# Why is the localization of a commutative Noetherian ring still Noetherian?

This is an unproven proposition I've come across in multiple places.

Suppose $A$ is a commutative Noetherian ring, and $S$ a multiplicative subset of $A$. Then $S^{-1}A$ is Noetherian.

Why is this? I thought about taking some chain of submodules $$S^{-1}M_1\subset S^{-1}M_2\subset\cdots$$ and pulling back to a chain $$M_1\subset M_2\subset\cdots$$ of submodules of $A$ which must eventually stablize. Is there more to it than this? I kind of wary of assuming all submodules of $S^{-1}A$ have form $S^{-1}M$ for $M\leq A$.

## 2 Answers

This is a standard property of localizations:

Theorem. Let $$R$$ be a commutative ring, and let $$S\neq\emptyset$$ be a multiplicative subset. Let $$M$$ be an $$R$$-module. Let $$\varphi\colon R\to S^{-1}R$$ be the canonical map ($$\varphi(r) = \frac{rs}{s}$$), and likewise, by abuse of notation, let $$\varphi\colon M\to S^{-1}M$$ be the natural map $$(\varphi(m) = \frac{sm}{s}$$ with $$s\in S$$).

1. For every submodule $$N$$ of $$M$$, $$S^{-1}N = \{\frac{a}{s}\mid a\in N\}$$ is a submodule of $$S^{-1}M$$.

2. If $$L$$ is a submodule of $$S^{-1}M$$, then $$\varphi^{-1}(L) = \{m\in M\mid \varphi(m)\in L\}$$ is a submodule of $$M$$.

3. If $$N$$ is a submodule of $$M$$, then $$N\subseteq \varphi^{-1}(S^{-1}N)$$. Moreover, if $$N=\varphi^{-1}(L)$$ for some submodule $$L$$ of $$S^{-1}M$$, then $$L=S^{-1}N$$.

Proof. $$S^{-1}N$$ is nonempty, as it contains $$\frac{0}{s}$$; it is closed under differences, since $$\frac{a}{s}-\frac{b}{t} = \frac{ta-sb}{st}\in S^{-1}N$$ if $$a,b\in N$$. And it is closed under scalar multiplication, since $$a\in N$$ implies $$ra\in N$$ for all $$r\in R$$, so $$\frac{r}{t}(\frac{a}{s}) = \frac{ra}{ts}\in S^{-1}N$$ if $$a\in N$$.

Now let $$L$$ be a submodule of $$S^{-1}M$$; since $$\varphi$$ is a module homomorphism, the pullback of a submodule is a submodule, so 2 is immediate.

Again, let $$N$$ be a submodule of $$M$$. Then for every $$a\in N$$ we have $$\varphi(a) = \frac{sa}{s}\in S^{-1}N$$, since $$sa\in N$$, hence $$a\in \varphi^{-1}(S^{-1}N)$$. Now assume that $$N=\varphi^{-1}(L)$$. If $$a\in N$$ and $$s\in S$$, then $$\frac{a}{s} = \frac{ssa}{sss} = \frac{s}{ss}\frac{sa}{s} =\frac{s}{ss}\varphi(a)\in L$$ (since $$\varphi(a)\in L$$), hence $$S^{-1}N\subseteq L$$. Conversely, let $$\frac{m}{t}\in L$$. Then $$\frac{tt}{t}\frac{m}{t} = \frac{(tt)m}{tt}=\varphi(m)\in L$$, hence $$m\in \varphi^{-1}(L) = N$$; thus, $$\frac{m}{t}\in S^{-1}N$$, proving that $$L\subseteq S^{-1}N$$. Therefore, $$L=S^{-1}N$$, as claimed. $$\Box$$

Corollary. Every submodule of $$S^{-1}M$$ is of the form $$S^{-1}N$$ for some submodule $$N$$ of $$M$$.

• Thanks, these properties and the detail are very helpful. – Buble Feb 18 '12 at 23:28
• @ArturoMagidin I swear to god I've learned so much from your posts on Math.SE :D – user38268 Feb 19 '12 at 1:00

Let $f\colon A \to S^{-1}A$ be the canonical homomorphism. It is true that every ideal $\mathfrak b$ of $S^{-1}A$ is of the form $S^{-1}\mathfrak a = f(\mathfrak a)(S^{-1}A)$ for some ideal $\mathfrak a$ of $A$. We can even take $\mathfrak a = f^{-1}(\mathfrak b)$, and this should help you prove that $S^{-1}A$ is Noetherian using increasing chains.

You can find related facts in and around Proposition 6.4 of Milne.

• Thanks for your help again, Dylan. btw, should it say some ideal $\mathfrak{a}$ of $A$? in the second line, not $S^{-1}A$? – Buble Feb 18 '12 at 23:27
• @Buble Yes indeed. I was toying with the wording and this mixup, inevitably, happened. – Dylan Moreland Feb 18 '12 at 23:47