Suppose R is a commutative Artinian ring then R is Noetherian.
I am aware of the proof which uses the idea of filtration. But I would like to prove this fact without that idea but haven't got far enough. Is there any reference for such a proof? or is anyone aware of one?
$\endgroup$ 03 Answers
$\begingroup$Here is a direct proof that the Jacobson radical $J$ of an Artinian ring is nilpotent.
We know that $J^n = J^{n+1}$ for some $n \geq 1$. Let $X = \{a \in A : aJ^n = 0\}$; it is enough to show $X = A$ as then $1 \in X$ and $J^n = 0$.
Now if $X < A$ then we can find a minimal non-zero submodule $Y/X$ of $A/X$ because $A$ is Artinian. Then $Y/X$ is a simple $A$-module, so the definition of the Jacobson radical as the simultaneous annihilator of all simple modules implies that $YJ \subseteq X$. Hence
$$YJ^n = YJ^{n+1} = YJ.J^n \subseteq XJ^n = 0$$
and hence $Y \subseteq X$. This contradicts the assumption that $Y/X$ is non-zero. Thus $X = A$ as required.
This proof also works in the non-commutative case, and shows that a right Artinian ring is right Noetherian.
$\endgroup$ 4 $\begingroup$This is a vague but subtle question. Here are some slides of Chris Conidis that take it very seriously.
To paraphrase, the question is this: All standard proofs that Artinian implies Noetherian prove the stronger statement that Artinian implies finite length. Is there a proof that manages to be simpler by failing to prove the latter?
Conidis gives an affirmative answer, for countable rings, by giving a proof which is valid in an axiomatic system too weak to prove finite length.
I don't have time to finish this line of exploration myself, but I thought it might be of interest (and it also pinpoints nilpotence of the Jacobson radical as the sticking point). There seem to be at least a few papers out there talking about this.
$\endgroup$ 4 $\begingroup$"Steps in Commutative Algebra" by "R. Y. Sharp"
8.39 LEMMA. Let R be a commutative Artinian ring. Then every prime
ideal of R is maximal.
8.40 LEMMA. Let R be a commutative Artinian ring. Then R has only
finitely many maximal ideals.
8.41 PROPOSITION. Let R be a commutative Artinian ring, and let N =
$\sqrt 0$, the nilradical of R. There exists t such that N^t = 0.
7.30 THEOREM. Let G be a module over the commutative ring R, and
assume that G is annihilated by the product of finitely many (not necessarily
distinct) maximal ideals of R, that is, there exist maximal ideals
M_i,..., M_n of R such that M_i... M_n G = 0. Then G is a Noetherian R-module if and only ifG is an Artinian R-module.
SO:
8.44 THEOREM. A commutative Artinian ring R is Noetherian.
