Local criterion for flatness#Miracle flatness theorem

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 $\backslash mathfrak\{a\}$, $\backslash bigcap\_\{n\; \backslash ge\; 1\}\; I^n(\backslash mathfrak\{a\}\; \backslash otimes\; M)\; =\; 0$ (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:{{harvnb|Matsumura|1989|loc=Theorem 22.3.}}

{{Ordered list

|M is a flat module.

|$M/I$ is flat over $A/I$ and $\backslash operatorname\{Tor\}\_1^A(A/I,\; M)\; =\; 0$.

|For each $n\; \backslash ge\; 0$, $M\_n\; =\; M/I^\{n+1\}\; M$ is flat over $A\_n\; =\; A/I^\{n+1\}$.

|In the notations of 3., $M\_0$ is $A\_0$-flat and the natural $\backslash operatorname\{gr\}\_I\; A$-module surjection

:$\backslash operatorname\{gr\}\_I\; A\; \backslash otimes\_\{A\_0\}\; M\_0\; \backslash to\; \backslash operatorname\{gr\}\_I\; M$

is an isomorphism; i.e., each $I^n/I^\{n+1\}\; \backslash otimes\_\{A\_0\}\; M\_0\; \backslash to\; I^nM\; /I^\{n+1\}\; M$ is an isomorphism.}}

The assumption that “A is a Noetherian ring” is used to invoke the Artin–Rees lemma and can be weakened; see {{harvnb|Fujiwara|Gabber|Kato|2011|loc=Proposition 2.2.1.}}

Following SGA 1, Exposé IV, we first prove a few lemmas, which are interesting themselves. (See also this [https://amathew.wordpress.com/2009/12/31/flatness-and-the-local-and-infinitesimal-criteria/ blog post] by Akhil Mathew for a proof of a special case.)

{{math_theorem|name=Lemma 1|Given a ring homomorphism $A\; \backslash to\; B$ and an $A$-module $M$, the following are equivalent.

1. For every $B$-module $X$, $\backslash operatorname\{Tor\}\_1^A(X,\; M)\; =\; 0.$
2. $M\; \backslash otimes\_A\; B$ is $B$-flat and $\backslash operatorname\{Tor\}\_1^A(B,\; M)\; =\; 0.$

Moreover, if $B\; =\; A/I$, the above two are equivalent to

{{ordered list|start=3|$\backslash operatorname\{Tor\}\_1^A(X,\; M)\; =\; 0$ for every $A$-module $X$ killed by some power of $I$.}}

}}

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 $N\; \backslash hookrightarrow\; N\text{'}$ is an injection of $B$-modules with cokernel C, then, as A-modules,

:$\backslash operatorname\{Tor\}^A\_1(C,\; M)\; =\; 0\; \backslash to\; N\; \backslash otimes\_A\; M\; \backslash to\; N\text{'}\; \backslash otimes\_A\; M$.

Since $N\; \backslash otimes\_A\; M\; \backslash simeq\; N\; \backslash otimes\_B\; (B\; \backslash otimes\_A\; N)$ and the same for $N\text{'}$, this proves 2. Conversely, considering $0\; \backslash to\; R\; \backslash to\; F\; \backslash to\; X\; \backslash to\; 0$ where F is B-free, we get:

:$\backslash operatorname\{Tor\}^A\_1(F,\; M)\; =\; 0\; \backslash to\; \backslash operatorname\{Tor\}^A\_1(X,\; M)\; \backslash to\; R\; \backslash otimes\_A\; M\; \backslash to\; F\; \backslash otimes\_A\; M$.

Here, the last map is injective by flatness and that gives us 1. To see the "Moreover" part, if 1. is valid, then $\backslash operatorname\{Tor\}\_1^A(I^n\; X/I^\{n+1\}X,\; M)\; =\; 0$ and so

:$\backslash operatorname\{Tor\}\_1^A(I^\{n+1\}\; X,\; M)\; \backslash to\; \backslash operatorname\{Tor\}\_1^A(I^n\; X,\; M)\; \backslash to\; 0.$

By descending induction, this implies 3. The converse is trivial. $\backslash square$

{{math_theorem|name=Lemma 2|Let $A$ be a ring and $M$ a module over it. If $\backslash operatorname\{Tor\}\_1^A(A/I^n,\; M)\; =\; 0$ for every $n\; >\; 0$, then the natural grade-preserving surjection

:$\backslash operatorname\{gr\}\_I(A)\; \backslash otimes\_\{A\_0\}\; M\_0\; \backslash to\; \backslash operatorname\{gr\}\_I\; M$

is an isomorphism. Moreover, when I is nilpotent,

:$M$ is flat if and only if $M\_0$ is flat over $A\_0$ and $\backslash operatorname\{gr\}\_I\; A\; \backslash otimes\_\{A\_0\}\; M\_0\; \backslash to\; \backslash operatorname\{gr\}\_I\; M$ is an isomorphism.}}

Proof: The assumption implies that $I^n\; \backslash otimes\; M\; =\; I^n\; M$ and so, since tensor product commutes with base extension,

:$\backslash operatorname\{gr\}\_I(A)\; \backslash otimes\_\{A\_0\}\; M\_0\; =\; \backslash oplus\_0^\{\backslash infty\}\; (I^n)\_0\; \backslash otimes\_\{A\_0\}\; M\_0\; =\; \backslash oplus\_0^\{\backslash infty\}\; (I^n\; \backslash otimes\_A\; M)\_0\; =\; \backslash oplus\_0^\{\backslash infty\}\; (I^n\; M)\_0\; =\; \backslash operatorname\{gr\}\_I\; M$.

For the second part, let $\backslash alpha\_i$ denote the exact sequence $0\; \backslash to\; \backslash operatorname\{Tor\}\_1^A(A/I^i,\; M)\; \backslash to\; I^i\; \backslash otimes\; M\; \backslash to\; I^i\; M\; \backslash to\; 0$ and $\backslash gamma\_i:\; 0\; \backslash to\; 0\; \backslash to\; I^i/I^\{i+1\}\; \backslash otimes\; M\; \backslash overset\{\backslash simeq\}\; \backslash to\; I^i\; M/I^\{i+1\}\; M\; \backslash to\; 0$. Consider the exact sequence of complexes:

:$\backslash alpha\_\{i+1\}\; \backslash to\; \backslash alpha\_i\; \backslash to\; \backslash gamma\_i.$

Then $\backslash operatorname\{Tor\}\_1^A(A/I^i,\; M)\; =\; 0,\; i\; >\; 0$ (it is so for large $i$ and then use descending induction). 3. of Lemma 1 then implies that $M$ is flat. $\backslash square$

Proof of the main statement.

The proof uses the cycle of implications $1.\backslash Rightarrow\; 4.\backslash Rightarrow\; 3.\backslash Rightarrow\; 2.\backslash Rightarrow\; 1.$

$2.\; \backslash Rightarrow\; 1.$: If $I$ is nilpotent, then, by Lemma 1, $\backslash operatorname\{Tor\}\_1^A(-,\; M)\; =\; 0$ and $M$ is flat over $A$. Thus, assume that the first assumption is valid. Let $\backslash mathfrak\{a\}\; \backslash subset\; A$ be an ideal and we shall show $\backslash mathfrak\{a\}\; \backslash otimes\; M\; \backslash to\; M$ is injective. For an integer $k\; >\; 0$, consider the exact sequence

:$0\; \backslash to\; \backslash mathfrak\{a\}/(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash to\; A/I^k\; \backslash to\; A/(\backslash mathfrak\{a\}\; +\; I^k)\; \backslash to\; 0.$

Since $\backslash operatorname\{Tor\}\_1^A(A/(\backslash mathfrak\{a\}\; +\; I^k),\; M)\; =\; 0$ by Lemma 1 (note $I^k$ kills $A/(\backslash mathfrak\{a\}\; +\; I^k)$), tensoring the above with $M$, we get:

:$0\; \backslash to\; \backslash mathfrak\{a\}/(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash otimes\; M\; \backslash to\; A/I^k\; \backslash otimes\; M\; =\; M/I^k\; M$.

Tensoring $M$ with $0\; \backslash to\; I^k\; \backslash cap\; \backslash mathfrak\{a\}\; \backslash to\; \backslash mathfrak\{a\}\; \backslash to\; \backslash mathfrak\{a\}/(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash to\; 0$, we also have:

:$(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash otimes\; M\; \backslash overset\{f\}\backslash to\; \backslash mathfrak\{a\}\; \backslash otimes\; M\; \backslash overset\{g\}\backslash to\; \backslash mathfrak\{a\}/(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash otimes\; M\; \backslash to\; 0.$

We combine the two to get the exact sequence:

:$(I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash otimes\; M\; \backslash overset\{f\}\backslash to\; \backslash mathfrak\{a\}\; \backslash otimes\; M\; \backslash overset\{g\text{'}\}\backslash to\; M/I^k\; M.$

Now, if $x$ is in the kernel of $\backslash mathfrak\{a\}\; \backslash otimes\; M\; \backslash to\; M$, then, a fortiori, $x$ is in $\backslash operatorname\{ker\}(g\text{'})\; =\; \backslash operatorname\{im\}(f)\; =\; (I^k\; \backslash cap\; \backslash mathfrak\{a\})\; \backslash otimes\; M$. By the Artin–Rees lemma, given $n\; >\; 0$, we can find $k\; >\; 0$ such that $I^k\; \backslash cap\; \backslash mathfrak\{a\}\; \backslash subset\; I^n\; \backslash mathfrak\{a\}$. Since $\backslash cap\_\{n\; \backslash ge\; 1\}\; I^n(\backslash mathfrak\{a\}\; \backslash otimes\; M)\; =\; 0$, we conclude $x\; =\; 0$.

$1.\; \backslash Rightarrow\; 4.$ follows from Lemma 2.

$4.\; \backslash Rightarrow\; 3.$: 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 $A\_n$.

$3.\; \backslash Rightarrow\; 2.$ Tensoring $0\; \backslash to\; I\; \backslash to\; A\; \backslash to\; A/I\; \backslash to\; 0$ with M, we see $\backslash operatorname\{Tor\}\_1^A(A/I,\; M)$ is the kernel of $I\; \backslash otimes\; M\; \backslash to\; M$. Thus, the implication is established by an argument similar to that of $2.\; \backslash Rightarrow\; 1.$$\backslash square$

The local criterion can be used to prove the following:

{{math_theorem|name=Proposition|math_statement=Given a morphism $f:\; X\; \backslash to\; Y$ of finite type between Noetherian schemes, $f$ is étale (flat and unramified) if and only if for each x in X, f is an analytically local isomorphism near x; i.e., with $y\; =\; f(x)$, $\backslash widehat\{f^*\_x\}\; :\; \backslash widehat\{\backslash mathcal\{O\}\_\{y,\; Y\}\}\; \backslash to\; \backslash widehat\{\backslash mathcal\{O\}\_\{x,\; X\}\}$ is an isomorphism.}}

Proof: Assume that $\backslash widehat\{\backslash mathcal\{O\}\_\{y,\; Y\}\}\; \backslash to\; \backslash widehat\{\backslash mathcal\{O\}\_\{x,\; X\}\}$ is an isomorphism and we show f is étale. First, since $\backslash mathcal\{O\}\_x\; \backslash to\; \backslash widehat\{\backslash mathcal\{O\}\_x\}$ is faithfully flat (in particular is a pure subring), we have:

:$\backslash mathfrak\{m\}\_y\; \backslash mathcal\{O\}\_x\; =\; \backslash mathfrak\{m\}\_y\; \backslash widehat\{\backslash mathcal\{O\}\_x\}\; \backslash cap\; \backslash mathcal\{O\}\_x\; =\; \backslash widehat\{\backslash mathfrak\{m\}\_y\}\; \backslash widehat\{\backslash mathcal\{O\}\_x\}\; \backslash cap\; \backslash mathcal\{O\}\_x\; =\; \backslash widehat\{\backslash mathfrak\{m\}\_x\}\; \backslash cap\; \backslash mathcal\{O\}\_x\; =\; \backslash mathfrak\{m\}\_x$.

Hence, $f$ is unramified (separability is trivial). Now, that $\backslash mathcal\{O\}\_y\; \backslash to\; \backslash mathcal\{O\}\_x$ 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 $\backslash mathfrak\{m\}\_y^n/\backslash mathfrak\{m\}\_y^\{n+1\}\; \backslash to\; \backslash mathfrak\{m\}\_x^n/\backslash mathfrak\{m\}\_x^\{n+1\}$

is an isomorphism. By induction and the five lemma, this implies $\backslash mathcal\{O\}\_y/\backslash mathfrak\{m\}\_y^n\; \backslash to\; \backslash mathcal\{O\}\_x/\backslash mathfrak\{m\}\_x^n$ is an isomorphism for each n. Passing to limit, we get the asserted isomorphism. $\backslash square$

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

B. Conrad calls the next theorem the miracle flatness theorem.Problem 10 in http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf

{{math theorem|Let $R\; \backslash to\; S$ be a local ring homomorphism between local Noetherian rings. If S is flat over R, then

:$\backslash dim\; S\; =\; \backslash dim\; R\; +\; \backslash dim\; S/\backslash mathfrak\{m\}\_R\; S$.

Conversely, if this dimension equality holds, if S is regular and if R is Cohen–Macaulay (e.g., regular), then R is flat over S.}}

{{reflist}}

• {{Citation | last1=Matsumura | first1=Hideyuki | title=Commutative ring theory | publisher=Cambridge University Press | edition=2nd | series=Cambridge Studies in Advanced Mathematics | isbn=978-0-521-36764-6 |mr=1011461 | year=1989 | volume=8}}
• Exposé IV of {{Citation | last1=Grothendieck | first1=Alexander | author1-link=Alexander Grothendieck | last2=Raynaud | first2=Michèle | title=Revêtements étales et groupe fondamental (SGA 1) | orig-date=1971 | arxiv=math/0206203 | publisher=Société Mathématique de France | location=Paris | series=Documents Mathématiques (Paris) [Mathematical Documents (Paris)] | isbn=978-2-85629-141-2 | mr=2017446 | year=2003 | volume=3| bibcode=2002math......6203G }}
• {{cite journal |last1=Fujiwara |first1=K. |last2=Gabber |first2=O. |last3=Kato |first3=F. |title=On Hausdorff completions of commutative rings in rigid geometry. |journal= Journal of Algebra |issue=322 |year=2011 |volume=332 |pages=293–321|doi=10.1016/j.jalgebra.2011.02.001 }}

• [https://amathew.wordpress.com/2009/12/31/flatness-and-the-local-and-infinitesimal-criteria/ blog post] by Akhil Mathew

Category:Commutative algebra

Category:Theorems in abstract algebra