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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 $M\leq A$.

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2 Answers 2

up vote 9 down vote accepted

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 $M$, 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 $M$.

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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.

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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

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