The Stacks project

Proof. The implication (2) $\Rightarrow $ (1) is immediate. Assume (1). By Nakayama's Lemma 10.20.1 the elements $x_\alpha $ generate $M$. Then one gets a short exact sequence

Tensoring with $R/\mathfrak m$ and using Lemma 10.39.12 we obtain $K/\mathfrak mK = 0$. By Nakayama's Lemma 10.20.1 we conclude $K = 0$. $\square$

Proof. Since any projective module is flat (as a direct summand of a free module) and every free module is projective, it suffices to prove that a flat module is free. Let $M$ be a flat module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be elements such that $\overline{x_\alpha } \in M/\mathfrak m M$ forms a basis over the residue field of $R$. By Lemma 10.101.1 the $x_\alpha $ are a basis for $M$ over $R$ and we win. $\square$

Proof. Let $R$, $I$, $M$, $\{ x_\alpha \} _{\alpha \in A}$ be as in the lemma and satisfy assumptions (1), (2), and (3). By Nakayama's Lemma 10.20.1 the elements $x_\alpha $ generate $M$ over $R$. The assumption $\text{Tor}_1^ R(R/I, M) = 0$ implies that we have a short exact sequence

Let $\sum f_\alpha x_\alpha = 0$ be a relation in $M$. By choice of $x_\alpha $ we see that $f_\alpha \in I$. Hence we conclude that $\sum f_\alpha \otimes x_\alpha = 0$ in $I \otimes _ R M$. The map $I \otimes _ R M \to I/I^2 \otimes _{R/I} M/IM$ and the fact that $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ implies that $f_\alpha \in I^2$! Hence we conclude that there are no relations among the images of the $x_\alpha $ in $M/I^2M$. In other words, we see that $M/I^2M$ is free with basis the images of the $x_\alpha $. Using the map $I \otimes _ R M \to I/I^3 \otimes _{R/I^2} M/I^2M$ we then conclude that $f_\alpha \in I^3$! And so on. Since $I^ n = 0$ for some $n$ by assumption (1) we win. $\square$

Proof. We may replace $R$, $M$, and $R'$ by $R/I_2$, $M/I_2M$, and $R'/\varphi (I)^2R'$. Then $I^2 = 0$ and $\varphi $ is injective. By Lemma 10.99.8 and the fact that $I^2 = 0$ it suffices to prove that $\text{Tor}^ R_1(R/I, M) = K = \mathop{\mathrm{Ker}}(I \otimes _ R M \to M)$ is zero. Set $M' = M \otimes _ R R'$ and $I' = IR'$. By assumption the map $I' \otimes _{R'} M' \to M'$ is injective. Hence $K$ maps to zero in

Then $I \to I'$ is an injective map of $R/I$-modules. Since $M/IM$ is flat over $R/I$ the map

is injective. This implies that $K$ is zero in $I \otimes _ R M = I \otimes _{R/I} M/IM$ as desired. $\square$

Proof. Define inductively $I_1 = I$ and $I_{n + 1} = \varphi ^{-1}(\varphi (I_ n)^2R')$ for $n \geq 1$. Note that by Lemma 10.101.4 we find that $M/I_ nM$ is flat over $R/I_ n$ for each $n \geq 1$. It is clear that $\varphi (I_ n) \subset \varphi (I)^{2^ n}R'$. Since $I$ is nilpotent we see that $\varphi (I_ n) = 0$ for some $n$. As $\varphi $ is injective we conclude that $I_ n = 0$ for some $n$ and we win. $\square$

Proof. The implication (1) $\Rightarrow $ (2) follows immediately from the definitions. Assume $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$. By Lemma 10.101.2 this implies that $M/IM$ is free over $R/I$. Pick a set $A$ and elements $x_\alpha \in M$ such that the images in $M/IM$ form a basis. By Lemma 10.101.3 we conclude that $M$ is free and in particular flat. $\square$

Proof. First proof: Let $I \subset R$ be the Jacobson radical of $R$. Then $I$ is nilpotent and $M/IM$ is flat over $R/I$ as $R/I$ is a product of fields, see Section 10.53. Hence $M$ is flat by an application of Lemma 10.101.5.

Second proof: By Lemma 10.53.6 we may write $R = \prod R_ i$ as a finite product of local Artinian rings. This induces similar product decompositions for both $R$ and $S$. Hence we reduce to the case where $R$ is local Artinian (details omitted).

Assume that $R \to S$, $M$ are as in the lemma satisfying (1), (2), and (3) and in addition that $R$ is local with maximal ideal $\mathfrak m$. Let $A$ be a set and $x_\alpha \in A$ be elements such that $\overline{x}_\alpha $ forms a basis for $M/\mathfrak mM$ over $R/\mathfrak m$. By Nakayama's Lemma 10.20.1 we see that the elements $x_\alpha $ generate $M$ as an $R$-module. Set $N = S \otimes _ R M$ and $I = \mathfrak mS$. Then $\{ 1 \otimes x_\alpha \} _{\alpha \in A}$ is a family of elements of $N$ which form a basis for $N/IN$. Moreover, since $N$ is flat over $S$ we have $\text{Tor}_1^ S(S/I, N) = 0$. Thus we conclude from Lemma 10.101.3 that $N$ is free on $\{ 1 \otimes x_\alpha \} _{\alpha \in A}$. The injectivity of $R \to S$ then guarantees that there cannot be a nontrivial relation among the $x_\alpha $ with coefficients in $R$. $\square$

Proof. As $M$ is flat over $R$ tensoring with the short exact sequence $0 \to I \to R \to R/I \to 0$ gives a short exact sequence

Note that $I \otimes _ R M \to IS \otimes _ S M$ is surjective. Combined with the above this means both maps in

are injective. Hence $\text{Tor}_1^ S(IS, M) = 0$ (see Remark 10.75.9) and we conclude that $M$ is a flat $S$-module by Lemma 10.99.8. To finish we need to show that $S_{\mathfrak q}$ is flat over $R$ for any prime $\mathfrak q \subset S$ such that $M \otimes _ S \kappa (\mathfrak q)$ is nonzero. This follows from Lemma 10.39.15 and 10.39.10. $\square$