Show that, if $\dim_{R/m}(m/m^2) = 1$, then there are no ideals $I$ of $R$ such that $m^{k + 1} \subsetneq I \subsetneq m^k$ .

Question

Let $R$ be a noetherian local ring and let $m$ be its maximal ideal.

I have already proved that $m^{k}/m^{k+1}$ is an $R/m$-vector space and that if $a_1, \ldots, a_n \in m$ generate $m/m^2$ as a vector space, then $a_1, \ldots, a_n$ generate $m$ as an ideal. Thus, by Nakayama's lemma $\dim_{R/m}(m/m^2)$ is the minimal number of generators of $m$. Moreover, if $\dim_{R/m}(m/m^2) = 1$, it is clear that $m$ is a principal ideal, since it is generated by one element.

However, it is not clear to me how to show that there are no ideals $I$ of $R$ such that $m^{k + 1} \subsetneq I \subsetneq m^k$ for any $k \in \mathbb{N}$.

Does somebody has an idea ?

Thanks for your help.

Answer 1

Suppose $m/m^2$ has dimension $1$, generated by $a+m^2$. Then for all $k\ge1$, $m^k/m^{k+1}$ is generated by $a^k+m^{k+1}$ as an $R/m$-vector space. So $\dim_{R/m}(m^k/m^{k+1})\le1$. If $I$ were an ideal strictly between $m^k$ and $m^{k+1}$ then $I/m^{k+1}$ would be a proper nonzero $R/m$-vector subspace of $m^k/m^{k+1}$: there aren't any.