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?



flatness condition for local noetherian ring without nilpotent elements
I assume you are referring to [page 78 here](http://arxiv.org/pdf/math/0206203v2.pdf). 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 $\mathfrak p$).


