Faithful module cancellation with maximal ideal 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.
 A: 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 $d<n$.
