# Localization of Noetherian and Artinian Modules

Theorem: Let $R$ be a commutative ring with unity, and $S\subset R$ be a multiplicatively closed subset. If $M$ is a Noetherian (Artinian) $R$-module then $S^{-1}M$ is Noetherian (Artinian) $S^{-1}R$-module.

I know the proof of the theorem, but my problem is that the converse is not true!
The main idea of the proof as I understand is that there is a 1-1 correspondence between submodules of $M$ and submodules of $S^{-1}M$, but if that the case shouldn't the converse be true also?!
Also can you please give me an example that the converse is not true. Thank you.

-

Let $R=k[x_1,x_2,\dots]$ be a ring of polynomials in infinitely many variables over a field and let $M=R$. Then set $S^{-1}=R \setminus \{0\}$ so that $S^{-1}R=k(x_1,x_2,\dots)$ the field of fractions of $R$ which is a field of rational functions in infinitely many variables. So $S^{-1}R$ is a field and thereby a Noetherian and Artinian $S^{-1}R$-module. The issue here is that localization is, well, local. In the ring theoretic setting we can think of the localization at a prime ideal $\mathfrak p \subset R$ as creating a ring whose ideal lattice consists of the ideals contained in $\mathfrak p$. So for instance when we localize at $(0)$ we always get a field when the ring is an integral domain. There is the notation though of local-to-global properties in commutative algebra rings such that for every localization property $p$ holds then it holds for $R$. Unfortunately Noetherian is not a local-to-global property with a little bit of work one can show that every localization at a prime ideal of $\prod_{n \in \mathbb N} k$ is Noetherian, but the ring itself is not.

-

Let $A$ be an integral domain of dimension greater than 1. Then $A_{(0)}$ is a field, so in particular it is artinian. But $A$ is not artinian.

-
Concretely let $k$ be a field then $A=k[x,y]$ and $A_{(0)}=k(x,y)$. – JSchlather Feb 3 '13 at 23:27
@JacobSchlather can you tell me the difference between k[x,y] and k(x,y). and why the proof does not go backwards?!, and how about the noetherian case? – i.a.m Feb 3 '13 at 23:32
$k[x,y]$ is the ring of polynomials in two variables, $k(x,y)$ is the field of fractions in two variables. – Damien L Feb 3 '13 at 23:34
@DamienL is it true that there is a 1-1 correspondance between submoduls of $M$ and submodules of $S^{-1}M$ see here – i.a.m Feb 3 '13 at 23:47
@JacobSchlather is it true that there is a 1-1 correspondance between submoduls of $M$ and submodules of $S^{-1}M$ see here – i.a.m Feb 3 '13 at 23:48

Submodules of $S^{-1} M$ are not in 1-1 correspondence with all submodules of $M$. They are only in 1-1 correspondence with submodules $N$ of $M$ satisfying the condition: If $s \in S$ and $m \in M$, then $sm \in N$ implies $m \in N$ (equivalently, the annihilator of any nonzero element of $M/N$ is disjoint from $S$).

In the example given by Damien L , you can see an extreme case of what goes wrong: If $A$ is domain regarded as a module over itself, and $I$ is a submodule (ideal), then the annihilator of $1 \in A/I$ is $I$. Thus in order to satisfy the condition above, $I$ must be disjoint from $S = A - \{0\}$, i.e., $I = \{0\}$.

-