MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


In Mumford's "Red Book of Varieties and Schemes", Chapter III, Paragraph 10 (entitled "Flat and smooth morphisms"), the following property is stated:

Let $M$ be a $B$-module, and $B$ an algebra over $A$. Let $f\in B$ have the property that for all maximal ideals $m \subset A$, multiplication by $f$ is injective in $M / m \cdot M$. Then $M$ flat over $A$ implies $M / f \cdot M$ flat over $A$.

My question is whether this statement is true as stated, without some finiteness assumptions (it is not clear, for example, what is the role of $B$ here if no finiteness is assumed), and, secondly, can someone indicate a proof, or an exact reference for a proof.

Thank you, Sasha

share|cite|improve this question
A reference (with finiteness condition) is Matsumura's "Commutative Ring Theory", Theorem 22.6. – Hailong Dao Oct 5 '11 at 21:46
Can you state this Theorem here? – Martin Brandenburg Oct 5 '11 at 23:57
up vote 4 down vote accepted

This is false without finiteness conditions: let $k$ be a field, $A=k[X,Y]$, $B=A_{(X)}$, $M=B$, $f=X$.

share|cite|improve this answer
Doesn't $x$ act like $0$ on $M/(x,y)M$? – Mattia Talpo Oct 5 '11 at 16:52
It does. However, $M/(x,y)M=0$. – user2035 Oct 5 '11 at 18:00
Isn't $M/xM = 0$? Zero module is flat! – Sasha Oct 5 '11 at 19:22
No, $M/xM\cong k(Y)$. – user2035 Oct 5 '11 at 19:44
Sorry, I mixed up $A_{(x)}$ and $A_x$. – Sasha Oct 5 '11 at 20:06

As Hailong Dao comments above, Matsumura's "Commutative Ring Theory", Theorem 22.6, implies the theorem in the body of the question for the case when $A$,$B$ are Noetherian, and $M$ is of finite type over $B$ (maybe almost what is written in the body of the question, since there one checks not for maximal ideals $m \subset A$ but for preimages of maximal ideals of $B$, which might be only prime).

Actually, there is also a version for the case when $B$ is of finite presentation over $A$, and $M$ is a $B$-module of finite presentation (which follows from the Noetherian case by the trick "finitely presented morphism is base change of morphism between Noetherian things". One can find it in Grothendieck's EGA IV, part 3, proposition 11.3.7 (in geometric formulation) (I do not understand the organization into chapters of EGA, so I hope I gave a correct reference). Thus there is a correct statement:

Let $i: A \to B$ be of finite presentation, $u: M \to N$ a morphism between two finitely-presented $B$-modules. Suppose that $N$ is flat over $A$. Then the following are equivalent:

a) $u$ is injective and $Coker(u)$ is flat over $A$.

b) For every $q \in Spec(B)$, the morphism $id\otimes u : k(p)\otimes_A M \to k(p)\otimes_A N$ is injective, where $p = i^{-1} (q)$ and $k(p)$ is the fraction field of $A / p$.

share|cite|improve this answer
The existing part of EGA consists of five chapters: 0, I, II, III, IV, distributed over the volumes as [0-1 + I], [II], [0-2 + III-1], [III-2], [0-3 + IV-1], [IV-2], [IV-3], [IV-4] where x-y is the y-th part of chapter x. The Springer re-edition is again [0-1 + I]. – user2035 Oct 7 '11 at 12:05
Thanks for the clarification! – Sasha Oct 7 '11 at 13:00

The purpose of this post is to analyze the situation while ignoring $B$ altogether. Fix a maximal ideal $m$ of $A$. By tensoring the exact sequence $0 \to M \to M \to M/fM \to 0$ with $A/m$ one deduces immediately that

$$Tor_i^A(M/fM, A/m) =0 $$ for $i>0$.

Flatness can be checked locally on maximal ideals, so we may as well let $R=A_m, k = R/mR$ and considering:

If $Tor_i^R(N,k) =0$ for $i>0$, when is $N$ flat?

In general, $N$ may not be flat, as a-fortiori's example show. However, it will be flat in many situations. The finiteness assumptions in EGA suffice. You can also check out the paper "A local flatness criterion ..." by Hans Schoutens (available on his website). For example, there he showed that $N$ will be flat if we assume it has projective dimension at most one. So if we know $M$ is projective to begin with, then $M/fM$ will be flat.

share|cite|improve this answer
$f$ is not known to be $M$-regular a priori, but we have $\mathrm{Tor}_1^A(M/fM,A/m)=0$ nevertheless. – user2035 Oct 7 '11 at 14:43
Dear a-fortiori: you are right, I forgot that $f$ is not necessarily $M$-regular. I will leave this answer since the Schoutens' paper might still be relevant. – Hailong Dao Oct 8 '11 at 21:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.