The Stacks project

Proof. See Algebra, Lemma 10.143.10. $\square$

Proof. This is true because $X' \to X_{S'}$ is étale as a morphism of schemes étale over $X$, see Morphisms, Lemma 29.36.18 and the base change of an étale morphism is étale, see Morphisms, Lemma 29.36.4. $\square$

Proof. As an étale morphism is a flat morphism (see Morphisms, Lemma 29.36.12) the implication “$\Leftarrow $” follows from Algebra, Lemma 10.39.4. For the converse assume that $\mathcal{F}$ is flat at $x$ over $S$. Denote $\tilde x \in X \times _ S T$ the point lying over $x$ in $X$ and over the image of $x$ in $T$ in $T$. Then $(X \times _ S T \to X)^*\mathcal{F}$ is flat at $\tilde x$ over $T$ via $\text{pr}_2 : X \times _ S T \to T$, see Morphisms, Lemma 29.25.7. The diagonal $\Delta _{T/S} : T \to T \times _ S T$ is an open immersion; combine Morphisms, Lemmas 29.35.13 and 29.36.5. So $X$ is identified with open subscheme of $X \times _ S T$, the restriction of $\text{pr}_2$ to this open is the given morphism $X \to T$, the point $\tilde x$ corresponds to the point $x$ in this open, and $(X \times _ S T \to X)^*\mathcal{F}$ restricted to this open is $\mathcal{F}$. Whence we see that $\mathcal{F}$ is flat at $x$ over $T$. $\square$

Proof. We may replace $S$ by an affine neighbourhood of $s$ and after that $T$ by an affine neighbourhood of $t$. Set $\mathcal{F} = (\mathop{\mathrm{Spec}}(\mathcal{O}_{T, t}) \to T)_*\widetilde M$. This is a quasi-coherent sheaf (see Schemes, Lemma 26.24.1 or argue directly) on $T$ whose stalk at $t$ is $M$ (details omitted). Apply Lemma 38.2.3. $\square$

Proof. By More on Algebra, Lemma 15.45.1 the local ring maps $\mathcal{O}_{S, s} \to \mathcal{O}_{S, s}^ h \to \mathcal{O}_{S, s}^{sh}$ are faithfully flat. Hence (3) $\Rightarrow $ (2) $\Rightarrow $ (1) and (5) $\Rightarrow $ (4) follow from Algebra, Lemma 10.39.4. By faithful flatness the equivalences (6) $\Leftrightarrow $ (5) and (5) $\Leftrightarrow $ (3) follow from Algebra, Lemma 10.39.8. Thus it suffices to show that (1) $\Rightarrow $ (2) $\Rightarrow $ (3) and (4) $\Rightarrow $ (5). To prove these we may assume $S$ is an affine scheme.

Assume (1). By Lemma 38.2.4 we see that $M^{sh}$ is flat over $\mathcal{O}_{T, t}$ for any étale neighbourhood $(T, t) \to (S, s)$. Since $\mathcal{O}_{S, s}^ h$ and $\mathcal{O}_{S, s}^{sh}$ are directed colimits of local rings of the form $\mathcal{O}_{T, t}$ (see Algebra, Lemmas 10.155.7 and 10.155.11) we conclude that $M^{sh}$ is flat over $\mathcal{O}_{S, s}^ h$ and $\mathcal{O}_{S, s}^{sh}$ by Algebra, Lemma 10.39.6. Thus (1) implies (2) and (3). Of course this implies also (2) $\Rightarrow $ (3) by replacing $\mathcal{O}_{S, s}$ by $\mathcal{O}_{S, s}^ h$. The same argument applies to prove (4) $\Rightarrow $ (5). $\square$

Proof. By More on Algebra, Lemma 15.45.1 the local ring maps $\mathcal{O}_{S, s} \to \mathcal{O}_{S, s}^ h \to \mathcal{O}_{S, s}^{sh}$ are faithfully flat. Hence (3) $\Rightarrow $ (2) $\Rightarrow $ (1) and (5) $\Rightarrow $ (4) follow from More on Algebra, Lemma 15.66.11. By faithful flatness the equivalences (6) $\Leftrightarrow $ (5) and (5) $\Leftrightarrow $ (3) follow from More on Algebra, Lemma 15.66.17. Thus it suffices to show that (1) $\Rightarrow $ (3), (2) $\Rightarrow $ (3), and (4) $\Rightarrow $ (5).

Assume (1). In particular $M^{sh}$ has vanishing cohomology in degrees $< a$ and $> b$. Hence we can represent $M^{sh}$ by a complex $P^\bullet $ of free $\mathcal{O}_{X, x}^{sh}$-modules with $P^ i = 0$ for $i > b$ (see for example the very general Derived Categories, Lemma 13.15.4). Note that $P^ n$ is flat over $\mathcal{O}_{S, s}$ for all $n$. Consider $\mathop{\mathrm{Coker}}(d_ P^{a - 1})$. By More on Algebra, Lemma 15.66.2 this is a flat $\mathcal{O}_{S, s}$-module. Hence by Lemma 38.2.5 this is a flat $\mathcal{O}_{S, s}^{sh}$-module. Thus $\tau _{\geq a}P^\bullet $ is a complex of flat $\mathcal{O}_{S, s}^{sh}$-modules representing $M^{sh}$ in $D(\mathcal{O}_{S, s}^{sh}$ and we find that $M^{sh}$ has tor amplitude in $[a, b]$, see More on Algebra, Lemma 15.66.3. Thus (1) implies (3). Of course this implies also (2) $\Rightarrow $ (3) by replacing $\mathcal{O}_{S, s}$ by $\mathcal{O}_{S, s}^ h$. The same argument applies to prove (4) $\Rightarrow $ (5). $\square$

Proof. The implication “$\Leftarrow $” follows immediately from Divisors, Lemma 31.6.4. Assume $t \in \text{WeakAss}(g^*\mathcal{G})$. Let $\mathop{\mathrm{Spec}}(A) \subset S$ be an affine open neighbourhood of $s$. Let $\mathcal{G}$ be the quasi-coherent sheaf associated to the $A$-module $M$. Let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. As $g$ is finite flat we have $g^{-1}(\mathop{\mathrm{Spec}}(A)) = \mathop{\mathrm{Spec}}(B)$ for some finite flat $A$-algebra $B$. Note that $g^*\mathcal{G}$ is the quasi-coherent $\mathcal{O}_{\mathop{\mathrm{Spec}}(B)}$-module associated to the $B$-module $M \otimes _ A B$ and $g_*g^*\mathcal{G}$ is the quasi-coherent $\mathcal{O}_{\mathop{\mathrm{Spec}}(A)}$-module associated to the $A$-module $M \otimes _ A B$. By Algebra, Lemma 10.78.5 we have $B_{\mathfrak p} \cong A_{\mathfrak p}^{\oplus n}$ for some integer $n \geq 0$. Note that $n \geq 1$ as we assumed there exists at least one point of $T$ lying over $s$. Hence we see by looking at stalks that

Now the assumption that $t \in \text{WeakAss}(g^*\mathcal{G})$ implies that $s \in \text{WeakAss}(g_*g^*\mathcal{G})$ by Divisors, Lemma 31.6.3 and hence by the above $s \in \text{WeakAss}(\mathcal{G})$. $\square$

Proof. After replacing $S$ and $U$ by affine neighbourhoods of $s$ and $u$ we may assume that $g$ is a standard étale morphism of affines, see Morphisms, Lemma 29.36.14. Thus we may assume $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(A[x, 1/g]/(f))$, where $f$ is monic and $f'$ is invertible in $A[x, 1/g]$. Note that $A[x, 1/g]/(f) = (A[x]/(f))_ g$ is also the localization of the finite free $A$-algebra $A[x]/(f)$. Hence we may think of $U$ as an open subscheme of the scheme $T = \mathop{\mathrm{Spec}}(A[x]/(f))$ which is finite locally free over $S$. This reduces us to Lemma 38.2.7 above. $\square$

Proof. The equivalence of (1) and (2) is the definition, see Divisors, Definition 31.5.1. The implications (2) $\Rightarrow $ (3) $\Rightarrow $ (4) follows from Divisors, Lemma 31.6.4 applied to the flat (More on Algebra, Lemma 15.45.1) morphisms

and the closed points. To prove (4) $\Rightarrow $ (2) we may replace $S$ by an affine neighbourhood. Suppose that $x \in \mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{S, s}^{sh}$ is an element whose annihilator has radical equal to $\mathfrak m_ s^{sh}$. (See Algebra, Lemma 10.66.2.) Since $\mathcal{O}_{S, s}^{sh}$ is equal to the limit of $\mathcal{O}_{U, u}$ over étale neighbourhoods $f : (U, u) \to (S, s)$ by Algebra, Lemma 10.155.11 we may assume that $x$ is the image of some $x' \in \mathcal{F}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{U, u}$. The local ring map $\mathcal{O}_{U, u} \to \mathcal{O}_{S, s}^{sh}$ is faithfully flat (as it is the strict henselization), hence universally injective (Algebra, Lemma 10.82.11). It follows that the annihilator of $x'$ is the inverse image of the annihilator of $x$. Hence the radical of this annihilator is equal to $\mathfrak m_ u$. Thus $u$ is a weakly associated point of $f^*\mathcal{F}$. By Lemma 38.2.8 we see that $s$ is a weakly associated point of $\mathcal{F}$. $\square$