Local criterion for flatness explained

In algebra, the local criterion for flatness gives conditions one can check to show flatness of a module.

Statement

Given a commutative ring A, an ideal I and an A-module M, suppose either

• A is a Noetherian ring and M is idealwise separated for I: for every ideal

ak{a} , c ap _{n} I ^{n(ak{a} ⊗} M) = 0 (for example, this is the case when A is a Noetherian local ring, I its maximal ideal and M finitely generated),or

• I is nilpotent.

Then the following are equivalent:

The assumption that “A is a Noetherian ring” is used to invoke the Artin–Rees lemma and can be weakened; see

Proof

Following SGA 1, Exposé IV, we first prove a few lemmas, which are interesting themselves. (See also this blog post by Akhil Mathew for a proof of a special case.)

Proof: The equivalence of the first two can be seen by studying the Tor spectral sequence. Here is a direct proof: if 1. is valid and

N \hookrightarrow N' is an injection of B -modules with cokernel C, then, as A-modules,\operatorname{Tor} ^{A} _{1 (C,} M) = 0 \to N ⊗ _{A} M \to N' ⊗ _{A} M .Since N ⊗ _{A} M \simeq N ⊗ _{B} (B ⊗ _{A} N) and the same for N' , this proves 2. Conversely, considering 0 \to R \to F \to X \to 0 where F is B-free, we get:\operatorname{Tor} ^{A} _{1 (F,} M) = 0 \to \operatorname{Tor} ^{A} _{1 (X,} M) \to R ⊗ _{A} M \to F ⊗ _{A} M .Here, the last map is injective by flatness and that gives us 1. To see the "Moreover" part, if 1. is valid, then \operatorname{Tor} ^{A(I} _{1} ^{n} X/I ^{n+1} X, M) = 0 and so\operatorname{Tor} ^{A(I} _{1} ^{n+1} X, M) \to \operatorname{Tor} ^{A(I} _{1} ^{n} X, M) \to 0 . By descending induction, this implies 3. The converse is trivial. \square

Proof: The assumption implies that

I ^{n} ⊗ M = I ^{n} M and so, since tensor product commutes with base extension,\operatorname{gr} _{I(A)} ⊗ _{A _{0}} M _{0} = ⊕ ^{inf ty} _{0} (I ^{n)} _{0} ⊗ _{A _{0}} M _{0} = ⊕ ^{inf ty} _{0} (I ^{n} ⊗ _{A} M) _{0} = ⊕ ^{inf ty} _{0} (I ^{n} M) _{0} = \operatorname{gr} _{I} M .

For the second part, let

\alpha _{i} denote the exact sequence 0 \to \operatorname{Tor} ^{A(A/I} _{1} ^{i,} M) \to I ^{i} ⊗ M \to I ^{i} M \to 0 and \gamma _{i:} 0 \to 0 \to I ^{i/I} ^{i+1} ⊗ M \overset{\simeq} \to I ^{i} M/I ^{i+1} M \to 0 . Consider the exact sequence of complexes:\alpha _{i+1} \to \alpha _{i} \to \gamma _{i.} Then \operatorname{Tor} ^{A(A/I} _{1} ^{i,} M) = 0 , i > 0 (it is so for large i and then use descending induction). 3. of Lemma 1 then implies that M is flat. \square

Proof of the main statement.

2 . ⇒ 1 .

If

I is nilpotent, then, by Lemma 1, \operatorname{Tor} ^{A( - ,} _{1} M) = 0 and M is flat over A . Thus, assume that the first assumption is valid. Let ak{a} \subset A be an ideal and we shall show ak{a} ⊗ M \to M is injective. For an integer k > 0 , consider the exact sequence0 \to ak{a}/(I ^{k} \cap ak{a}) \to A/I ^{k} \to A/( ak{a} + I ^{k)} \to 0 . Since \operatorname{Tor} ^{A(A/( ak{a} +} _{1} I ^{k),} M) = 0 by Lemma 1 (note I ^{k} kills A/( ak{a} + I ^{k)} ), tensoring the above with M , we get:0 \to ak{a}/(I ^{k} \cap ak{a}) ⊗ M \to A/I ^{k} ⊗ M = M/I ^{k} M .Tensoring M with 0 \to I ^{k} \cap ak{a} \to ak{a} \to ak{a}/(I ^{k} \cap ak{a}) \to 0 , we also have:(I ^{k} \cap ak{a}) ⊗ M \overset{f}\to ak{a} ⊗ M \overset{g}\to ak{a}/(I ^{k} \cap ak{a}) ⊗ M \to 0 . We combine the two to get the exact sequence:(I ^{k} \cap ak{a}) ⊗ M \overset{f}\to ak{a} ⊗ M \overset{g'}\to M/I ^{k} M.

Now, if

x is in the kernel of ak{a} ⊗ M \to M , then, a fortiori, x is in \operatorname{ker}(g') = \operatorname{im}(f) = (I ^{k} \cap ak{a}) ⊗ M . By the Artin–Rees lemma, given n > 0 , we can find k > 0 such that I ^{k} \cap ak{a} \subset I ^{n} ak{a} . Since \cap _{n} I ^{n(ak{a} ⊗} M) = 0 , we conclude x = 0 .1 . ⇒ 4 . follows from Lemma 2.4 . ⇒ 3 .

Since

(A _{n)} _{0} = A _{0} , the condition 4. is still valid with M, A replaced by M _{n,} A _{n} . Then Lemma 2 says that M _{n} is flat over A _{n} .3 . ⇒ 2 . Tensoring 0 \to I \to A \to A/I \to 0 with M, we see \operatorname{Tor} ^{A(A/I,} _{1} M) is the kernel of I ⊗ M \to M . Thus, the implication is established by an argument similar to that of 2 . ⇒ 1 . \square

Application: characterization of an étale morphism

The local criterion can be used to prove the following:

Proof: Assume that

\widehat{ l{O} _{y,} } \to \widehat is an isomorphism and we show f is étale. First, since l{O} _{x} \to \widehat{ l{O} _{x}} is faithfully flat (in particular is a pure subring), we have:ak{m} _{y} l{O} _{x} = ak{m} _{y} \widehat{ l{O} _{x}} \cap l{O} _{x} = \widehat{ ak{m} _{y}} \widehat{ l{O} _{x}} \cap l{O} _{x} = \widehat{ ak{m} _{x}} \cap l{O} _{x} = ak{m} _{x} .Hence, f is unramified (separability is trivial). Now, that l{O} _{y} \to l{O} _{x} is flat follows from (1) the assumption that the induced map on completion is flat and (2) the fact that flatness descends under faithfully flat base change (it shouldn’t be hard to make sense of (2)).

Next, we show the converse: by the local criterion, for each n, the natural map

ak{m} ^{n +1} _{y} \to ak{m} ^{n +1} _{x} is an isomorphism. By induction and the five lemma, this implies l{O} _{y/ak{m}} ^{n} ^{y} \to l{O} _{x/ak{m}} ^{n} ^{x} is an isomorphism for each n. Passing to limit, we get the asserted isomorphism. \square

Mumford’s Red Book gives an extrinsic proof of the above fact (Ch. III, § 5, Theorem 3).

Miracle flatness theorem

B. Conrad calls the next theorem the miracle flatness theorem.^[1]

References

• • Exposé IV of
• Fujiwara . K. . Gabber . O. . Kato . F. . On Hausdorff completions of commutative rings in rigid geometry. . Journal of Algebra . 322 . 2011 . 293—321.

External links

• blog post by Akhil Mathew

Notes and References

1. Problem 10 in http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf