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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.

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  • $\begingroup$ 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. $\endgroup$
    – Mohan
    Apr 9, 2016 at 22:33
  • $\begingroup$ 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. $\endgroup$ Apr 9 at 8:45

1 Answer 1

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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.

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  • $\begingroup$ Is the R/m-vector space in your answer generated by the generators of m (finitely many due to Noetherianness of R)? $\endgroup$
    – Divide1918
    Mar 30, 2021 at 14:50
  • $\begingroup$ No. It is a quotient of a finitely generated ideal of R. $\endgroup$
    – user26857
    Mar 30, 2021 at 15:25

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