38.4 One step dévissage

In this section we explain what is a one step dévissage of a module. A one step dévissage exist étale locally on base and target. We discuss base change, Zariski shrinking and étale localization of a one step dévissage.

Definition 38.4.1. Let $S$ be a scheme. Let $X$ be locally of finite type over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $s \in S$ be a point. A one step dévissage of $\mathcal{F}/X/S$ over $s$ is given by morphisms of schemes over $S$

$\xymatrix{ X & Z \ar[l]_ i \ar[r]^\pi & Y }$

and a quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{G}$ of finite type such that

1. $X$, $S$, $Z$ and $Y$ are affine,

2. $i$ is a closed immersion of finite presentation,

3. $\mathcal{F} \cong i_*\mathcal{G}$,

4. $\pi$ is finite, and

5. the structure morphism $Y \to S$ is smooth with geometrically irreducible fibres of dimension $\dim (\text{Supp}(\mathcal{F}_ s))$.

In this case we say $(Z, Y, i, \pi , \mathcal{G})$ is a one step dévissage of $\mathcal{F}/X/S$ over $s$.

Note that such a one step dévissage can only exist if $X$ and $S$ are affine. In the definition above we only require $X$ to be (locally) of finite type over $S$ and we continue working in this setting below. In [GruRay] the authors use consistently the setup where $X \to S$ is locally of finite presentation and $\mathcal{F}$ quasi-coherent $\mathcal{O}_ X$-module of finite type. The advantage of this choice is that it “makes sense” to ask for $\mathcal{F}$ to be of finite presentation as an $\mathcal{O}_ X$-module, whereas in our setting it “does not make sense”. Please see More on Morphisms, Section 37.52 for a discussion; the observations made there show that in our setup we may consider the condition of $\mathcal{F}$ being “locally of finite presentation relative to $S$”, and we could work consistently with this notion. Instead however, we will rely on the results of Lemma 38.3.3 and the observations in Remark 38.6.3 to deal with this issue in an ad hoc fashion whenever it comes up.

Definition 38.4.2. Let $S$ be a scheme. Let $X$ be locally of finite type over $S$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $x \in X$ be a point with image $s$ in $S$. A one step dévissage of $\mathcal{F}/X/S$ at $x$ is a system $(Z, Y, i, \pi , \mathcal{G}, z, y)$, where $(Z, Y, i, \pi , \mathcal{G})$ is a one step dévissage of $\mathcal{F}/X/S$ over $s$ and

1. $\dim _ x(\text{Supp}(\mathcal{F}_ s)) = \dim (\text{Supp}(\mathcal{F}_ s))$,

2. $z \in Z$ is a point with $i(z) = x$ and $\pi (z) = y$,

3. we have $\pi ^{-1}(\{ y\} ) = \{ z\}$,

4. the extension $\kappa (s) \subset \kappa (y)$ is purely transcendental.

A one step dévissage of $\mathcal{F}/X/S$ at $x$ can only exist if $X$ and $S$ are affine. Condition (1) assures us that $Y \to S$ has relative dimension equal to $\dim _ x(\text{Supp}(\mathcal{F}_ s))$ via condition (5) of Definition 38.4.1.

Lemma 38.4.3. Let $f : X \to S$ be morphism of schemes which is locally of finite type. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$ with image $s = f(x)$ in $S$. Then there exists a commutative diagram of pointed schemes

$\xymatrix{ (X, x) \ar[d]_ f & (X', x') \ar[l]^ g \ar[d] \\ (S, s) & (S', s') \ar[l] \\ }$

such that $(S', s') \to (S, s)$ and $(X', x') \to (X, x)$ are elementary étale neighbourhoods, and such that $g^*\mathcal{F}/X'/S'$ has a one step dévissage at $x'$.

Proof. This is immediate from Definition 38.4.2 and Lemma 38.3.2. $\square$

Lemma 38.4.4. Let $S$, $X$, $\mathcal{F}$, $s$ be as in Definition 38.4.1. Let $(Z, Y, i, \pi , \mathcal{G})$ be a one step dévissage of $\mathcal{F}/X/S$ over $s$. Let $(S', s') \to (S, s)$ be any morphism of pointed schemes. Given this data let $X', Z', Y', i', \pi '$ be the base changes of $X, Z, Y, i, \pi$ via $S' \to S$. Let $\mathcal{F}'$ be the pullback of $\mathcal{F}$ to $X'$ and let $\mathcal{G}'$ be the pullback of $\mathcal{G}$ to $Z'$. If $S'$ is affine, then $(Z', Y', i', \pi ', \mathcal{G}')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ over $s'$.

Proof. Fibre products of affines are affine, see Schemes, Lemma 26.17.2. Base change preserves closed immersions, morphisms of finite presentation, finite morphisms, smooth morphisms, morphisms with geometrically irreducible fibres, and morphisms of relative dimension $n$, see Morphisms, Lemmas 29.2.4, 29.21.4, 29.44.6, 29.34.5, 29.29.2, and More on Morphisms, Lemma 37.25.2. We have $i'_*\mathcal{G}' \cong \mathcal{F}'$ because pushforward along the finite morphism $i$ commutes with base change, see Cohomology of Schemes, Lemma 30.5.1. We have $\dim (\text{Supp}(\mathcal{F}_ s)) = \dim (\text{Supp}(\mathcal{F}'_{s'}))$ by Morphisms, Lemma 29.28.3 because

$\text{Supp}(\mathcal{F}_ s) \times _ s s' = \text{Supp}(\mathcal{F}'_{s'}).$

This proves the lemma. $\square$

Lemma 38.4.5. Let $S$, $X$, $\mathcal{F}$, $x$, $s$ be as in Definition 38.4.2. Let $(Z, Y, i, \pi , \mathcal{G}, z, y)$ be a one step dévissage of $\mathcal{F}/X/S$ at $x$. Let $(S', s') \to (S, s)$ be a morphism of pointed schemes which induces an isomorphism $\kappa (s) = \kappa (s')$. Let $(Z', Y', i', \pi ', \mathcal{G}')$ be as constructed in Lemma 38.4.4 and let $x' \in X'$ (resp. $z' \in Z'$, $y' \in Y'$) be the unique point mapping to both $x \in X$ (resp. $z \in Z$, $y \in Y$) and $s' \in S'$. If $S'$ is affine, then $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ at $x'$.

Proof. By Lemma 38.4.4 $(Z', Y', i', \pi ', \mathcal{G}')$ is a one step dévissage of $\mathcal{F}'/X'/S'$ over $s'$. Properties (1) – (4) of Definition 38.4.2 hold for $(Z', Y', i', \pi ', \mathcal{G}', z', y')$ as the assumption that $\kappa (s) = \kappa (s')$ insures that the fibres $X'_{s'}$, $Z'_{s'}$, and $Y'_{s'}$ are isomorphic to $X_ s$, $Z_ s$, and $Y_ s$. $\square$

Definition 38.4.6. Let $S$, $X$, $\mathcal{F}$, $x$, $s$ be as in Definition 38.4.2. Let $(Z, Y, i, \pi , \mathcal{G}, z, y)$ be a one step dévissage of $\mathcal{F}/X/S$ at $x$. Let us define a standard shrinking of this situation to be given by standard opens $S' \subset S$, $X' \subset X$, $Z' \subset Z$, and $Y' \subset Y$ such that $s \in S'$, $x \in X'$, $z \in Z'$, and $y \in Y'$ and such that

$(Z', Y', i|_{Z'}, \pi |_{Z'}, \mathcal{G}|_{Z'}, z, y)$

is a one step dévissage of $\mathcal{F}|_{X'}/X'/S'$ at $x$.

Lemma 38.4.7. With assumption and notation as in Definition 38.4.6 we have:

1. If $S' \subset S$ is a standard open neighbourhood of $s$, then setting $X' = X_{S'}$, $Z' = Z_{S'}$ and $Y' = Y_{S'}$ we obtain a standard shrinking.

2. Let $W \subset Y$ be a standard open neighbourhood of $y$. Then there exists a standard shrinking with $Y' = W \times _ S S'$.

3. Let $U \subset X$ be an open neighbourhood of $x$. Then there exists a standard shrinking with $X' \subset U$.

Proof. Part (1) is immediate from Lemma 38.4.5 and the fact that the inverse image of a standard open under a morphism of affine schemes is a standard open, see Algebra, Lemma 10.17.4.

Let $W \subset Y$ as in (2). Because $Y \to S$ is smooth it is open, see Morphisms, Lemma 29.34.10. Hence we can find a standard open neighbourhood $S'$ of $s$ contained in the image of $W$. Then the fibres of $W_{S'} \to S'$ are nonempty open subschemes of the fibres of $Y \to S$ over $S'$ and hence geometrically irreducible too. Setting $Y' = W_{S'}$ and $Z' = \pi ^{-1}(Y')$ we see that $Z' \subset Z$ is a standard open neighbourhood of $z$. Let $\overline{h} \in \Gamma (Z, \mathcal{O}_ Z)$ be a function such that $Z' = D(\overline{h})$. As $i : Z \to X$ is a closed immersion, we can find a function $h \in \Gamma (X, \mathcal{O}_ X)$ such that $i^\sharp (h) = \overline{h}$. Take $X' = D(h) \subset X$. In this way we obtain a standard shrinking as in (2).

Let $U \subset X$ be as in (3). We may after shrinking $U$ assume that $U$ is a standard open. By More on Morphisms, Lemma 37.43.4 there exists a standard open $W \subset Y$ neighbourhood of $y$ such that $\pi ^{-1}(W) \subset i^{-1}(U)$. Apply (2) to get a standard shrinking $X', S', Z', Y'$ with $Y' = W_{S'}$. Since $Z' \subset \pi ^{-1}(W) \subset i^{-1}(U)$ we may replace $X'$ by $X' \cap U$ (still a standard open as $U$ is also standard open) without violating any of the conditions defining a standard shrinking. Hence we win. $\square$

Lemma 38.4.8. Let $S$, $X$, $\mathcal{F}$, $x$, $s$ be as in Definition 38.4.2. Let $(Z, Y, i, \pi , \mathcal{G}, z, y)$ be a one step dévissage of $\mathcal{F}/X/S$ at $x$. Let

$\xymatrix{ (Y, y) \ar[d] & (Y', y') \ar[l] \ar[d] \\ (S, s) & (S', s') \ar[l] }$

be a commutative diagram of pointed schemes such that the horizontal arrows are elementary étale neighbourhoods. Then there exists a commutative diagram

$\xymatrix{ & & (X'', x'') \ar[lld] \ar[d] & (Z'', z'') \ar[l] \ar[lld] \ar[d] \\ (X, x) \ar[d] & (Z, z) \ar[l] \ar[d] & (S'', s'') \ar[lld] & (Y'', y'') \ar[lld] \ar[l] \\ (S, s) & (Y, y) \ar[l] }$

of pointed schemes with the following properties:

1. $(S'', s'') \to (S', s')$ is an elementary étale neighbourhood and the morphism $S'' \to S$ is the composition $S'' \to S' \to S$,

2. $Y''$ is an open subscheme of $Y' \times _{S'} S''$,

3. $Z'' = Z \times _ Y Y''$,

4. $(X'', x'') \to (X, x)$ is an elementary étale neighbourhood, and

5. $(Z'', Y'', i'', \pi '', \mathcal{G}'', z'', y'')$ is a one step dévissage at $x''$ of the sheaf $\mathcal{F}''$.

Here $\mathcal{F}''$ (resp. $\mathcal{G}''$) is the pullback of $\mathcal{F}$ (resp. $\mathcal{G}$) via the morphism $X'' \to X$ (resp. $Z'' \to Z$) and $i'' : Z'' \to X''$ and $\pi '' : Z'' \to Y''$ are as in the diagram.

Proof. Let $(S'', s'') \to (S', s')$ be any elementary étale neighbourhood with $S''$ affine. Let $Y'' \subset Y' \times _{S'} S''$ be any affine open neighbourhood containing the point $y'' = (y', s'')$. Then we obtain an affine $(Z'', z'')$ by (3). Moreover $Z_{S''} \to X_{S''}$ is a closed immersion and $Z'' \to Z_{S''}$ is an étale morphism. Hence Lemma 38.2.1 applies and we can find an étale morphism $X'' \to X_{S'}$ of affines such that $Z'' \cong X'' \times _{X_{S'}} Z_{S'}$. Denote $i'' : Z'' \to X''$ the corresponding closed immersion. Setting $x'' = i''(z'')$ we obtain a commutative diagram as in the lemma. Properties (1), (2), (3), and (4) hold by construction. Thus it suffices to show that (5) holds for a suitable choice of $(S'', s'') \to (S', s')$ and $Y''$.

We first list those properties which hold for any choice of $(S'', s'') \to (S', s')$ and $Y''$ as in the first paragraph. As we have $Z'' = X'' \times _ X Z$ by construction we see that $i''_*\mathcal{G}'' = \mathcal{F}''$ (with notation as in the statement of the lemma), see Cohomology of Schemes, Lemma 30.5.1. Set $n = \dim (\text{Supp}(\mathcal{F}_ s)) = \dim _ x(\text{Supp}(\mathcal{F}_ s))$. The morphism $Y'' \to S''$ is smooth of relative dimension $n$ (because $Y' \to S'$ is smooth of relative dimension $n$ as the composition $Y' \to Y_{S'} \to S'$ of an étale and smooth morphism of relative dimension $n$ and because base change preserves smooth morphisms of relative dimension $n$). We have $\kappa (y'') = \kappa (y)$ and $\kappa (s) = \kappa (s'')$ hence $\kappa (y'')$ is a purely transcendental extension of $\kappa (s'')$. The morphism of fibres $X''_{s''} \to X_ s$ is an étale morphism of affine schemes over $\kappa (s) = \kappa (s'')$ mapping the point $x''$ to the point $x$ and pulling back $\mathcal{F}_ s$ to $\mathcal{F}''_{s''}$. Hence

$\dim (\text{Supp}(\mathcal{F}''_{s''})) = \dim (\text{Supp}(\mathcal{F}_ s)) = n = \dim _ x(\text{Supp}(\mathcal{F}_ s)) = \dim _{x''}(\text{Supp}(\mathcal{F}''_{s''}))$

because dimension is invariant under étale localization, see Descent, Lemma 35.18.2. As $\pi '' : Z'' \to Y''$ is the base change of $\pi$ we see that $\pi ''$ is finite and as $\kappa (y) = \kappa (y'')$ we see that $\pi ^{-1}(\{ y''\} ) = \{ z''\}$.

At this point we have verified all the conditions of Definition 38.4.1 except we have not verified that $Y'' \to S''$ has geometrically irreducible fibres. Of course in general this is not going to be true, and it is at this point that we will use that $\kappa (s) \subset \kappa (y)$ is purely transcendental. Namely, let $T \subset Y'_{s'}$ be the irreducible component of $Y'_{s'}$ containing $y' = (y, s')$. Note that $T$ is an open subscheme of $Y'_{s'}$ as this is a smooth scheme over $\kappa (s')$. By Varieties, Lemma 33.7.14 we see that $T$ is geometrically connected because $\kappa (s') = \kappa (s)$ is algebraically closed in $\kappa (y') = \kappa (y)$. As $T$ is smooth we see that $T$ is geometrically irreducible. Hence More on Morphisms, Lemma 37.42.4 applies and we can find an elementary étale morphism $(S'', s'') \to (S', s')$ and an affine open $Y'' \subset Y'_{S''}$ such that all fibres of $Y'' \to S''$ are geometrically irreducible and such that $T = Y''_{s''}$. After shrinking (first $Y''$ and then $S''$) we may assume that both $Y''$ and $S''$ are affine. This finishes the proof of the lemma. $\square$

Lemma 38.4.9. Let $S$, $X$, $\mathcal{F}$, $s$ be as in Definition 38.4.1. Let $(Z, Y, i, \pi , \mathcal{G})$ be a one step dévissage of $\mathcal{F}/X/S$ over $s$. Let $\xi \in Y_ s$ be the (unique) generic point. Then there exists an integer $r > 0$ and an $\mathcal{O}_ Y$-module map

$\alpha : \mathcal{O}_ Y^{\oplus r} \longrightarrow \pi _*\mathcal{G}$

such that

$\alpha : \kappa (\xi )^{\oplus r} \longrightarrow (\pi _*\mathcal{G})_\xi \otimes _{\mathcal{O}_{Y, \xi }} \kappa (\xi )$

is an isomorphism. Moreover, in this case we have

$\dim (\text{Supp}(\mathop{\mathrm{Coker}}(\alpha )_ s)) < \dim (\text{Supp}(\mathcal{F}_ s)).$

Proof. By assumption the schemes $S$ and $Y$ are affine. Write $S = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. As $\pi$ is finite the $\mathcal{O}_ Y$-module $\pi _*\mathcal{G}$ is a finite type quasi-coherent $\mathcal{O}_ Y$-module. Hence $\pi _*\mathcal{G} = \widetilde{N}$ for some finite $B$-module $N$. Let $\mathfrak p \subset B$ be the prime ideal corresponding to $\xi$. To obtain $\alpha$ set $r = \dim _{\kappa (\mathfrak p)} N \otimes _ B \kappa (\mathfrak p)$ and pick $x_1, \ldots , x_ r \in N$ which form a basis of $N \otimes _ B \kappa (\mathfrak p)$. Take $\alpha : B^{\oplus r} \to N$ to be the map given by the formula $\alpha (b_1, \ldots , b_ r) = \sum b_ ix_ i$. It is clear that $\alpha : \kappa (\mathfrak p)^{\oplus r} \to N \otimes _ B \kappa (\mathfrak p)$ is an isomorphism as desired. Finally, suppose $\alpha$ is any map with this property. Then $N' = \mathop{\mathrm{Coker}}(\alpha )$ is a finite $B$-module such that $N' \otimes \kappa (\mathfrak p) = 0$. By Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $N'_{\mathfrak p} = 0$. Since the fibre $Y_ s$ is geometrically irreducible of dimension $n$ with generic point $\xi$ and since we have just seen that $\xi$ is not in the support of $\mathop{\mathrm{Coker}}(\alpha )$ the last assertion of the lemma holds. $\square$

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