TITLE: Quotient of flat module is flat - a property in Mumford's Red book
QUESTION [8 upvotes]: Hi,
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

REPLY [2 votes]: 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.