Lemma 38.3.1. Let $f : X \to S$ be a finite type morphism of affine schemes. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module. Let $x \in X$ with image $s = f(x)$ in $S$. Set $\mathcal{F}_ s = \mathcal{F}|_{X_ s}$. Then there exist a closed immersion $i : Z \to X$ of finite presentation, and a quasi-coherent finite type $\mathcal{O}_ Z$-module $\mathcal{G}$ such that $i_*\mathcal{G} = \mathcal{F}$ and $Z_ s = \text{Supp}(\mathcal{F}_ s)$.

## 38.3 The local structure of a finite type module

The key technical lemma that makes a lot of the arguments in this chapter work is the geometric Lemma 38.3.2.

**Proof.**
Say the morphism $f : X \to S$ is given by the ring map $A \to B$ and that $\mathcal{F}$ is the quasi-coherent sheaf associated to the $B$-module $M$. By Morphisms, Lemma 29.15.2 we know that $A \to B$ is a finite type ring map, and by Properties, Lemma 28.16.1 we know that $M$ is a finite $B$-module. In particular the support of $\mathcal{F}$ is the closed subscheme of $\mathop{\mathrm{Spec}}(B)$ cut out by the annihilator $I = \{ x \in B \mid xm = 0\ \forall m \in M\} $ of $M$, see Algebra, Lemma 10.40.5. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$ and let $\mathfrak p \subset A$ be the prime ideal corresponding to $s$. Note that $X_ s = \mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ and that $\mathcal{F}_ s$ is the quasi-coherent sheaf associated to the $B \otimes _ A \kappa (\mathfrak p)$ module $M \otimes _ A \kappa (\mathfrak p)$. By Morphisms, Lemma 29.5.3 the support of $\mathcal{F}_ s$ is equal to $V(I(B \otimes _ A \kappa (\mathfrak p)))$. Since $B \otimes _ A \kappa (\mathfrak p)$ is of finite type over $\kappa (\mathfrak p)$ there exist finitely many elements $f_1, \ldots , f_ m \in I$ such that

Denote $i : Z \to X$ the closed subscheme cut out by $(f_1, \ldots , f_ m)$, in a formula $Z = \mathop{\mathrm{Spec}}(B/(f_1, \ldots , f_ m))$. Since $M$ is annihilated by $I$ we can think of $M$ as an $B/(f_1, \ldots , f_ m)$-module. In other words, $\mathcal{F}$ is the pushforward of a finite type module on $Z$. As $Z_ s = \text{Supp}(\mathcal{F}_ s)$ by construction, this proves the lemma. $\square$

Lemma 38.3.2. 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 $S$. Set $\mathcal{F}_ s = \mathcal{F}|_{X_ s}$ and $n = \dim _ x(\text{Supp}(\mathcal{F}_ s))$. Then we can construct

elementary étale neighbourhoods $g : (X', x') \to (X, x)$, $e : (S', s') \to (S, s)$,

a commutative diagram

\[ \xymatrix{ X \ar[dd]_ f & X' \ar[dd] \ar[l]^ g & Z' \ar[l]^ i \ar[d]^\pi \\ & & Y' \ar[d]^ h \\ S & S' \ar[l]_ e & S' \ar@{=}[l] } \]a point $z' \in Z'$ with $i(z') = x'$, $y' = \pi (z')$, $h(y') = s'$,

a finite type quasi-coherent $\mathcal{O}_{Z'}$-module $\mathcal{G}$,

such that the following properties hold

$X'$, $Z'$, $Y'$, $S'$ are affine schemes,

$i$ is a closed immersion of finite presentation,

$i_*(\mathcal{G}) \cong g^*\mathcal{F}$,

$\pi $ is finite and $\pi ^{-1}(\{ y'\} ) = \{ z'\} $,

the extension $\kappa (y')/\kappa (s')$ is purely transcendental,

$h$ is smooth of relative dimension $n$ with geometrically integral fibres.

**Proof.**
Let $V \subset S$ be an affine neighbourhood of $s$. Let $U \subset f^{-1}(V)$ be an affine neighbourhood of $x$. Then it suffices to prove the lemma for $f|_ U : U \to V$ and $\mathcal{F}|_ U$. Hence in the rest of the proof we assume that $X$ and $S$ are affine.

First, suppose that $X_ s = \text{Supp}(\mathcal{F}_ s)$, in particular $n = \dim _ x(X_ s)$. Apply More on Morphisms, Lemmas 37.47.2 and 37.47.3. This gives us a commutative diagram

and point $x' \in X'$. We set $Z' = X'$, $i = \text{id}$, and $\mathcal{G} = g^*\mathcal{F}$ to obtain a solution in this case.

In general choose a closed immersion $Z \to X$ and a sheaf $\mathcal{G}$ on $Z$ as in Lemma 38.3.1. Applying the result of the previous paragraph to $Z \to S$ and $\mathcal{G}$ we obtain a diagram

and point $z' \in Z'$ satisfying all the required properties. We will use Lemma 38.2.1 to embed $Z'$ into a scheme étale over $X$. We cannot apply the lemma directly as we want $X'$ to be a scheme over $S'$. Instead we consider the morphisms

The first morphism is étale by Morphisms, Lemma 29.36.18. The second is a closed immersion as a base change of a closed immersion. Finally, as $X$, $S$, $S'$, $Z$, $Z'$ are all affine we may apply Lemma 38.2.1 to get an étale morphism of affine schemes $X' \to X \times _ S S'$ such that

As $Z \to X$ is a closed immersion of finite presentation, so is $Z' \to X'$. Let $x' \in X'$ be the point corresponding to $z' \in Z'$. Then the completed diagram

is a solution of the original problem. $\square$

Lemma 38.3.3. Assumptions and notation as in Lemma 38.3.2. If $f$ is locally of finite presentation then $\pi $ is of finite presentation. In this case the following are equivalent

$\mathcal{F}$ is an $\mathcal{O}_ X$-module of finite presentation in a neighbourhood of $x$,

$\mathcal{G}$ is an $\mathcal{O}_{Z'}$-module of finite presentation in a neighbourhood of $z'$, and

$\pi _*\mathcal{G}$ is an $\mathcal{O}_{Y'}$-module of finite presentation in a neighbourhood of $y'$.

Still assuming $f$ locally of finite presentation the following are equivalent to each other

$\mathcal{F}_ x$ is an $\mathcal{O}_{X, x}$-module of finite presentation,

$\mathcal{G}_{z'}$ is an $\mathcal{O}_{Z', z'}$-module of finite presentation, and

$(\pi _*\mathcal{G})_{y'}$ is an $\mathcal{O}_{Y', y'}$-module of finite presentation.

**Proof.**
Assume $f$ locally of finite presentation. Then $Z' \to S$ is locally of finite presentation as a composition of such, see Morphisms, Lemma 29.21.3. Note that $Y' \to S$ is also locally of finite presentation as a composition of a smooth and an étale morphism. Hence Morphisms, Lemma 29.21.11 implies $\pi $ is locally of finite presentation. Since $\pi $ is finite we conclude that it is also separated and quasi-compact, hence $\pi $ is actually of finite presentation.

To prove the equivalence of (1), (2), and (3) we also consider: (4) $g^*\mathcal{F}$ is a $\mathcal{O}_{X'}$-module of finite presentation in a neighbourhood of $x'$. The pullback of a module of finite presentation is of finite presentation, see Modules, Lemma 17.11.4. Hence (1) $\Rightarrow $ (4). The étale morphism $g$ is open, see Morphisms, Lemma 29.36.13. Hence for any open neighbourhood $U' \subset X'$ of $x'$, the image $g(U')$ is an open neighbourhood of $x$ and the map $\{ U' \to g(U')\} $ is an étale covering. Thus (4) $\Rightarrow $ (1) by Descent, Lemma 35.7.3. Using Descent, Lemma 35.7.10 and some easy topological arguments (see More on Morphisms, Lemma 37.47.4) we see that (4) $\Leftrightarrow $ (2) $\Leftrightarrow $ (3).

To prove the equivalence of (a), (b), (c) consider the ring maps

The first ring map is faithfully flat. Hence $\mathcal{F}_ x$ is of finite presentation over $\mathcal{O}_{X, x}$ if and only if $g^*\mathcal{F}_{x'}$ is of finite presentation over $\mathcal{O}_{X', x'}$, see Algebra, Lemma 10.83.2. The second ring map is surjective (hence finite) and finitely presented by assumption, hence $g^*\mathcal{F}_{x'}$ is of finite presentation over $\mathcal{O}_{X', x'}$ if and only if $\mathcal{G}_{z'}$ is of finite presentation over $\mathcal{O}_{Z', z'}$, see Algebra, Lemma 10.36.23. Because $\pi $ is finite, of finite presentation, and $\pi ^{-1}(\{ y'\} ) = \{ x'\} $ the ring homomorphism $\mathcal{O}_{Y', y'} \leftarrow \mathcal{O}_{Z', z'}$ is finite and of finite presentation, see More on Morphisms, Lemma 37.47.4. Hence $\mathcal{G}_{z'}$ is of finite presentation over $\mathcal{O}_{Z', z'}$ if and only if $\pi _*\mathcal{G}_{y'}$ is of finite presentation over $\mathcal{O}_{Y', y'}$, see Algebra, Lemma 10.36.23. $\square$

Lemma 38.3.4. Assumptions and notation as in Lemma 38.3.2. The following are equivalent

$\mathcal{F}$ is flat over $S$ in a neighbourhood of $x$,

$\mathcal{G}$ is flat over $S'$ in a neighbourhood of $z'$, and

$\pi _*\mathcal{G}$ is flat over $S'$ in a neighbourhood of $y'$.

The following are equivalent also

$\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$,

$\mathcal{G}_{z'}$ is flat over $\mathcal{O}_{S', s'}$, and

$(\pi _*\mathcal{G})_{y'}$ is flat over $\mathcal{O}_{S', s'}$.

**Proof.**
To prove the equivalence of (1), (2), and (3) we also consider: (4) $g^*\mathcal{F}$ is flat over $S$ in a neighbourhood of $x'$. We will use Lemma 38.2.3 to equate flatness over $S$ and $S'$ without further mention. The étale morphism $g$ is flat and open, see Morphisms, Lemma 29.36.13. Hence for any open neighbourhood $U' \subset X'$ of $x'$, the image $g(U')$ is an open neighbourhood of $x$ and the map $U' \to g(U')$ is surjective and flat. Thus (4) $\Leftrightarrow $ (1) by Morphisms, Lemma 29.25.13. Note that

Hence the flatness of $g^*\mathcal{F}$, $\mathcal{G}$ and $\pi _*\mathcal{G}$ over $S'$ are all equivalent (this uses that $X'$, $Z'$, $Y'$, and $S'$ are all affine). Some omitted topological arguments (compare More on Morphisms, Lemma 37.47.4) regarding affine neighbourhoods now show that (4) $\Leftrightarrow $ (2) $\Leftrightarrow $ (3).

To prove the equivalence of (a), (b), (c) consider the commutative diagram of local ring maps

We will use Lemma 38.2.4 to equate flatness over $\mathcal{O}_{S, s}$ and $\mathcal{O}_{S', s'}$ without further mention. The map $\gamma $ is faithfully flat. Hence $\mathcal{F}_ x$ is flat over $\mathcal{O}_{S, s}$ if and only if $g^*\mathcal{F}_{x'}$ is flat over $\mathcal{O}_{S', s'}$, see Algebra, Lemma 10.39.9. As $\mathcal{O}_{S', s'}$-modules the modules $g^*\mathcal{F}_{x'}$, $\mathcal{G}_{z'}$, and $\pi _*\mathcal{G}_{y'}$ are all isomorphic, see More on Morphisms, Lemma 37.47.4. This finishes the proof. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)