TITLE: Why is the localization of a commutative Noetherian ring still Noetherian?
QUESTION [33 upvotes]: 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 $M\leq A$.

REPLY [2 votes]: To verify that a commutative ring $R$ is Noetherian, it suffices to check that the prime ideals of $R$ are finitely-generated. There is a bijective correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ not intersecting $S$ which is given by the map $I \mapsto I \cap R$ with an inverse $J \mapsto \{\frac{a}{s} | s \in S\}$. A prime ideal in $S^{-1} R$ is thus finitely-generated with generators $\{g_i/1\}$, where $g_i$ are the generators of the corresponding prime ideal in $R$.