## 38.13 Flat finite type modules, Part II

We will need the following lemma.

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 $\mathfrak r \not\subset \overline{\mathfrak q}$.

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

$\mathfrak r_1 \subset \overline{\mathfrak q}, \ldots , \mathfrak r_ r \subset \overline{\mathfrak q}, \quad \mathfrak r_{r + 1} \not\subset \overline{\mathfrak q}, \ldots , \mathfrak r_ n \not\subset \overline{\mathfrak q}$

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 sligthly 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

$\mathfrak q \in \text{WeakAss}_ S(M \otimes _ R N) \Leftrightarrow \Big( \mathfrak p \in \text{WeakAss}_ R(M) \text{ and } \overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N}) \Big)$

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

$\xymatrix{ S \ar[r] & S_ g \ar[r] & S' \\ & R \ar[lu] \ar[u] \ar[r] & R' \ar[u] } \quad \quad \xymatrix{ \mathfrak q \ar@{-}[r] \ar@{-}[rd] & \mathfrak qS_ g \ar@{-}[d] \ar@{-}[r] & \mathfrak q' \ar@{-}[d] \\ & \mathfrak p \ar@{-}[r] & \mathfrak p' }$

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

$\overline{N}' = N' \otimes _{R'} \kappa (\mathfrak p') = \overline{N} \otimes _{\overline{S}} \overline{S}'$

By Lemma 38.2.8 we have

\begin{align*} \text{WeakAss}_{S'}(M' \otimes _{R'} N') & = (\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S))^{-1}\text{WeakAss}_ S(M \otimes _ R N) \\ \text{WeakAss}_{R'}(M') & = (\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R))^{-1}\text{WeakAss}_ R(M) \\ \text{Ass}_{\overline{S}'}(\overline{N}') & = (\mathop{\mathrm{Spec}}(\overline{S}') \to \mathop{\mathrm{Spec}}(\overline{S}))^{-1} \text{Ass}_{\overline{S}}(\overline{N}) \end{align*}

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

\begin{align*} \mathfrak q \in \text{WeakAss}_ S(M \otimes _ R N) & \Leftrightarrow \mathfrak q' \in \text{WeakAss}_{S'}(M' \otimes _{R'} N') \\ \mathfrak p \in \text{WeakAss}_ R(M) & \Leftrightarrow \mathfrak p' \in \text{WeakAss}_{R'}(M') \\ \overline{\mathfrak q} \in \text{Ass}_{\overline{S}}(\overline{N}) & \Leftrightarrow \overline{\mathfrak q}' \in \text{WeakAss}_{\overline{S}'}(\overline{N}') \end{align*}

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

38.13.2.1
$$\label{flat-equation-key-observation} \text{if }\mathfrak r' \in \text{Ass}_{\overline{S}'}(\overline{N}') \text{ lies over }\mathfrak r \subset \overline{S}\text{ then } \mathfrak r \subset \overline{\mathfrak q}$$

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

$z' \in (M' \otimes _{R'} N')_{\mathfrak p'} = M'_{\mathfrak p'} \otimes _{R'_{\mathfrak p'}} N'_{\mathfrak p'}$

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

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

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

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

4. $\overline{\mathfrak q}$ 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,

1. $R$ 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

$\Psi : N/IN \longrightarrow (N/IN)^{\oplus n}, \quad z \longmapsto (g_1z, \ldots , g_ nz).$

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

$x \in \text{WeakAss}_ X(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}) \Leftrightarrow s \in \text{WeakAss}_ S(\mathcal{G}) \text{ and } x \in \text{Ass}_{X_ s}(\mathcal{F}_ s).$

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

$i_*(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}) = i_*\mathcal{F} \otimes _{\mathcal{O}_{\mathbf{A}^ n_ S}} p^*\mathcal{G}$

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

$\text{WeakAss}_ S(M \otimes _ R N) = \bigcup \nolimits _{\mathfrak p \in \text{WeakAss}_ R(M)} \text{Ass}_{S \otimes _ R \kappa (\mathfrak p)}(N \otimes _ R \kappa (\mathfrak p))$

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

$\text{WeakAss}_ X(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}) = \bigcup \nolimits _{s \in \text{WeakAss}_ S(\mathcal{G})} \text{Ass}_{X_ s}(\mathcal{F}_ s)$

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

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

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

3. 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

$\text{WeakAss}_ X(\mathcal{F}') \subset \bigcup \nolimits _{s \in \text{WeakAss}(S)} \text{Ass}_{X_ s}(\mathcal{F}'_ s)$

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

$U = \{ \mathfrak q \subset S \mid M_{\mathfrak q}\text{ flat over }R\}$

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.

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

2. 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. $\square$

Remark 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

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

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

3. 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 37.52.1). If we ever need this result in the Stacks project we will convert this remark into a lemma with a proof.

Remark 38.13.12 (Algebra version of Remark 38.13.11). 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

$U = \{ \mathfrak q \subset S \mid M_{\mathfrak q}\text{ flat over }R\}$

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 15.79.2). If we ever need this result in the Stacks project we will convert this remark into a lemma with a proof.

Comment #1511 by jojo on

This page (the whole web page not just the math) is broken for me, basically there is math where the left menu should be and the menu is not at the right place.

I feel maybe it's the last remark that's causing problem (at least this last remark isn't rendering properly).

Comment #1512 by on

Thanks for telling us about it! I added it to the list of issues and I hope to get to another round of bug fixing soon. I think I know what's causing the problem and it should be easily fixable.

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