Let $k$ be a field of characteristic $0$ and $R = k[[x\_1, \dotsc, x\_n]]$. Suppose that $M$ is a faithful, finitely generated $R$-module and $\mathfrak{a} < R$ is an ideal such that $\mathfrak{a} M = \mathfrak{m} M$, where $\mathfrak{m} = (x\_1, \dotsc, x\_n)$. Is it true that $\mathfrak{a} = \mathfrak{m}$?


I know that $\sqrt{a} = \mathfrak{m}$ and the conclusion is false if $\mathfrak{m}$ is not maximal. For example $(x\_1^2, x\_2^2)(x\_1, x\_2) = (x\_1^2, x\_1 x\_2, x\_2^2) (x\_1, x\_2)$ in the ring $k[[x\_1, x\_2]]$. However, I have not been able to find a counter-example with $\mathfrak{m}$ maximal.



Faithful module cancellation with maximal ideal
Consider first the case where $\mathfrak{a}\subseteq \mathfrak{m}^2$. In this case we have $\mathfrak{a}\cdot M\subseteq \mathfrak{m}^2\cdot M\subseteq \mathfrak{m}\cdot M=\mathfrak{a}\cdot M$. In particulal, we have an equality $\mathfrak{m}\cdot (\mathfrak{m}\cdot M) = \mathfrak{m}\cdot M $. By Nakayama, we know that this implies that $\mathfrak{m}\cdot M=0$, contradicting the fact that the action is faithful.


This means that $\mathfrak{a}\nsubseteq \mathfrak{m}^2$. Write $d=dim\_k(\mathfrak{(a + m^2)}/\mathfrak{m^2})$. If $d=n$ we are done, so assume that this is not the case. Without loss of generality we can assume that $\mathfrak{a} = (x\_{n-d+1},\ldots, x\_n) + (\mathfrak{a\cap m^2})$. Write $I=(x\_{n-d+1},\ldots, x\_n)$, and consider the ring $R\_1:=R/I$, the module $M\_1:=M/IM$, and the image $\mathfrak{a\_1}$ of $\mathfrak{a}$ in $R\_1$. It holds that $\mathfrak{a}\_1\cdot M\_1 = \mathfrak{m}\_1\cdot M\_1$, and also that $\mathfrak{a\_1}\subseteq \mathfrak{m}\_1^2$, where $\mathfrak{m}\_1$ is the maximal ideal of $R\_1$. We conclude that $\mathfrak{m}\_1\cdot M\_1=0$, again using Nakayama. Going back to submodules of $M$, this means that $\mathfrak{m}\cdot M \subseteq IM$. Using the faithfulness of $M$ and Proposition 2.4. in Atiyah-Macdonald (with $\phi=$ multiplication by $x\_i$, $i\leq n-d$), we see that some power of $x\_i$ is contained in $I$. But this contradicts the assumption $d<n$.


