# Ideal of definition of local ring

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.

• You are not able to use the fact that the Jacobson radical of an Artinian ring is nilpotent? That's what the "first part" amounts to. Commented Mar 20, 2023 at 11:40
• thanks a lot for hint Commented Mar 20, 2023 at 12:06
• I've edited your post to include some more MathJax as well as to resolve some typos and grammatical issues. Please take a look at the edit and consider making some of these improvements in your posts going forward. Commented Mar 20, 2023 at 16:34

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.

• Thanks for replying. can you please give hint for first part also Commented Mar 20, 2023 at 11:02
• @ANarode: it is rather elementary. I added a couple of sentences. Commented Mar 20, 2023 at 17:12