If a maximal ideal $\mathfrak m$ is flat, then $\dim_k (\mathfrak m/\mathfrak m^2) \leq 1$ This is an exercise that bothers me a lot:

Let $R$ be a commutative ring with $1$. Let $\mathfrak{m}$ be a maximal ideal in $R$.
  If $\mathfrak m$ is flat as an $R$-module then the vector space dimension $\dim_k(\mathfrak{m}/\mathfrak{m}^2) \leq 1$ (where $k= R/\mathfrak{m}$).

I've tried to work with the dual space of $\mathfrak{m}/\mathfrak{m}^2$ and with the identity $\mathfrak{m}/\mathfrak{m}^2 = R/\mathfrak{m} \otimes_R \mathfrak{m}$. I've tried to show that for $\dim_k(\mathfrak{m}/\mathfrak{m}^2) > 1$ it is $0= Hom_R(\mathfrak{m}, R/\mathfrak{m}) \cong \mathfrak{m}/\mathfrak{m}^2$. Which would be a contradiction. However, I'm not even sure this is true.
Also, until now I think I got a solution for finitely generated $\mathfrak{m}$. It doesn't really make me happy though: after localizing at $\mathfrak{m}$ we can assume that $(R, \mathfrak{m})$ is local. For finitely generated modules over local rings it holds flat $\Rightarrow$ free (Matsumura) and then the $\mathfrak{m}$ is a free $R$-module of dimension $1$, therefore also $\dim_k(\mathfrak{m}/\mathfrak{m}^2) \leq 1$. However: this only works for finitely generated $\mathfrak{m}$ and also we haven't had the Theorem of Matsumura in the lecture so far.
My Question is: are there any ideas how to do this? I'm not even looking for a full solution, sadly I'm out of creative ideas. (Note that we haven't introduced TOR in our lecture yet)
 A: 
If $(A,m)$ is a local ring and $M$ is a flat $A$-module, then any family of elements $x_1,\dots,x_n\in M$ such that their images in $M/mM$ are linearly independent over $A/m$ is also linearly independent over $A$. (Matsumura, Commutative Ring Theory, Theorem 7.10.)

If $M$ is an ideal of $A$, then there is no pair $x,y$ of elements in $M$ linearly independent over $A$ ($xy+y(-x)=0$), so $\dim_{A/m}M/mM\le1$. 
A: Here is some intuition for the statement, at least:
Imagine for a moment that $R$ were actually a local domain, and that $\mathfrak m$
were finitely generated.  Then if $\mathfrak m$ is flat, it is free, and so has a rank.  But when we extend scalars to the fraction field $F$ of $R$, the inclusion
$\mathfrak m \subset R$ becomes an equality, and so this free rank must be one.
Thus $\mathfrak m/\mathfrak m^2$ would then be at most one-dimensional.
(I know that you already said that you could prove the result when $\mathfrak m$ is f.g. using that f.g. flat $\implies$ free, so maybe this is useless; but it is
the intuitive picture that came to mind when I read the question, which I would try to build on to prove the general result.)
