The Stacks project

10.100 Flatness criteria over Artinian rings

We discuss some flatness criteria for modules over Artinian rings. Note that an Artinian local ring has a nilpotent maximal ideal so that the following two lemmas apply to Artinian local rings.

Lemma 10.100.1. Let $(R, \mathfrak m)$ be a local ring with nilpotent maximal ideal $\mathfrak m$. Let $M$ be a flat $R$-module. If $A$ is a set and $x_\alpha \in M$, $\alpha \in A$ is a collection of elements of $M$, then the following are equivalent:

  1. $\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for the vector space $M/\mathfrak mM$ over $R/\mathfrak m$, and

  2. $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M$ over $R$.

Proof. The implication (2) $\Rightarrow $ (1) is immediate. Assume (1). By Nakayama's Lemma 10.19.1 the elements $x_\alpha $ generate $M$. Then one gets a short exact sequence

\[ 0 \to K \to \bigoplus \nolimits _{\alpha \in A} R \to M \to 0 \]

Tensoring with $R/\mathfrak m$ and using Lemma 10.38.12 we obtain $K/\mathfrak mK = 0$. By Nakayama's Lemma 10.19.1 we conclude $K = 0$. $\square$

Lemma 10.100.2. Let $R$ be a local ring with nilpotent maximal ideal. Let $M$ be an $R$-module. The following are equivalent

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

  2. $M$ is a free $R$-module, and

  3. $M$ is a projective $R$-module.

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 10.100.1 the $x_\alpha $ are a basis for $M$ over $R$ and we win. $\square$

Lemma 10.100.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be a collection of elements of $M$. Assume

  1. $I$ is nilpotent,

  2. $\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ over $R/I$, and

  3. $\text{Tor}_1^ R(R/I, M) = 0$.

Then $M$ is free on $\{ x_\alpha \} _{\alpha \in A}$ over $R$.

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 10.19.1 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

\[ 0 \to I \otimes _ R M \to M \to M/IM \to 0. \]

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. $\square$

Lemma 10.100.4. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

  1. $M/IM$ is flat over $R/I$, and

  2. $R' \otimes _ R M$ is flat over $R'$.

Set $I_2 = \varphi ^{-1}(\varphi (I^2)R')$. Then $M/I_2M$ is flat over $R/I_2$.

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 10.98.8 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

\[ I' \otimes _{R'} M' = I' \otimes _ R M = I' \otimes _{R/I} M/IM. \]

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

\[ I \otimes _{R/I} M/IM \longrightarrow I' \otimes _{R/I} M/IM \]

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

Lemma 10.100.5. Let $\varphi : R \to R'$ be a ring map. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

  1. $I$ is nilpotent,

  2. $R \to R'$ is injective,

  3. $M/IM$ is flat over $R/I$, and

  4. $R' \otimes _ R M$ is flat over $R'$.

Then $M$ is flat over $R$.

Proof. Define inductively $I_1 = I$ and $I_{n + 1} = \varphi ^{-1}(\varphi (I_ n)^2R')$ for $n \geq 1$. Note that by Lemma 10.100.4 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. $\square$

Here is the local Artinian version of the local criterion for flatness.

Lemma 10.100.6. Let $R$ be an Artinian local ring. Let $M$ be an $R$-module. Let $I \subset R$ be a proper ideal. The following are equivalent

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

  2. $M/IM$ is flat over $R/I$ and $\text{Tor}_1^ R(R/I, M) = 0$.

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 10.100.2 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 10.100.3 we conclude that $M$ is free and in particular flat. $\square$

It turns out that flatness descends along injective homomorphism whose source is an Artinian ring.

Lemma 10.100.7. Let $R \to S$ be a ring map. Let $M$ be an $R$-module. Assume

  1. $R$ is Artinian

  2. $R \to S$ is injective, and

  3. $M \otimes _ R S$ is a flat $S$-module.

Then $M$ is a flat $R$-module.

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 10.52. Hence $M$ is flat by an application of Lemma 10.100.5.

Second proof: By Lemma 10.52.6 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 10.19.1 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 10.100.3 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 $R$. $\square$

Please compare the lemma below to Lemma 10.98.15 (the case of Noetherian local rings), Lemma 10.127.8 (the case of finitely presented algebras), and Lemma 10.127.10 (the case of locally nilpotent ideals).

Lemma 10.100.8 (Critère de platitude par fibres: Nilpotent case). Let

\[ \xymatrix{ S \ar[rr] & & S' \\ & R \ar[lu] \ar[ru] } \]

be a commutative diagram in the category of rings. Let $I \subset R$ be a nilpotent ideal and $M$ an $S'$-module. Assume

  1. The module $M/IM$ is a flat $S/IS$-module.

  2. The module $M$ is a flat $R$-module.

Then $M$ is a flat $S$-module and $S_{\mathfrak q}$ is flat over $R$ for every $\mathfrak q \subset S$ such that $M \otimes _ S \kappa (\mathfrak q)$ is nonzero.

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

\[ 0 \to I \otimes _ R M \to M \to M/IM \to 0. \]

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

\[ I \otimes _ R M \to IS \otimes _ S M \to M \]

are injective. Hence $\text{Tor}_1^ S(IS, M) = 0$ (see Remark 10.74.9) and we conclude that $M$ is a flat $S$-module by Lemma 10.98.8. 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 10.38.15 and 10.38.10. $\square$


Comments (2)

Comment #2551 by Stella Gastineau on

I think you can remove the condition that is flat from Lemma 10.100.1 for a much cleaner proof: If is any -module and if forms a basis on , then let be the submodule of generated by the . Then the composition is surjective and so ie . Then by nilpotency of , we have that and so the form a generating set for . (These are the standard techniques for the different forms of Nakayama's lemma, but we can remove the requirement that is finitely generated because we have for all -modules when is local Artinian. You still need to use Axiom of choice to find such a , but that is used anyways in Lemma 10.100.2.)

Comment #2552 by on

Dear Stella, Lemma 10.100.1 is not true when the module isn't flat. Please read carefully. Everybody: if you have a comment about a specific lemma, then please leave the comment on the tag page of the lemma.


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