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.
Best Answer
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.