# Noetherian local ring is Artinian iff maximal ideal is nilpotent

I am browsing through some old lecture notes, and I am trying to prove the following:

Let $$A$$ be a Noetherian local ring with maximal ideal $$\mathfrak m$$. Show that the following are equivalent:

(a) $$\mathfrak m^{q}=0$$ for some $$q\in \mathbb N.$$

(b) $$A$$ is Artinian.

I am only interested in $$(a)\implies (b)$$. So suppose that $$I_0 \supset I_1 \supset \cdots$$ is a descending chain of ideals. One is easily able to show that there exists an $$n_0$$, such that:

$$\dfrac{I_n\cap \mathfrak m^r}{I_n\cap \mathfrak m^{r+1}}=\dfrac{I_{n+1}\cap \mathfrak m^r}{I_{n+1}\cap \mathfrak m^{r+1}}, \forall n\ge n_0,r.$$

Now the proof claims that, the following chain of inclusions holds:

$$I_{n}\subset I_{n+1}+(I_n\cap \mathfrak m)\subset I_{n+1}+(I_n\cap \mathfrak m^2)\subset \cdots \subset I_{n+1}+(I_n\cap \mathfrak m^q)=I_{n+1}\forall n\ge n_0,$$

but I don't understand why this is true and I need some help here.

• Take $r=0$ to get the first inclusion, take $r=1$ to get second etc. For example, with $r=0$, you have $\frac{I_n}{I_n\cap\mathfrak{m}}=\frac{I_{n+1}}{I_{n+1}\cap\mathfrak{m}}$, which translates into the first inclusion. Apr 9, 2016 at 22:33
• Note that the result may be stated in a stronger way: a Noetherian ring is Artin local if and only if there is a nilpotent maximal ideal. Apr 9 at 8:45

The equality you mentioned comes from the inclusion $$\frac{I_{n+1}\cap\mathfrak m^r}{I_{n+1}\cap\mathfrak m^{r+1}}\subseteq\frac{I_n\cap\mathfrak m^r}{I_n\cap\mathfrak m^{r+1}},$$ so for $$x\in I_n\cap\mathfrak m^r$$ there is $$y\in I_{n+1}\cap\mathfrak m^r$$ such that $$x-y\in I_n\cap\mathfrak m^{r+1}$$. This shows that $$I_n\cap\mathfrak m^r\subseteq I_{n+1}\cap\mathfrak m^r+I_n\cap\mathfrak m^{r+1}$$. Now consider $$r=0$$, then $$r=1$$, and so on.
Edit. In order to give a full proof of $$(a)\implies(b)$$ let me mention that for each $$n,r\ge 0$$ $$\frac{I_n\cap\mathfrak m^r}{I_n\cap\mathfrak m^{r+1}}$$ is a finitely generated $$R/\mathfrak m$$-vector space, and therefore there is no such strictly descending chain.