Is $A / \mathfrak{m}$ flat if $A$ is a local ring?

Question

I'd like to prove the following: if $A$ is a local ring and $\mathfrak{m} \subset A$ its maximal ideal, then $A / \mathfrak{m}$ is a flat $A$-module. How can I do this? I've tried to find a suitable isomorphism for the tensor-product of $A / \mathfrak{m} \otimes_A N$ for an $A$-module $N$, - but I do not see where to go.

Thanks for any help!

Answer 1

This is never true unless $\mathfrak{m}=0$, i.e. unless $A$ is a field. Let $x\in\mathfrak{m}$, and consider the short exact sequence $0\to (x)\to A\to A/(x)\to 0$. If $A/\mathfrak{m}$ is flat, we can tensor it with the short exact sequence to get a short exact sequence $0\to (x)/\mathfrak{m}(x)\to A/\mathfrak{m}\to A/\mathfrak{m}\to 0$. Exactness of this sequence means that in fact $(x)/\mathfrak{m}(x)=0$, i.e. that $(x)=\mathfrak{m}(x)$. By Nakayama's lemma, this implies $(x)=0$, so $x=0$. Since $x\in\mathfrak{m}$ was arbitrary, this implies $\mathfrak{m}=0$.

Answer 2

Consider the example where $A=k[x]/(x^2)$, with $k$ a field. Does that work?

Answer 3

This is definitely false: $A/\mathfrak m$ is a finitely generated $A$-module, and if it were flat it would be free since finitely generated flat modules over local rings are free, and this is not true unless $\mathfrak m=0$, i. e. unless $A$ is a field.