Vtgyjfy

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
$$
S^{-1}M_1subset S^{-1}M_2subsetcdots
$$
and pulling back to a chain
$$
M_1subset M_2subsetcdots
$$
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 $Mleq A$.

This is a standard property of localizations:

Theorem. Let $R$ be a commutative ring, and let $Sneqemptyset$ be a multiplicative subset. Let $M$ be an $R$-module. Let $varphicolon Rto S^{-1}R$ be the canonical map ($varphi(r) = frac{rs}{s}$), and likewise, by abuse of notation, let $varphicolon Mto S^{-1}M$ be the natural map $(varphi(m) = frac{sm}{s}$ with $sin S$).

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




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




3. If $N$ is a submodule of $M$, then $Nsubseteq varphi^{-1}(S^{-1}N)$. 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 $a,bin N$. And it is closed under scalar multiplication, since $ain N$ implies $rain N$ for all $rin R$, so $frac{r}{t}(frac{a}{s}) = frac{ra}{ts}in S^{-1}N$ if $ain N$.

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 $ain N$ we have
$varphi(a) = frac{sa}{s}in S^{-1}N$, since $sain N$, hence $ain varphi^{-1}(S^{-1}N)$. Now assume that $N=varphi^{-1}(L)$. If $ain N$ and $sin S$, 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 $S^{-1}Nsubseteq L$. Conversely, let $frac{m}{t}in L$. Then $frac{tt}{t}frac{m}{t} = frac{(tt)m}{tt}=varphi(m)in L$, hence $min varphi^{-1}(L) = N$; thus, $frac{m}{t}in S^{-1}N$, proving that $Lsubseteq S^{-1}N$. 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 $M$.

Let $fcolon A to S^{-1}A$ 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.