## 10.39 Flat modules and flat ring maps

One often used result is that if $M = \mathop{\mathrm{colim}}\nolimits _{i\in \mathcal{I}} M_ i$ is a colimit of $R$-modules and if $N$ is an $R$-module then

$M \otimes N = \mathop{\mathrm{colim}}\nolimits _{i\in \mathcal{I}} M_ i \otimes _ R N,$

see Lemma 10.12.9. This property is usually expressed by saying that $\otimes$ commutes with colimits. Another often used result is that if $0 \to N_1 \to N_2 \to N_3 \to 0$ is an exact sequence and if $M$ is any $R$-module, then

$M \otimes _ R N_1 \to M \otimes _ R N_2 \to M \otimes _ R N_3 \to 0$

is still exact, see Lemma 10.12.10. Both of these properties tell us that the functor $N \mapsto M \otimes _ R N$ is right exact. See Categories, Section 4.23 and Homology, Section 12.7. An $R$-module $M$ is flat if $N \mapsto N \otimes _ R M$ is also left exact, i.e., if it is exact. Here is the precise definition.

Definition 10.39.1. Let $R$ be a ring.

1. An $R$-module $M$ is called flat if whenever $N_1 \to N_2 \to N_3$ is an exact sequence of $R$-modules the sequence $M \otimes _ R N_1 \to M \otimes _ R N_2 \to M \otimes _ R N_3$ is exact as well.

2. An $R$-module $M$ is called faithfully flat if the complex of $R$-modules $N_1 \to N_2 \to N_3$ is exact if and only if the sequence $M \otimes _ R N_1 \to M \otimes _ R N_2 \to M \otimes _ R N_3$ is exact.

3. A ring map $R \to S$ is called flat if $S$ is flat as an $R$-module.

4. A ring map $R \to S$ is called faithfully flat if $S$ is faithfully flat as an $R$-module.

Here is an example of how you can use the flatness condition.

Lemma 10.39.2. Let $R$ be a ring. Let $I, J \subset R$ be ideals. Let $M$ be a flat $R$-module. Then $IM \cap JM = (I \cap J)M$.

Proof. Consider the exact sequence $0 \to I \cap J \to R \to R/I \oplus R/J$. Tensoring with the flat module $M$ we obtain an exact sequence

$0 \to (I \cap J) \otimes _ R M \to M \to M/IM \oplus M/JM$

Since the kernel of $M \to M/IM \oplus M/JM$ is equal to $IM \cap JM$ we conclude. $\square$

Lemma 10.39.3. Let $R$ be a ring. Let $\{ M_ i, \varphi _{ii'}\}$ be a directed system of flat $R$-modules. Then $\mathop{\mathrm{colim}}\nolimits _ i M_ i$ is a flat $R$-module.

Proof. This follows as $\otimes$ commutes with colimits and because directed colimits are exact, see Lemma 10.8.8. $\square$

Lemma 10.39.4. A composition of (faithfully) flat ring maps is (faithfully) flat. If $R \to R'$ is (faithfully) flat, and $M'$ is a (faithfully) flat $R'$-module, then $M'$ is a (faithfully) flat $R$-module.

Proof. The first statement of the lemma is a particular case of the second, so it is clearly enough to prove the latter. Let $R \to R'$ be a flat ring map, and $M'$ a flat $R'$-module. We need to prove that $M'$ is a flat $R$-module. Let $N_1 \to N_2 \to N_3$ be an exact complex of $R$-modules. Then, the complex $R' \otimes _ R N_1 \to R' \otimes _ R N_2 \to R' \otimes _ R N_3$ is exact (since $R'$ is flat as an $R$-module), and so the complex $M' \otimes _{R'} \left(R' \otimes _ R N_1\right) \to M' \otimes _{R'} \left(R' \otimes _ R N_2\right) \to M' \otimes _{R'} \left(R' \otimes _ R N_3\right)$ is exact (since $M'$ is a flat $R'$-module). Since $M' \otimes _{R'} \left(R' \otimes _ R N\right) \cong \left(M' \otimes _{R'} R'\right) \otimes _ R N \cong M' \otimes _ R N$ for any $R$-module $N$ functorially (by Lemmas 10.12.7 and 10.12.3), this complex is isomorphic to the complex $M' \otimes _ R N_1 \to M' \otimes _ R N_2 \to M' \otimes _ R N_3$, which is therefore also exact. This shows that $M'$ is a flat $R$-module. Tracing this argument backwards, we can show that if $R \to R'$ is faithfully flat, and if $M'$ is faithfully flat as an $R'$-module, then $M'$ is faithfully flat as an $R$-module. $\square$

Lemma 10.39.5. Let $M$ be an $R$-module. The following are equivalent:

1. $M$ is flat over $R$.

2. for every injection of $R$-modules $N \subset N'$ the map $N \otimes _ R M \to N'\otimes _ R M$ is injective.

3. for every ideal $I \subset R$ the map $I \otimes _ R M \to R \otimes _ R M = M$ is injective.

4. for every finitely generated ideal $I \subset R$ the map $I \otimes _ R M \to R \otimes _ R M = M$ is injective.

Proof. The implications (1) implies (2) implies (3) implies (4) are all trivial. Thus we prove (4) implies (1). Suppose that $N_1 \to N_2 \to N_3$ is exact. Let $K = \mathop{\mathrm{Ker}}(N_2 \to N_3)$ and $Q = \mathop{\mathrm{Im}}(N_2 \to N_3)$. Then we get maps

$N_1 \otimes _ R M \to K \otimes _ R M \to N_2 \otimes _ R M \to Q \otimes _ R M \to N_3 \otimes _ R M$

Observe that the first and third arrows are surjective. Thus if we show that the second and fourth arrows are injective, then we are done1. Hence it suffices to show that $- \otimes _ R M$ transforms injective $R$-module maps into injective $R$-module maps.

Assume $K \to N$ is an injective $R$-module map and let $x \in \mathop{\mathrm{Ker}}(K \otimes _ R M \to N \otimes _ R M)$. We have to show that $x$ is zero. The $R$-module $K$ is the union of its finite $R$-submodules; hence, $K \otimes _ R M$ is the colimit of $R$-modules of the form $K_ i \otimes _ R M$ where $K_ i$ runs over all finite $R$-submodules of $K$ (because tensor product commutes with colimits). Thus, for some $i$ our $x$ comes from an element $x_ i \in K_ i \otimes _ R M$. Thus we may assume that $K$ is a finite $R$-module. Assume this. We regard the injection $K \to N$ as an inclusion, so that $K \subset N$.

The $R$-module $N$ is the union of its finite $R$-submodules that contain $K$. Hence, $N \otimes _ R M$ is the colimit of $R$-modules of the form $N_ i \otimes _ R M$ where $N_ i$ runs over all finite $R$-submodules of $N$ that contain $K$ (again since tensor product commutes with colimits). Notice that this is a colimit over a directed system (since the sum of two finite submodules of $N$ is again finite). Hence, (by Lemma 10.8.4) the element $x \in K \otimes _ R M$ maps to zero in at least one of these $R$-modules $N_ i \otimes _ R M$ (since $x$ maps to zero in $N \otimes _ R M$). Thus we may assume $N$ is a finite $R$-module.

Assume $N$ is a finite $R$-module. Write $N = R^{\oplus n}/L$ and $K = L'/L$ for some $L \subset L' \subset R^{\oplus n}$. For any $R$-submodule $G \subset R^{\oplus n}$, we have a canonical map $G \otimes _ R M \to M^{\oplus n}$ obtained by composing $G \otimes _ R M \to R^ n \otimes _ R M = M^{\oplus n}$. It suffices to prove that $L \otimes _ R M \to M^{\oplus n}$ and $L' \otimes _ R M \to M^{\oplus n}$ are injective. Namely, if so, then we see that $K \otimes _ R M = L' \otimes _ R M/L \otimes _ R M \to M^{\oplus n}/L \otimes _ R M$ is injective too2.

Thus it suffices to show that $L \otimes _ R M \to M^{\oplus n}$ is injective when $L \subset R^{\oplus n}$ is an $R$-submodule. We do this by induction on $n$. The base case $n = 1$ we handle below. For the induction step assume $n > 1$ and set $L' = L \cap R \oplus 0^{\oplus n - 1}$. Then $L'' = L/L'$ is a submodule of $R^{\oplus n - 1}$. We obtain a diagram

$\xymatrix{ & L' \otimes _ R M \ar[r] \ar[d] & L \otimes _ R M \ar[r] \ar[d] & L'' \otimes _ R M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & M \ar[r] & M^{\oplus n} \ar[r] & M^{\oplus n - 1} \ar[r] & 0 }$

By induction hypothesis and the base case the left and right vertical arrows are injective. The rows are exact. It follows that the middle vertical arrow is injective too.

The base case of the induction above is when $L \subset R$ is an ideal. In other words, we have to show that $I \otimes _ R M \to M$ is injective for any ideal $I$ of $R$. We know this is true when $I$ is finitely generated. However, $I = \bigcup I_\alpha$ is the union of the finitely generated ideals $I_\alpha$ contained in it. In other words, $I = \mathop{\mathrm{colim}}\nolimits I_\alpha$. Since $\otimes$ commutes with colimits we see that $I \otimes _ R M = \mathop{\mathrm{colim}}\nolimits I_\alpha \otimes _ R M$ and since all the morphisms $I_\alpha \otimes _ R M \to M$ are injective by assumption, the same is true for $I \otimes _ R M \to M$. $\square$

Lemma 10.39.6. Let $\{ R_ i, \varphi _{ii'}\}$ be a system of rings over the directed set $I$. Let $R = \mathop{\mathrm{colim}}\nolimits _ i R_ i$. Let $M$ be an $R$-module such that $M$ is flat as an $R_ i$-module for all $i$. Then $M$ is flat as an $R$-module.

Proof. Let $\mathfrak a \subset R$ be a finitely generated ideal. By Lemma 10.39.5 it suffices to show that $\mathfrak a \otimes _ R M \to M$ is injective. We can find an $i \in I$ and a finitely generated ideal $\mathfrak a' \subset R_ i$ such that $\mathfrak a = \mathfrak a'R$. Then $\mathfrak a = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} \mathfrak a'R_{i'}$. Hence the map $\mathfrak a \otimes _ R M \to M$ is the colimit of the maps

$\mathfrak a'R_{i'} \otimes _{R_{i'}} M \longrightarrow M$

which are all injective by assumption. Since $\otimes$ commutes with colimits and since colimits over $I$ are exact by Lemma 10.8.8 we win. $\square$

Lemma 10.39.7. Suppose that $M$ is (faithfully) flat over $R$, and that $R \to R'$ is a ring map. Then $M \otimes _ R R'$ is (faithfully) flat over $R'$.

Proof. For any $R'$-module $N$ we have a canonical isomorphism $N \otimes _{R'} (R'\otimes _ R M) = N \otimes _ R M$. Hence the desired exactness properties of the functor $-\otimes _{R'}(R'\otimes _ R M)$ follow from the corresponding exactness properties of the functor $-\otimes _ R M$. $\square$

Lemma 10.39.8. Let $R \to R'$ be a faithfully flat ring map. Let $M$ be a module over $R$, and set $M' = R' \otimes _ R M$. Then $M$ is flat over $R$ if and only if $M'$ is flat over $R'$.

Proof. By Lemma 10.39.7 we see that if $M$ is flat then $M'$ is flat. For the converse, suppose that $M'$ is flat. Let $N_1 \to N_2 \to N_3$ be an exact sequence of $R$-modules. We want to show that $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3 \otimes _ R M$ is exact. We know that $N_1 \otimes _ R R' \to N_2 \otimes _ R R' \to N_3 \otimes _ R R'$ is exact, because $R \to R'$ is flat. Flatness of $M'$ implies that $N_1 \otimes _ R R' \otimes _{R'} M' \to N_2 \otimes _ R R' \otimes _{R'} M' \to N_3 \otimes _ R R' \otimes _{R'} M'$ is exact. We may write this as $N_1 \otimes _ R M \otimes _ R R' \to N_2 \otimes _ R M \otimes _ R R' \to N_3 \otimes _ R M \otimes _ R R'$. Finally, faithful flatness implies that $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3 \otimes _ R M$ is exact. $\square$

Lemma 10.39.9. Let $R$ be a ring. Let $S \to S'$ be a flat map of $R$-algebras. Let $M$ be a module over $S$, and set $M' = S' \otimes _ S M$.

1. If $M$ is flat over $R$, then $M'$ is flat over $R$.

2. If $S \to S'$ is faithfully flat, then $M$ is flat over $R$ if and only if $M'$ is flat over $R$.

Proof. Let $N \to N'$ be an injection of $R$-modules. By the flatness of $S \to S'$ we have

$\mathop{\mathrm{Ker}}(N \otimes _ R M \to N' \otimes _ R M) \otimes _ S S' = \mathop{\mathrm{Ker}}(N \otimes _ R M' \to N' \otimes _ R M')$

If $M$ is flat over $R$, then the left hand side is zero and we find that $M'$ is flat over $R$ by the second characterization of flatness in Lemma 10.39.5. If $M'$ is flat over $R$ then we have the vanishing of the right hand side and if in addition $S \to S'$ is faithfully flat, this implies that $\mathop{\mathrm{Ker}}(N \otimes _ R M \to N' \otimes _ R M)$ is zero which in turn shows that $M$ is flat over $R$. $\square$

Lemma 10.39.10. Let $R \to S$ be a ring map. Let $M$ be an $S$-module. If $M$ is flat as an $R$-module and faithfully flat as an $S$-module, then $R \to S$ is flat.

Proof. Let $N_1 \to N_2 \to N_3$ be an exact sequence of $R$-modules. By assumption $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3 \otimes _ R M$ is exact. We may write this as

$N_1 \otimes _ R S \otimes _ S M \to N_2 \otimes _ R S \otimes _ S M \to N_3 \otimes _ R S \otimes _ S M.$

By faithful flatness of $M$ over $S$ we conclude that $N_1 \otimes _ R S \to N_2 \otimes _ R S \to N_3 \otimes _ R S$ is exact. Hence $R \to S$ is flat. $\square$

Let $R$ be a ring. Let $M$ be an $R$-module. Let $\sum f_ i x_ i = 0$ be a relation in $M$. We say the relation $\sum f_ i x_ i$ is trivial if there exist an integer $m \geq 0$, elements $y_ j \in M$, $j = 1, \ldots , m$, and elements $a_{ij} \in R$, $i = 1, \ldots , n$, $j = 1, \ldots , m$ such that

$x_ i = \sum \nolimits _ j a_{ij} y_ j, \forall i, \quad \text{and}\quad 0 = \sum \nolimits _ i f_ ia_{ij}, \forall j.$

Proof. Assume $M$ is flat and let $\sum f_ i x_ i = 0$ be a relation in $M$. Let $I = (f_1, \ldots , f_ n)$, and let $K = \mathop{\mathrm{Ker}}(R^ n \to I, (a_1, \ldots , a_ n) \mapsto \sum _ i a_ i f_ i)$. So we have the short exact sequence $0 \to K \to R^ n \to I \to 0$. Then $\sum f_ i \otimes x_ i$ is an element of $I \otimes _ R M$ which maps to zero in $R \otimes _ R M = M$. By flatness $\sum f_ i \otimes x_ i$ is zero in $I \otimes _ R M$. Thus there exists an element of $K \otimes _ R M$ mapping to $\sum e_ i \otimes x_ i \in R^ n \otimes _ R M$ where $e_ i$ is the $i$th basis element of $R^ n$. Write this element as $\sum k_ j \otimes y_ j$ and then write the image of $k_ j$ in $R^ n$ as $\sum a_{ij} e_ i$ to get the result.

Assume every relation is trivial, let $I$ be a finitely generated ideal, and let $x = \sum f_ i \otimes x_ i$ be an element of $I \otimes _ R M$ mapping to zero in $R \otimes _ R M = M$. This just means exactly that $\sum f_ i x_ i$ is a relation in $M$. And the fact that it is trivial implies easily that $x$ is zero, because

$x = \sum f_ i \otimes x_ i = \sum f_ i \otimes \left(\sum a_{ij}y_ j\right) = \sum \left(\sum f_ i a_{ij}\right) \otimes y_ j = 0$
$\square$

Lemma 10.39.12. Suppose that $R$ is a ring, $0 \to M'' \to M' \to M \to 0$ a short exact sequence, and $N$ an $R$-module. If $M$ is flat then $N \otimes _ R M'' \to N \otimes _ R M'$ is injective, i.e., the sequence

$0 \to N \otimes _ R M'' \to N \otimes _ R M' \to N \otimes _ R M \to 0$

is a short exact sequence.

Proof. Let $R^{(I)} \to N$ be a surjection from a free module onto $N$ with kernel $K$. The result follows from the snake lemma applied to the following diagram

$\begin{matrix} & & 0 & & 0 & & 0 & & \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes _ R N & \to & M' \otimes _ R N & \to & M \otimes _ R N & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ 0 & \to & (M'')^{(I)} & \to & (M')^{(I)} & \to & M^{(I)} & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M''\otimes _ R K & \to & M' \otimes _ R K & \to & M \otimes _ R K & \to & 0 \\ & & & & & & \uparrow & & \\ & & & & & & 0 & & \end{matrix}$

with exact rows and columns. The middle row is exact because tensoring with the free module $R^{(I)}$ is exact. $\square$

Lemma 10.39.13. Suppose that $0 \to M' \to M \to M'' \to 0$ is a short exact sequence of $R$-modules. If $M'$ and $M''$ are flat so is $M$. If $M$ and $M''$ are flat so is $M'$.

Proof. We will use the criterion that a module $N$ is flat if for every ideal $I \subset R$ the map $N \otimes _ R I \to N$ is injective, see Lemma 10.39.5. Consider an ideal $I \subset R$. Consider the diagram

$\begin{matrix} 0 & \to & M' & \to & M & \to & M'' & \to & 0 \\ & & \uparrow & & \uparrow & & \uparrow & & \\ & & M'\otimes _ R I & \to & M \otimes _ R I & \to & M''\otimes _ R I & \to & 0 \end{matrix}$

with exact rows. This immediately proves the first assertion. The second follows because if $M''$ is flat then the lower left horizontal arrow is injective by Lemma 10.39.12. $\square$

Lemma 10.39.14. Let $R$ be a ring. Let $M$ be an $R$-module. The following are equivalent

1. $M$ is faithfully flat, and

2. $M$ is flat and for all $R$-module homomorphisms $\alpha : N \to N'$ we have $\alpha = 0$ if and only if $\alpha \otimes \text{id}_ M = 0$.

Proof. If $M$ is faithfully flat, then $0 \to \mathop{\mathrm{Ker}}(\alpha ) \to N \to N'$ is exact if and only if the same holds after tensoring with $M$. This proves (1) implies (2). For the other, assume (2). Let $N_1 \to N_2 \to N_3$ be a complex, and assume the complex $N_1 \otimes _ R M \to N_2 \otimes _ R M \to N_3\otimes _ R M$ is exact. Take $x \in \mathop{\mathrm{Ker}}(N_2 \to N_3)$, and consider the map $\alpha : R \to N_2/\mathop{\mathrm{Im}}(N_1)$, $r \mapsto rx + \mathop{\mathrm{Im}}(N_1)$. By the exactness of the complex $-\otimes _ R M$ we see that $\alpha \otimes \text{id}_ M$ is zero. By assumption we get that $\alpha$ is zero. Hence $x$ is in the image of $N_1 \to N_2$. $\square$

Lemma 10.39.15. Let $M$ be a flat $R$-module. The following are equivalent:

1. $M$ is faithfully flat,

2. for every nonzero $R$-module $N$, then tensor product $M \otimes _ R N$ is nonzero,

3. for all $\mathfrak p \in \mathop{\mathrm{Spec}}(R)$ the tensor product $M \otimes _ R \kappa (\mathfrak p)$ is nonzero, and

4. for all maximal ideals $\mathfrak m$ of $R$ the tensor product $M \otimes _ R \kappa (\mathfrak m) = M/{\mathfrak m}M$ is nonzero.

Proof. Assume $M$ faithfully flat and $N \not= 0$. By Lemma 10.39.14 the nonzero map $1 : N \to N$ induces a nonzero map $M \otimes _ R N \to M \otimes _ R N$, so $M \otimes _ R N \not= 0$. Thus (1) implies (2). The imlpications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) are immediate.

Assume (4). Suppose that $N_1 \to N_2 \to N_3$ is a complex and suppose that $N_1 \otimes _ R M \to N_2\otimes _ R M \to N_3\otimes _ R M$ is exact. Let $H$ be the cohomology of the complex, so $H = \mathop{\mathrm{Ker}}(N_2 \to N_3)/\mathop{\mathrm{Im}}(N_1 \to N_2)$. To finish the proof we will show $H = 0$. By flatness we see that $H \otimes _ R M = 0$. Take $x \in H$ and let $I = \{ f \in R \mid fx = 0 \}$ be its annihilator. Since $R/I \subset H$ we get $M/IM \subset H \otimes _ R M = 0$ by flatness of $M$. If $I \not= R$ we may choose a maximal ideal $I \subset \mathfrak m \subset R$. This immediately gives a contradiction. $\square$

Lemma 10.39.16. Let $R \to S$ be a flat ring map. The following are equivalent:

1. $R \to S$ is faithfully flat,

2. the induced map on $\mathop{\mathrm{Spec}}$ is surjective, and

3. any closed point $x \in \mathop{\mathrm{Spec}}(R)$ is in the image of the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$.

Proof. This follows quickly from Lemma 10.39.15, because we saw in Remark 10.17.8 that $\mathfrak p$ is in the image if and only if the ring $S \otimes _ R \kappa (\mathfrak p)$ is nonzero. $\square$

Lemma 10.39.17. A flat local ring homomorphism of local rings is faithfully flat.

Proof. Immediate from Lemma 10.39.16. $\square$

Flatness meshes well with localization.

Lemma 10.39.18. Let $R$ be a ring. Let $S \subset R$ be a multiplicative subset.

1. The localization $S^{-1}R$ is a flat $R$-algebra.

2. If $M$ is an $S^{-1}R$-module, then $M$ is a flat $R$-module if and only if $M$ is a flat $S^{-1}R$-module.

3. Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak p}$ is a flat $R_{\mathfrak p}$-module for all primes $\mathfrak p$ of $R$.

4. Suppose $M$ is an $R$-module. Then $M$ is a flat $R$-module if and only if $M_{\mathfrak m}$ is a flat $R_{\mathfrak m}$-module for all maximal ideals $\mathfrak m$ of $R$.

5. Suppose $R \to A$ is a ring map, $M$ is an $A$-module, and $g_1, \ldots , g_ m \in A$ are elements generating the unit ideal of $A$. Then $M$ is flat over $R$ if and only if each localization $M_{g_ i}$ is flat over $R$.

6. Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p$ the prime of $R$ lying under $\mathfrak q$) for all primes $\mathfrak q$ of $A$.

7. Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Then $M$ is a flat $R$-module if and only if the localization $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module (with $\mathfrak p = R \cap \mathfrak m$) for all maximal ideals $\mathfrak m$ of $A$.

Proof. Let us prove the last statement of the lemma. In the proof we will use repeatedly that localization is exact and commutes with tensor product, see Sections 10.9 and 10.12.

Suppose $R \to A$ is a ring map, and $M$ is an $A$-module. Assume that $M_{\mathfrak m}$ is a flat $R_{\mathfrak p}$-module for all maximal ideals $\mathfrak m$ of $A$ (with $\mathfrak p = R \cap \mathfrak m$). Let $I \subset R$ be an ideal. We have to show the map $I \otimes _ R M \to M$ is injective. We can think of this as a map of $A$-modules. By assumption the localization $(I \otimes _ R M)_{\mathfrak m} \to M_{\mathfrak m}$ is injective because $(I \otimes _ R M)_{\mathfrak m} = I_{\mathfrak p} \otimes _{R_{\mathfrak p}} M_{\mathfrak m}$. Hence the kernel of $I \otimes _ R M \to M$ is zero by Lemma 10.23.1. Hence $M$ is flat over $R$.

Conversely, assume $M$ is flat over $R$. Pick a prime $\mathfrak q$ of $A$ lying over the prime $\mathfrak p$ of $R$. Suppose that $I \subset R_{\mathfrak p}$ is an ideal. We have to show that $I \otimes _{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is injective. We can write $I = J_{\mathfrak p}$ for some ideal $J \subset R$. Then the map $I \otimes _{R_{\mathfrak p}} M_{\mathfrak q} \to M_{\mathfrak q}$ is just the localization (at $\mathfrak q$) of the map $J \otimes _ R M \to M$ which is injective. Since localization is exact we see that $M_{\mathfrak q}$ is a flat $R_{\mathfrak p}$-module.

This proves (7) and (6). The other statements follow in a straightforward way from the last statement (proofs omitted). $\square$

Lemma 10.39.19. Let $R \to S$ be flat. Let $\mathfrak p \subset \mathfrak p'$ be primes of $R$. Let $\mathfrak q' \subset S$ be a prime of $S$ mapping to $\mathfrak p'$. Then there exists a prime $\mathfrak q \subset \mathfrak q'$ mapping to $\mathfrak p$.

Proof. By Lemma 10.39.18 the local ring map $R_{\mathfrak p'} \to S_{\mathfrak q'}$ is flat. By Lemma 10.39.17 this local ring map is faithfully flat. By Lemma 10.39.16 there is a prime mapping to $\mathfrak p R_{\mathfrak p'}$. The inverse image of this prime in $S$ does the job. $\square$

The property of $R \to S$ described in the lemma is called the “going down property”. See Definition 10.41.1.

Lemma 10.39.20. Let $R$ be a ring. Let $\{ S_ i, \varphi _{ii'}\}$ be a directed system of faithfully flat $R$-algebras. Then $S = \mathop{\mathrm{colim}}\nolimits _ i S_ i$ is a faithfully flat $R$-algebra.

Proof. By Lemma 10.39.3 we see that $S$ is flat. Let $\mathfrak m \subset R$ be a maximal ideal. By Lemma 10.39.16 none of the rings $S_ i/\mathfrak m S_ i$ is zero. Hence $S/\mathfrak mS = \mathop{\mathrm{colim}}\nolimits S_ i/\mathfrak mS_ i$ is nonzero as well because $1$ is not equal to zero. Thus the image of $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ contains $\mathfrak m$ and we see that $R \to S$ is faithfully flat by Lemma 10.39.16. $\square$

[1] Here is the argument in more detail: Assume that we know that the second and fourth arrows are injective. Lemma 10.12.10 (applied to the exact sequence $K \to N_2 \to Q \to 0$) yields that the sequence $K \otimes _ R M \to N_2 \otimes _ R M \to Q \otimes _ R M \to 0$ is exact. Hence, $\mathop{\mathrm{Ker}}\left(N_2 \otimes _ R M \to Q \otimes _ R M\right) = \mathop{\mathrm{Im}}\left(K \otimes _ R M \to N_2 \otimes _ R M\right)$. Since $\mathop{\mathrm{Im}}\left(K \otimes _ R M \to N_2 \otimes _ R M\right) = \mathop{\mathrm{Im}}\left(N_1 \otimes _ R M \to N_2 \otimes _ R M\right)$ (due to the surjectivity of $N_1 \otimes _ R M \to K \otimes _ R M$) and $\mathop{\mathrm{Ker}}\left(N_2 \otimes _ R M \to Q \otimes _ R M\right) = \mathop{\mathrm{Ker}}\left(N_2 \otimes _ R M \to N_3 \otimes _ R M\right)$ (due to the injectivity of $Q \otimes _ R M \to N_3 \otimes _ R M$), this becomes $\mathop{\mathrm{Ker}}\left(N_2 \otimes _ R M \to N_3 \otimes _ R M\right) = \mathop{\mathrm{Im}}\left(N_1 \otimes _ R M \to N_2 \otimes _ R M\right)$, which shows that the functor $- \otimes _ R M$ is exact, whence $M$ is flat.
[2] This becomes obvious if we identify $L' \otimes _ R M$ and $L \otimes _ R M$ with submodules of $M^{\oplus n}$ (which is legitimate since the maps $L \otimes _ R M \to M^{\oplus n}$ and $L' \otimes _ R M \to M^{\oplus n}$ are injective and commute with the obvious map $L' \otimes _ R M \to L \otimes _ R M$).

Comment #391 by Fan on

The first sentence is a bit ambiguous: "and if $N$ is another then ..." Another what? $R$-module or colimit of $R$-modules? It seems that it refers to an $R$-module

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