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In SGA 1, chapter 4, there is a corollary describing a flatness condition for a module of finite type over a local noetherian integral ring. The corollary is number 4.4, and states:

Suppose that $A$ is a local noetherian integral ring with maximal ideal $I$ and residue field $k = A/I$ and field of fractions $K$. Let $M$ be a module of finite type over $A$. Then saying that $M$ is flat (note: SGA actually cites here a previous proposition with equivalent conditions of flatness which in this case is satisfied) is equivalent to saying that $M\otimes_A K$ and $M\otimes_A k$ are vector spaces of the same dimension.

There is a remark immediately following this corollary saying the reader is left to generalize this to the case where $A$ is only assumed to be a ring without nilpotent elements. How does one show this more general case?

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Your title seems to be incomplete. – Mariano Suárez-Alvarez Feb 18 '12 at 23:56
Ah, fixed--sorry about that :) – Lucy Feb 19 '12 at 5:15
Now there seems to be something more than what you intended! :D – Mariano Suárez-Alvarez Feb 19 '12 at 5:26
Hahaha, fail >.< – Lucy Feb 19 '12 at 5:55
up vote 6 down vote accepted

I assume you are referring to page 78 here. I think the author is only implying that you may drop the condition of $A$ being an integral domain (in favor of it being a ring without nilpotent elements). You should still assume that $A$ is noetherian and local with maximal ideal $I$ (since in the formulation of the corollary left to the reader, the author still refers to the field $k=A/I$, which would be ambiguous otherwise; also there are conditions on $A$ and $I$ in place throughout the entire section which prohibit you from working with an arbitrary ring).

So you can apply the criterion for flatness given in Corollary 4.4 to the $A/{\mathfrak p}$-module $M/{\mathfrak p} M$ for each minimal prime $\mathfrak p$ of $A$ (since $A/\mathfrak p$ will be an integral domain again). What you want to show then is that $M$ is flat as an $A$-module if and only if $M/{\mathfrak p}M$ is flat as an $A/{\mathfrak p}$-module for each minimal prime $\mathfrak p$. By the foregoing Corollary 4.3, this is the same as showing that $M$ is free if and only if $M/\mathfrak p M$ is free for any minimal prime $\mathfrak p$ (however, the "and only if"-part is trivial).

Here is how I think the proof goes: Since $A$ is local, we may lift a $k$-basis of $k\otimes M$ to a minimal $A$-generating set of $M$ (by Nakayama's lemma). We get an exact sequence $$ A^i \stackrel{\varphi}{\longrightarrow} A^{dim_k k\otimes M} \longrightarrow M \longrightarrow 0 $$ for some $i\in\mathbb Z_{\geq 0}$ and some $\varphi\in A^{dim_kk\otimes M \times i}$. If $\varphi$ is zero then $M$ is free. So assume $\varphi$ is not zero and $x$ is some non-zero entry of the matrix. Because $A$ has no nilpotent elements there is some minimal prime $\mathfrak p$ such that $x \notin \mathfrak p$. Now, by tensoring $$ (A/\mathfrak p)^i \stackrel{\varphi}{\longrightarrow} (A/\mathfrak p)^{dim_k k\otimes M} \longrightarrow M/\mathfrak p M \longrightarrow 0 $$ with the fraction field $K$ of $A/\mathfrak p$, we see that $dim_K K\otimes (M/\mathfrak p M) < dim_k k \otimes M= dim_k k\otimes (M/\mathfrak p M)$, i.e. $M/\mathfrak p M$ is not flat/free/projective by Corollary 4.4 (i.e., we have shown: $M$ not free $\Longrightarrow$ $M/\mathfrak p M$ not free for some minimal prime $\mathfrak p$).

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For a generalization to arbitrary noetherian reduced rings, see EGA III, 7.6.9. Applied to a projective resolution $P_*\to M$ for a finitely generated $A$-module $M$, it says in particular that $d(x)\colon x\mapsto\dim_{k(x)}(M\otimes k(x))$ is semi-continuous on $\mathrm{Spec}(A)$, and $M$ is flat iff $d$ is locally constant. The statement in SGA is a special case since semi-continuity together with $\dim_K(M\otimes K)=\dim_k(M\otimes k)$ already implies that $d$ is constant. A more elementary statement can be found in Mumford's treatment of semi-continuity in his Abelian Varieties (Lemma 1 in section II.5).

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