# 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)

• Hmm, good question: I can only do what you've already said. Where does the question come from? Was it assigned in your course? – Pete L. Clark May 1 '13 at 2:23
• @Louis: If you look at the proof in Matsumura, the conclusion is : if $M$ is any flat $R$-module ($R$ local ring), then any family of elements $x_1, \dots, x_n\in M$ which are linearly independent in $M/\mathfrak mM$ are actually $R$-linearly independent. Now if $M$ is any ideal of $R$, there is no pair $f,g$ of $R$-linearly independent elements in $M$ ($g.f+(-f).g=0$, so $M/\mathfrak mM$ has dimension $\le 1$. – user18119 May 1 '13 at 5:22
• Yes, it was assigned in my Algebra I lecture course. However, later in the exercise we will use this statement to construct a torsionfree module that is not flat by a maximal ideal of $K[X_1, ..., X_n]$. So the finitely generated case already satisfied me a bit. – Louis May 1 '13 at 9:01
• @QiL'8: sadly I don't have the book of Matsumura. I see how from what you said the solution follows, but do you maybe have a link to the proof of the theorem (i.e. how does $M$ flat imply that a set of $x_1,..., x_n$ lineary independent in $M/\mathfrak{m}$ is linearly independent in $M$. Replacing flat by projective, this seems more likely for me to be done on my own, but for flat, I don't see an idea right now. (if I already know the theorem holds, I could probably just look at the submodule generated by $x_1,..., x_n$ and apply the theorem for this submodule). – Louis May 1 '13 at 9:23
• @Louis: sorry I don't have a link. Try to find the book in your library, the proof I alluded to is in Prop. 3G, p. 21. Projectivity is not needed. And yes, a submodule of a flat module is not necessarily flat. But the proof doesn't rely on this assumption. – user18119 May 1 '13 at 16:28

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.)
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 $\dim_{A/m}M/mM\le1$.