Are the powers of the maximal ideal in the local Noetherian ring of formal power series in $n$ variables free modules

Question

Let $\mathbb C[[t_1, \dots, t_n]]$ be the ring of formal power series in $n$ indeterminates. Let $\mathfrak{m}=(t_1, ..., t_n)$ be the maximal ideal in $\mathbb C[[t_1, ..., t_n]]$. Is $\mathfrak{m}^p$ a free module? By definition, $\mathfrak{m}^p=\{\sum_{i_1+\dotsi_n=p}a_{i_1\dots i_n}t_1^{i_1}\cdots t_n^{i_n}: a_{i_1\dots i_n}\in\mathbb C[[t_1, ..., t_n]]\}$, but is it true that this is isomorphic to the direct sum $\bigoplus_{i_1+...+i_n=p}\mathbb C[[t_1, ..., t_n]]t_1^{i_1}\cdots t_n^{i_n}$? I think so as, the generators $t_1^{i_1}\cdots t_n^{i_n}$ are linearly independent over $\mathbb C[[t_1, ..., t_n]]$.

Answer 1

The answer is no, except in the trivial case $n=1$. The reason why the direct sum decomposition does not hold is the generating monomials are not linearly independent over $\mathbb{C}[[t_1,\dots,t_n]]$. For example, $t_1$ and $t_2$ are linearly dependent because $$t_2\cdot t_1 - t_1\cdot t_2=0.$$ More generally, any two elements $f,g\in\mathfrak{m}^p$ are linearly dependent for the same reason, which proves that the ideal is not a free module (in the $n\geq 2$ case).

In fact, an ideal is a free module if and only if it is principal, see If ideal $I$ of domain $R$ is free $R$-module, then $I$ is principal ideal..