The Stacks project

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

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

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

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. $\square$

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

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

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 $I \otimes _ R M \to M$. $\square$

Proof. Part (1) is a special case of part (2) with $M_ i = M$ for all $i$ and $f_{i i'} = \text{id}_ M$. Proof of (2). 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 $\mathfrak a = \mathop{\mathrm{colim}}\nolimits _{i' \geq i} \mathfrak a'R_{i'}$. Since $\otimes $ commutes with colimits the map $\mathfrak a \otimes _ R M \to M$ is the colimit of the maps

These maps are all injective by assumption. Since colimits over $I$ are exact by Lemma 10.8.8 we win. $\square$

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 $-\otimes _ R M$. $\square$

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. $\square$

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

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 $R$. $\square$

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

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. $\square$

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

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

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

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

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. $\square$

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 $N_1 \to N_2$. $\square$

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 implications (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. $\square$

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. $\square$

Proof. Immediate from Lemma 10.39.16. $\square$

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). $\square$

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. $\square$

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. $\square$