# Equivalence between artinian ring and module whose length is finite

In Matsumura's commutative algebra, proposition 2.C, one says :

"A ring $A$ is artinian iff the length of $A$ as $A$-module is finite".

In the proof we have a descending chain : $A \supseteq p_1 \supseteq p_1p_2 \supseteq .... \supseteq p_1... p_{r-1} \supseteq I \supseteq Ip_1 \supseteq Ip_1p_2 \supseteq ... \supseteq I^s =(0).$ where $p_1$ ...$p_r$ the maximal ideals of A and $I=p_1...p_r$. Can someone explain me the following sentence : "Each factor module of this chain is a vector space over the field $A/p_i=k_i$ for some $i$, and its subspaces correspond bijectvely to the intermediate ideals." ... Thank you.

• In general if $R$ is a ring, $I$ an ideal an $M$ a $R$-module, then $M/IM$ is a $R/I$-module and submodules of $M/IM$ correspond to intermediate modules between $IM$ and $M$. In your case, you have $R=A$, $M$ is one of the ideals occuring in the descending chain and $I=p_i$ is the factor, which is added in the next term of the chain. – MooS Sep 27 '17 at 11:52
• @MooS Please consider posting as a solution... you seem to have addressed the question asked directly and completely. – rschwieb Sep 27 '17 at 12:38
• Ok thank you Moos for answering. So now, the proof is complete for me if I can show that every ideal in A is include between two elements of the chain above. So I try to show that if $I$ is an ideal of A, and let's say $p_1...p_j$ the ideals maximum in which it includes, we have $p_1...p_{j+1} \subset I$. In other terms, that ideals are effectively between two elements of the chain above. I would like to find this by myself but I come back if I don't succeed. – Axel S Sep 27 '17 at 12:56
• Your latter claim is false. Clearly $p_2$ does not satisfy this. It is not included in $p_1$, neither does it include $p_1$. If you think that you need this to complete the proof, you are seemingly still missing the logic of the proof. – MooS Sep 27 '17 at 13:51
• That's right I noticed it after I wrote it. But if $J$ is an ideal of $A$ it's true that we can reorder the indexes as $J$ is between two terms of the chain above ? That's what I want to show as, if it's the case, all sub-modules of A are effectively intermediate ideals and A has a finite length. I don't think ( hope) it's a difficult fact to to prove ( if my assertion is not false) and I am going to try it a bit later in the day – Axel S Sep 27 '17 at 14:40