Lemma 38.13.1. Let $R \to S$ be a ring map of finite presentation. Let $N$ be a finitely presented $S$-module. Let $\mathfrak q \subset S$ be a prime ideal lying over $\mathfrak p \subset R$. Set $\overline{S} = S \otimes _ R \kappa (\mathfrak p)$, $\overline{\mathfrak q} = \mathfrak q \overline{S}$, and $\overline{N} = N \otimes _ R \kappa (\mathfrak p)$. Then we can find a $g \in S$ with $g \not\in \mathfrak q$ such that $\overline{g} \in \mathfrak r$ for all $\mathfrak r \in \text{Ass}_{\overline{S}}(\overline{N})$ such that $\mathfrak r \not\subset \overline{\mathfrak q}$.

## 38.13 Flat finite type modules, Part II

We will need the following lemma.

**Proof.**
Namely, if $\text{Ass}_{\overline{S}}(\overline{N}) = \{ \mathfrak r_1, \ldots , \mathfrak r_ n\} $ (finiteness by Algebra, Lemma 10.63.5), then after renumbering we may assume that

Since $\overline{\mathfrak q}$ is a prime ideal we see that the product $\mathfrak r_{r + 1} \ldots \mathfrak r_ n$ is not contained in $\overline{\mathfrak q}$ and hence we can pick an element $a$ of $\overline{S}$ contained in $\mathfrak r_{r + 1}, \ldots , \mathfrak r_ n$ but not in $\overline{\mathfrak q}$. If there exists $g \in S$ mapping to $a$, then $g$ works. In general we can find a nonzero element $\lambda \in \kappa (\mathfrak p)$ such that $\lambda a$ is the image of a $g \in S$. $\square$

The following lemma has a slightly stronger variant Lemma 38.13.4 below.

Lemma 38.13.2. Let $R \to S$ be a ring map of finite presentation. Let $N$ be a finitely presented $S$-module which is flat as an $R$-module. Let $M$ be an $R$-module. Let $\mathfrak q$ be a prime of $S$ lying over $\mathfrak p \subset R$. Then

Here $\overline{S} = S \otimes _ R \kappa (\mathfrak p)$, $\overline{\mathfrak q} = \mathfrak q \overline{S}$, and $\overline{N} = N \otimes _ R \kappa (\mathfrak p)$.

**Proof.**
Pick $g \in S$ as in Lemma 38.13.1. Apply Proposition 38.12.4 to the morphism of schemes $\mathop{\mathrm{Spec}}(S_ g) \to \mathop{\mathrm{Spec}}(R)$, the quasi-coherent module associated to $N_ g$, and the points corresponding to the primes $\mathfrak qS_ g$ and $\mathfrak p$. Translating into algebra we obtain a commutative diagram of rings

endowed with primes as shown, the horizontal arrows are étale, and $N \otimes _ S S'$ is projective as an $R'$-module. Set $N' = N \otimes _ S S'$, $M' = M \otimes _ R R'$, $\overline{S}' = S' \otimes _{R'} \kappa (\mathfrak q')$, $\overline{\mathfrak q}' = \mathfrak q' \overline{S}'$, and

By Lemma 38.2.8 we have

Use Algebra, Lemma 10.66.9 for $\overline{N}$ and $\overline{N}'$. In particular we have

Our careful choice of $g$ and the formula for $\text{Ass}_{\overline{S}'}(\overline{N}')$ above shows that

This will be a key observation later in the proof. We will use the characterization of weakly associated primes given in Algebra, Lemma 10.66.2 without further mention.

Suppose that $\overline{\mathfrak q} \not\in \text{Ass}_{\overline{S}}(\overline{N})$. Then $\overline{\mathfrak q}' \not\in \text{Ass}_{\overline{S}'}(\overline{N}')$. By Algebra, Lemmas 10.63.9, 10.63.5, and 10.15.2 there exists an element $\overline{a}' \in \overline{\mathfrak q}'$ which is not a zerodivisor on $\overline{N}'$. After replacing $\overline{a}'$ by $\lambda \overline{a}'$ for some nonzero $\lambda \in \kappa (\mathfrak p)$ we can find $a' \in \mathfrak q'$ mapping to $\overline{a}'$. By Lemma 38.7.6 the map $a' : N'_{\mathfrak p'} \to N'_{\mathfrak p'}$ is $R'_{\mathfrak p'}$-universally injective. In particular we see that $a' : M' \otimes _{R'} N' \to M' \otimes _{R'} N'$ is injective after localizing at $\mathfrak p'$ and hence after localizing at $\mathfrak q'$. Clearly this implies that $\mathfrak q' \not\in \text{WeakAss}_{S'}(M' \otimes _{R'} N')$. We conclude that $\mathfrak q \in \text{WeakAss}_ S(M \otimes _ R N)$ implies $\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$.

Assume $\mathfrak q \in \text{WeakAss}_ S(M \otimes _ R N)$. We want to show $\mathfrak p \in \text{WeakAss}_ S(M)$. Let $z \in M \otimes _ R N$ be an element such that $\mathfrak q$ is minimal over $J = \text{Ann}_ S(z)$. Let $f_ i \in \mathfrak p$, $i \in I$ be a set of generators of the ideal $\mathfrak p$. Since $\mathfrak q$ lies over $\mathfrak p$, for every $i$ we can choose an $n_ i \geq 1$ and $g_ i \in S$, $g_ i \not\in \mathfrak q$ with $g_ i f_ i^{n_ i} \in J$, i.e., $g_ i f_ i^{n_ i} z = 0$. Let $z' \in (M' \otimes _{R'} N')_{\mathfrak p'}$ be the image of $z$. Observe that $z'$ is nonzero because $z$ has nonzero image in $(M \otimes _ R N)_\mathfrak q$ and because $S_\mathfrak q \to S'_{\mathfrak q'}$ is faithfully flat. We claim that $f_ i^{n_ i} z' = 0$.

Proof of the claim: Let $g'_ i \in S'$ be the image of $g_ i$. By the key observation (38.13.2.1) we find that the image $\overline{g}'_ i \in \overline{S}'$ is not contained in $\mathfrak r'$ for any $\mathfrak r' \in \text{Ass}_{\overline{S}'}(\overline{N})$. Hence by Lemma 38.7.6 we see that $g'_ i : N'_{\mathfrak p'} \to N'_{\mathfrak p'}$ is $R'_{\mathfrak p'}$-universally injective. In particular we see that $g'_ i : M' \otimes _{R'} N' \to M' \otimes _{R'} N'$ is injective after localizating at $\mathfrak p'$. The claim follows because $g_ i f_ i^{n_ i} z' = 0$.

Our claim shows that the annihilator of $z'$ in $R'_{\mathfrak p'}$ contains the elements $f_ i^{n_ i}$. As $R \to R'$ is étale we have $\mathfrak p'R'_{\mathfrak p'} = \mathfrak pR'_{\mathfrak p'}$ by Algebra, Lemma 10.143.5. Hence the annihilator of $z'$ in $R'_{\mathfrak p'}$ has radical equal to $\mathfrak p' R_{\mathfrak p'}$ (here we use $z'$ is not zero). On the other hand

The module $N'_{\mathfrak p'}$ is projective over the local ring $R'_{\mathfrak p'}$ and hence free (Algebra, Theorem 10.85.4). Thus we can find a finite free direct summand $F' \subset N'_{\mathfrak p'}$ such that $z' \in M'_{\mathfrak p'} \otimes _{R'_{\mathfrak p'}} F'$. If $F'$ has rank $n$, then we deduce that $\mathfrak p' R'_{\mathfrak p'} \in \text{WeakAss}_{R'_{\mathfrak p'}}({M'_{\mathfrak p'}}^{\oplus n})$. This implies $\mathfrak p'R'_{\mathfrak p'} \in \text{WeakAss}(M'_{\mathfrak p'})$ for example by Algebra, Lemma 10.66.4. Then $\mathfrak p' \in \text{WeakAss}_{R'}(M')$ which in turn gives $\mathfrak p \in \text{WeakAss}_ R(M)$. This finishes the proof of the implication “$\Rightarrow $” of the equivalence of the lemma.

Assume that $\mathfrak p \in \text{WeakAss}_ R(M)$ and $\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$. We want to show that $\mathfrak q$ is weakly associated to $M \otimes _ R N$. Note that $\overline{\mathfrak q}'$ is a maximal element of $\text{Ass}_{\overline{S}'}(\overline{N}')$. This is a consequence of (38.13.2.1) and the fact that there are no inclusions among the primes of $\overline{S}'$ lying over $\overline{\mathfrak q}$ (as fibres of étale morphisms are discrete Morphisms, Lemma 29.36.7). Thus, after replacing $R, S, \mathfrak p, \mathfrak q, M, N$ by $R', S', \mathfrak p', \mathfrak q', M', N'$ we may assume, in addition to the assumptions of the lemma, that

$\mathfrak p \in \text{WeakAss}_ R(M)$,

$\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$,

$N$ is projective as an $R$-module, and

$\overline{\mathfrak q}$ is maximal in $\text{Ass}_{\overline{S}}(\overline{N})$.

There is one more reduction, namely, we may replace $R, S, M, N$ by their localizations at $\mathfrak p$. This leads to one more condition, namely,

$R$ is a local ring with maximal ideal $\mathfrak p$.

We will finish by showing that (1) – (5) imply $\mathfrak q \in \text{WeakAss}(M \otimes _ R N)$.

Since $R$ is local and $\mathfrak p \in \text{WeakAss}_ R(M)$ we can pick a $y \in M$ whose annihilator $I$ has radical equal to $\mathfrak p$. Write $\overline{\mathfrak q} = (\overline{g}_1, \ldots , \overline{g}_ n)$ for some $\overline{g}_ i \in \overline{S}$. Choose $g_ i \in S$ mapping to $\overline{g}_ i$. Then $\mathfrak q = \mathfrak pS + g_1S + \ldots + g_ nS$. Consider the map

This is a homomorphism of projective $R/I$-modules. The local ring $R/I$ is auto-associated (More on Algebra, Definition 15.15.1) as $\mathfrak p/I$ is locally nilpotent. The map $\Psi \otimes \kappa (\mathfrak p)$ is not injective, because $\overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N})$. Hence More on Algebra, Lemma 15.15.4 implies $\Psi $ is not injective. Pick $z \in N/IN$ nonzero in the kernel of $\Psi $. The annihilator $J = \text{Ann}_ S(z)$ contains $IS$ and $g_ i$ by construction. Thus $\sqrt{J} \subset S$ contains $\mathfrak q$. Let $\mathfrak s \subset S$ be a prime minimal over $J$. Then $\mathfrak q \subset \mathfrak s$, $\mathfrak s$ lies over $\mathfrak p$, and $\mathfrak s \in \text{WeakAss}_ S(N/IN)$. The last fact by definition of weakly associated primes. Apply the “$\Rightarrow $” part of the lemma (which we've already proven) to the ring map $R \to S$ and the modules $R/I$ and $N$ to conclude that $\overline{\mathfrak s} \in \text{Ass}_{\overline{S}}(\overline{N})$. Since $\overline{\mathfrak q} \subset \overline{\mathfrak s}$ the maximality of $\overline{\mathfrak q}$, see condition (4) above, implies that $\overline{\mathfrak q} = \overline{\mathfrak s}$. This shows that $\mathfrak q = \mathfrak s$ and we conlude what we want. $\square$

Lemma 38.13.3. Let $S$ be a scheme. Let $f : X \to S$ be locally of finite type. Let $x \in X$ with image $s \in S$. Let $\mathcal{F}$ be a finite type quasi-coherent sheaf on $X$. Let $\mathcal{G}$ be a quasi-coherent sheaf on $S$. If $\mathcal{F}$ is flat at $x$ over $S$, then

**Proof.**
In this paragraph we reduce to $f$ being of finite presentation. The question is local on $X$ and $S$, hence we may assume $X$ and $S$ are affine. Write $X = \mathop{\mathrm{Spec}}(B)$, $S = \mathop{\mathrm{Spec}}(A)$ and write $B = A[x_1, \ldots , x_ n]/I$. In other words we obtain a closed immersion $i : X \to \mathbf{A}^ n_ S$ over $S$. Denote $t = i(x) \in \mathbf{A}^ n_ S$. Note that $i_*\mathcal{F}$ is a finite type quasi-coherent sheaf on $\mathbf{A}^ n_ S$ which is flat at $t$ over $S$ and note that

where $p : \mathbf{A}^ n_ S \to S$ is the projection. Note that $t$ is a weakly associated point of $i_*(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G})$ if and only if $x$ is a weakly associated point of $\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}$, see Divisors, Lemma 31.6.3. Similarly $x \in \text{Ass}_{X_ s}(\mathcal{F}_ s)$ if and only if $t \in \text{Ass}_{\mathbf{A}^ n_ s}((i_*\mathcal{F})_ s)$ (see Algebra, Lemma 10.63.14). Hence it suffices to prove the lemma in case $X = \mathbf{A}^ n_ S$. Thus we may assume that $X \to S$ is of finite presentation.

In this paragraph we reduce to $\mathcal{F}$ being of finite presentation and flat over $S$. Choose an elementary étale neighbourhood $e : (S', s') \to (S, s)$ and an open $V \subset X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ as in Proposition 38.10.3. Let $x' \in X' = X \times _ S S'$ be the unique point mapping to $x$ and $s'$. Then it suffices to prove the statement for $X' \to S'$, $x'$, $s'$, $(X' \to X)^*\mathcal{F}$, and $e^*\mathcal{G}$, see Lemma 38.2.8. Let $v \in V$ the unique point mapping to $x'$ and let $s' \in \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$ be the closed point. Then $\mathcal{O}_{V, v} = \mathcal{O}_{X', x'}$ and $\mathcal{O}_{\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}), s'} = \mathcal{O}_{S', s'}$ and similarly for the stalks of pullbacks of $\mathcal{F}$ and $\mathcal{G}$. Also $V_{s'} \subset X'_{s'}$ is an open subscheme. Since the condition of being a weakly associated point depend only on the stalk of the sheaf, we may replace $X' \to S'$, $x'$, $s'$, $(X' \to X)^*\mathcal{F}$, and $e^*\mathcal{G}$ by $V \to \mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'})$, $v$, $s'$, $(V \to X)^*\mathcal{F}$, and $(\mathop{\mathrm{Spec}}(\mathcal{O}_{S', s'}) \to S)^*\mathcal{G}$. Thus we may assume that $f$ is of finite presentation and $\mathcal{F}$ of finite presentation and flat over $S$.

Assume $f$ is of finite presentation and $\mathcal{F}$ of finite presentation and flat over $S$. After shrinking $X$ and $S$ to affine neighbourhoods of $x$ and $s$, this case is handled by Lemma 38.13.2. $\square$

Lemma 38.13.4. Let $R \to S$ be a ring map which is essentially of finite type. Let $N$ be a localization of a finite $S$-module flat over $R$. Let $M$ be an $R$-module. Then

**Proof.**
This lemma is a translation of Lemma 38.13.3 into algebra. Details of translation omitted.
$\square$

Lemma 38.13.5. Let $f : X \to S$ be a morphism which is locally of finite type. Let $\mathcal{F}$ be a finite type quasi-coherent sheaf on $X$ which is flat over $S$. Let $\mathcal{G}$ be a quasi-coherent sheaf on $S$. Then we have

**Proof.**
Immediate consequence of Lemma 38.13.3.
$\square$

Theorem 38.13.6. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

$X \to S$ is locally of finite presentation,

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

the set of weakly associated points of $S$ is locally finite in $S$.

Then $U = \{ x \in X \mid \mathcal{F}\text{ flat at }x\text{ over }S\} $ is open in $X$ and $\mathcal{F}|_ U$ is an $\mathcal{O}_ U$-module of finite presentation and flat over $S$.

**Proof.**
Let $x \in X$ be such that $\mathcal{F}$ is flat at $x$ over $S$. We have to find an open neighbourhood of $x$ such that $\mathcal{F}$ restricts to a $S$-flat finitely presented module on this neighbourhood. The problem is local on $X$ and $S$, hence we may assume that $X$ and $S$ are affine. As $\mathcal{F}_ x$ is a finitely presented $\mathcal{O}_{X, x}$-module by Lemma 38.10.9 we conclude from Algebra, Lemma 10.126.5 there exists a finitely presented $\mathcal{O}_ X$-module $\mathcal{F}'$ and a map $\varphi : \mathcal{F}' \to \mathcal{F}$ which induces an isomorphism $\varphi _ x : \mathcal{F}'_ x \to \mathcal{F}_ x$. In particular we see that $\mathcal{F}'$ is flat over $S$ at $x$, hence by openness of flatness More on Morphisms, Theorem 37.15.1 we see that after shrinking $X$ we may assume that $\mathcal{F}'$ is flat over $S$. As $\mathcal{F}$ is of finite type after shrinking $X$ we may assume that $\varphi $ is surjective, see Modules, Lemma 17.9.4 or alternatively use Nakayama's lemma (Algebra, Lemma 10.20.1). By Lemma 38.13.5 we have

As $\text{WeakAss}(S)$ is finite by assumption and since $\text{Ass}_{X_ s}(\mathcal{F}'_ s)$ is finite by Divisors, Lemma 31.2.5 we conclude that $\text{WeakAss}_ X(\mathcal{F}')$ is finite. Using Algebra, Lemma 10.15.2 we may, after shrinking $X$ once more, assume that $\text{WeakAss}_ X(\mathcal{F}')$ is contained in the generalization of $x$. Now consider $\mathcal{K} = \mathop{\mathrm{Ker}}(\varphi )$. We have $\text{WeakAss}_ X(\mathcal{K}) \subset \text{WeakAss}_ X(\mathcal{F}')$ (by Divisors, Lemma 31.5.4) but on the other hand, $\varphi _ x$ is an isomorphism, also $\varphi _{x'}$ is an isomorphism for all $x' \leadsto x$. We conclude that $\text{WeakAss}_ X(\mathcal{K}) = \emptyset $ whence $\mathcal{K} = 0$ by Divisors, Lemma 31.5.5. $\square$

Lemma 38.13.7. Let $R \to S$ be a ring map of finite presentation. Let $M$ be a finite $S$-module. Assume $\text{WeakAss}_ S(S)$ is finite. Then

is open in $\mathop{\mathrm{Spec}}(S)$ and for every $g \in S$ such that $D(g) \subset U$ the localization $M_ g$ is a finitely presented $S_ g$-module flat over $R$.

**Proof.**
Follows immediately from Theorem 38.13.6.
$\square$

Lemma 38.13.8. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Assume the set of weakly associated points of $S$ is locally finite in $S$. Then the set of points $x \in X$ where $f$ is flat is an open subscheme $U \subset X$ and $U \to S$ is flat and locally of finite presentation.

**Proof.**
The problem is local on $X$ and $S$, hence we may assume that $X$ and $S$ are affine. Then $X \to S$ corresponds to a finite type ring map $A \to B$. Choose a surjection $A[x_1, \ldots , x_ n] \to B$ and consider $B$ as an $A[x_1, \ldots , x_ n]$-module. An application of Lemma 38.13.7 finishes the proof.
$\square$

Lemma 38.13.9. Let $f : X \to S$ be a morphism of schemes which is locally of finite type and flat. If $S$ is integral, then $f$ is locally of finite presentation.

**Proof.**
Special case of Lemma 38.13.8.
$\square$

Proposition 38.13.10. Let $R$ be a domain. Let $R \to S$ be a ring map of finite type. Let $M$ be a finite $S$-module.

If $S$ is flat over $R$, then $S$ is a finitely presented $R$-algebra.

If $M$ is flat as an $R$-module, then $M$ is finitely presented as an $S$-module.

**Proof.**
Part (1) is a special case of Lemma 38.13.9. For Part (2) choose a surjection $R[x_1, \ldots , x_ n] \to S$. By Lemma 38.13.7 we find that $M$ is finitely presented as an $R[x_1, \ldots , x_ n]$-module. We conclude by Algebra, Lemma 10.6.4.
$\square$

Lemma 38.13.11 (Finite type version of Theorem 38.13.6). Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Assume

$X \to S$ is locally of finite type,

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

the set of weakly associated points of $S$ is locally finite in $S$.

Then $U = \{ x \in X \mid \mathcal{F}\text{ flat at }x\text{ over }S\} $ is open in $X$ and $\mathcal{F}|_ U$ is flat over $S$ and locally finitely presented relative to $S$ (see More on Morphisms, Definition 37.58.1).

**Proof.**
The question is local on $X$ and $S$. Thus we may assume $X$ and $S$ are affine. Then we may choose a closed immersion $i : X \to \mathbf{A}^ n_ S$. We apply Theorem 38.13.6 to $X' = \mathbf{A}^ n_ S \to S$ and the quasi-coherent module $\mathcal{F}' = i_*\mathcal{F}$ of finite type and we find that

is open in $X'$ and that $\mathcal{F}'|_{U'}$ is of finite presentation. Since $\mathcal{F}'$ restricts to zero on $X' \setminus i(X)$ and since $\mathcal{F}'_{i(x)} \cong \mathcal{F}_ x$ for all $x \in X$ we see that

Hence $U = i^{-1}(U')$ is open. Moreover, it is clear that $\mathcal{F}'|_{U'} = (i|_ U)_*(\mathcal{F}|_ U)$. Hence we conclude that $\mathcal{F}|_ U$ is finitely presented relative to $S$ by More on Morphisms, Lemmas 37.58.3 and 37.58.4. $\square$

Lemma 38.13.12. Let $R \to S$ be a ring map of finite type. Let $M$ be a finite $S$-module. Assume $\text{WeakAss}_ R(R)$ is finite. Then

is open in $\mathop{\mathrm{Spec}}(S)$ and for every $g \in S$ such that $D(g) \subset U$ the localization $M_ g$ is flat over $R$ and an $S_ g$-module finitely presented relative to $R$ (see More on Algebra, Definition 15.80.2).

**Proof.**
This is Lemma 38.13.11 translated into algebra.
$\square$

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