## 38.23 Flattening a map

Theorem 38.23.3 is the key to further flattening statements.

Lemma 38.23.1. Let $S$ be a scheme. Let $g : X' \to X$ be a flat morphism of schemes over $S$ with $X$ locally of finite type over $S$. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ X$-module which is flat over $S$. If $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$ then the canonical map

\[ \mathcal{F} \longrightarrow g_*g^*\mathcal{F} \]

is injective, and remains injective after any base change.

**Proof.**
The final assertion means that $\mathcal{F}_ T \to (g_ T)_*g_ T^*\mathcal{F}_ T$ is injective for any morphism $T \to S$. The assumption $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$ is preserved by base change, see Divisors, Lemma 31.7.3 and Remark 31.7.4. The same holds for the assumption of flatness and finite type. Hence it suffices to prove the injectivity of the displayed arrow. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{F} \to g_*g^*\mathcal{F})$. 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 assumption any point $x$ of $\text{Ass}_{X/S}(\mathcal{F})$ is the image of some $x' \in X'$. Since $g$ is flat the local ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X', x'}$ is faithfully flat, hence the map

\[ \mathcal{F}_ x \longrightarrow g^*\mathcal{F}_{x'} = \mathcal{F}_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X', x'} \]

is injective (see Algebra, Lemma 10.82.11). This implies that $\mathcal{K}_ x = 0$ as desired.
$\square$

Lemma 38.23.2. Let $A$ be a ring. Let $u : M \to N$ be a surjective map of $A$-modules. If $M$ is projective as an $A$-module, then there exists an ideal $I \subset A$ such that for any ring map $\varphi : A \to B$ the following are equivalent

$u \otimes 1 : M \otimes _ A B \to N \otimes _ A B$ is an isomorphism, and

$\varphi (I) = 0$.

**Proof.**
As $M$ is projective we can find a projective $A$-module $C$ such that $F = M \oplus C$ is a free $A$-module. By replacing $u$ by $u \oplus 1 : F = M \oplus C \to N \oplus C$ we see that we may assume $M$ is free. In this case let $I$ be the ideal of $A$ generated by coefficients of all the elements of $\mathop{\mathrm{Ker}}(u)$ with respect to some (fixed) basis of $M$. The reason this works is that, since $u$ is surjective and $\otimes _ A B$ is right exact, $\mathop{\mathrm{Ker}}(u \otimes 1)$ is the image of $\mathop{\mathrm{Ker}}(u) \otimes _ A B$ in $M \otimes _ A B$.
$\square$

Theorem 38.23.3. In Situation 38.20.1 assume

$f$ is of finite presentation,

$\mathcal{F}$ is of finite presentation, flat over $S$, and pure relative to $S$, and

$u$ is surjective.

Then $F_{iso}$ is representable by a closed immersion $Z \to S$. Moreover $Z \to S$ is of finite presentation if $\mathcal{G}$ is of finite presentation.

**Proof.**
We will use without further mention that $\mathcal{F}$ is universally pure over $S$, see Lemma 38.18.3. By Lemma 38.20.2 and Descent, Lemmas 35.37.2 and 35.39.1 the question is local for the étale topology on $S$. Hence it suffices to prove, given $s \in S$, that there exists an étale neighbourhood of $(S, s)$ so that the theorem holds.

Using Lemma 38.12.5 and after replacing $S$ by an elementary étale neighbourhood of $s$ we may assume there exists a commutative diagram

\[ \xymatrix{ X \ar[dr] & & X' \ar[ll]^ g \ar[ld] \\ & S & } \]

of schemes of finite presentation over $S$, where $g$ is étale, $X_ s \subset g(X')$, the schemes $X'$ and $S$ are affine, $\Gamma (X', g^*\mathcal{F})$ a projective $\Gamma (S, \mathcal{O}_ S)$-module. Note that $g^*\mathcal{F}$ 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

\[ E = \{ t \in S \mid \text{Ass}_{X_ t}(\mathcal{F}_ t) \subset g(X') \} . \]

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 a smaller affine neighbourhood of $s$ we may assume that $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$.

Since we have assumed that $u$ is surjective we have $F_{iso} = F_{inj}$. From Lemma 38.23.1 it follows that $u : \mathcal{F} \to \mathcal{G}$ is injective if and only if $g^*u : g^*\mathcal{F} \to g^*\mathcal{G}$ is injective, and the same remains true after any base change. Hence we have reduced to the case where, in addition to the assumptions in the theorem, $X \to S$ is a morphism of affine schemes and $\Gamma (X, \mathcal{F})$ is a projective $\Gamma (S, \mathcal{O}_ S)$-module. This case follows immediately from Lemma 38.23.2.

To see that $Z$ is of finite presentation if $\mathcal{G}$ is of finite presentation, combine Lemma 38.20.2 part (4) with Limits, Remark 32.6.2.
$\square$

Lemma 38.23.4. Let $f:X\to S$ be a morphism of schemes which is of finite presentation, flat, and pure. Let $Y$ be a closed subscheme of $X$. Let $F=f_*Y$ be the Weil restriction functor of $Y$ along $f$, defined by

\[ F : (\mathit{Sch}/S)^{opp} \to \textit{Sets}, \quad T \mapsto \left\{ \begin{matrix} \{ *\}
& \text{if}
& Y_ T\to X_ T \text{ is an isomorphism, }
\\ \emptyset
& \text{else.}
& \end{matrix} \right. \]

Then $F$ is representable by a closed immersion $Z\to S$. Moreover $Z\to S$ is of finite presentation if $Y\to S$ is.

**Proof.**
Let $\mathcal{I}$ be the ideal sheaf defining $Y$ in $X$ and let $u:\mathcal{O}_ X\to \mathcal{O}_ X/\mathcal{I}$ be the surjection. Then for an $S$-scheme $T$, the closed immersion $Y_ T\to X_ T$ is an isomorphism if and only if $u_ T$ is an isomorphism. Hence the result follows from Theorem 38.23.3.
$\square$

## Comments (2)

Comment #7167 by Zongzhu Lin on

Comment #7307 by Johan on