Definition. Let $R$ be a Noetherian local ring with maximal ideal $\mathfrak{m}$. An ideal $I$ is called an ideal of definition of $R$ if $\mathfrak{m}^n \subset I \subset \mathfrak{m}$ for some $n\geq1$.
Prove that $I$ is an ideal of definition of $R$ if and only if $R/I$ is an artinian ring.
I got stuck while proving this theorem. Suppose $R/I$ is an artinian ring, then consider descending chain $\mathfrak{m}/I\supseteq \mathfrak{m}^2/I \supseteq \mathfrak{m}^3/I \dots $. Since $R/I$ is an artinian ring, this chain will terminate, i.e there exists $n$ such that $\mathfrak{m}^n/I=\mathfrak{m}^{n+1}/I=\mathfrak{m}^{n+2}/I=\dots$. From this, how can I conclude $ \mathfrak{m}^n\subset I\subset \mathfrak{m}^{n+1} $?
Also for converse part suppose $\mathfrak{m}^n \subset I \subset \mathfrak{m}$ for some $n\geq1$, then I was trying to prove $R/I$ is Noetherian and the dimension of $R/I$ is zero, but I could not prove that the dimension of $R/I$ is zero.
Please give some hints tom prove these. Thanks in advance.
In order to prove that $R/I$ has dimension $0$ you need to prove that there is no prime ideal $\frak p\neq\frak m$ such that $$ {\frak m}^n\subset I\subset\frak p\subset \frak m. \qquad (\ast) $$ That's because "dimension $0$" means that there are no proper inclusions between prime ideals (i.e. every prime ideal is maximal) and the fact that pulling back under the natural quotient map $R\rightarrow R/I$ sets up a bijection between the ideals of $R/I$ and the ideals of $R$ containing $I$, a bijection that preserves prime ideals.
Said that the argument is rather simple. Suppose there's a prime $\frak p$ as in $(\ast)$ and let $x\in\frak m\setminus\frak p$. Then $x^n\in\frak m^n$ contradicts the definition of $\frak p$ being prime.