X')$, and with $\Gamma (X', g^*\mathcal{F})$ a projective $\Gamma (S, \mathcal{O}_ S)$-module. Note that in this case $g^*\mathcal{F}$ is universally pure over $S$, see Lemma 38.17.4.

Let $U \subset g(X')$ be an affine open neighbourhood of $x$. We claim that $\mathcal{F}|_ U$ is pure along $U_ s$. If we prove this, then the lemma follows because $\mathcal{F}|_ U$ will be pure relative to $S$ after shrinking $S$, see Lemma 38.18.6, whereupon the projectivity follows from Proposition 38.19.2. To prove the claim we have to show, after replacing $(S, s)$ by an arbitrary elementary étale neighbourhood, that any point $\xi $ of $\text{Ass}_{U/S}(\mathcal{F}|_ U)$ lying over some $s' \in S$, $s' \leadsto s$ specializes to a point of $U_ s$. Since $U \subset g(X')$ we can find a $\xi ' \in X'$ with $g(\xi ') = \xi $. Because $g^*\mathcal{F}$ is pure over $S$, using Lemma 38.18.1, we see there exists a specialization $\xi ' \leadsto x'$ with $x' \in \text{Ass}_{X'_ s}(g^*\mathcal{F}_ s)$. Then $g(x') \in \text{Ass}_{X_ s}(\mathcal{F}_ s)$ (see for example Lemma 38.2.8 applied to the étale morphism $X'_ s \to X_ s$ of Noetherian schemes) and hence $g(x') \leadsto x$ by our choice of $X$ above! Since $x \in U$ we conclude that $g(x') \in U$. Thus $\xi = g(\xi ') \leadsto g(x') \in U_ s$ as desired. $\square$

Proof. The question is Zariski local on $X$ and $S$. Hence we may assume that $X$ and $S$ are affine. Then we can find a closed immersion $i : X \to \mathbf{A}^ n_ S$ over $S$. It is clear that it suffices to prove the lemma for the sheaf $i_*\mathcal{F}$ on $\mathbf{A}^ n_ S$ and the point $i(x)$. In this way we reduce to the case where $X \to S$ is of finite presentation. After replacing $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ and $X$ by an open of $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ we may assume that $\mathcal{F}$ is of finite presentation, see Proposition 38.10.3. In this case we may appeal to Lemma 38.19.3 and Algebra, Theorem 10.85.4 to conclude. $\square$

Proof. We can find a morphism of finite type $X \to S = \mathop{\mathrm{Spec}}(A)$ and a point $x \in X$ lying over the closed point $s$ of $S$ and a finite type quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ such that $\mathcal{F}_ x \cong N$ as an $A$-module. After shrinking $X$ we may assume that each point of $\text{Ass}_{X_ s}(\mathcal{F}_ s)$ specializes to $x$. By Lemma 38.19.4 we see that there exists a fundamental system of affine open neighbourhoods $U_ i \subset X$ of $x$ such that $\Gamma (U_ i, \mathcal{F})$ is a free $A$-module $F_ i$. Note that if $U_{i'} \subset U_ i$, then

is injective because a section of the kernel would be supported at a closed subset of $X_ s$ not meeting $x$ which is a contradiction to our choice of $X$ above. Since the maps $F_ i \to F_{i'}$ are $A$-universally injective (Lemma 38.7.5) it follows that $N$ is Mittag-Leffler by Algebra, Lemma 10.89.9. $\square$

Proof. The final statement follows from the fact that $I$ is a finitely generated ideal by More on Algebra, Lemma 15.24.2 and Algebra, Lemma 10.50.15.

Proof of existence of $I$. Let $A \subset A^ h$ be the henselization. Let $B'$ be the localization of $B \otimes _ A A^ h$ at the maximal ideal $\mathfrak m_ B \otimes A^ h + B \otimes \mathfrak m_{A^ h}$. Then $B \to B'$ is flat, hence faithfully flat. Let $N' = N \otimes _ B B'$. Let $x \in N$ and let $x' \in N'$ be the image. We claim that for an ideal $I \subset A$ we have $x \in IN \Leftrightarrow x' \in IN'$. Namely, $N/IN \to N'/IN'$ is the tensor product of $B \to B'$ with $N/IN$ and $B \to B'$ is universally injective by Algebra, Lemma 10.82.11. By More on Algebra, Lemma 15.123.6 and Algebra, Lemma 10.50.17 the map $A \to A^ h$ defines an inclusion preserving bijection $I \mapsto IA^ h$ on sets of ideals. We conclude that $x$ has a content ideal in $A$ if and only if $x'$ has a content ideal in $A^ h$. The assertion for $x' \in N'$ follows from Lemma 38.19.5 and Algebra, Lemma 10.89.6. $\square$

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 $X$ is 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 $a \in A$ and let $I = (\pi ) \subset R$ be its content ideal, see Lemma 38.19.6. Then $a = \pi a'$ and $a'$ maps to a nonzero element of $A/\mathfrak mA$ where $\mathfrak m \subset R$ is the maximal ideal. 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 $A/\mathfrak m A = \mathcal{O}_{X_ s, x'}$. This is a contradiction unless $A$ is reduced, which is what we wanted to show.

Of course, if $X$ is reduced, so is the generic fibre of $X$ over $R$. If $\mathfrak p \subset R$ is a prime ideal, then $R/\mathfrak p$ is a valuation ring by Algebra, Lemma 10.50.9. Hence redoing the argument with the base change of $X$ to $R/\mathfrak p$ proves the fibre over $\mathfrak p$ is reduced. $\square$