Noetherian localization of a commutative ring

Question

I realize this question has been answered multiple times (here for example), but I still can't figure out this problem even after looking at every solution given.

The problem statement is: Prove that If $R$ is a commutative Noetherian ring with unity and $S \subset R$ is a multiplicative set, then so is $S^{-1}R$.

The following I did on my own: Let $(J_k)_{k=1}^{\infty} \subset S^{-1}R$ be an increasing sequence of ideals in $S^{-1}R$. Then for every $k$, $J_k = S^{-1}I_k$ for $I_k \subset R$ an ideal. Letting $f: R \longrightarrow S^{-1}R$ be the canonical homomorphism $x \mapsto \frac{xs}{s}$, the pullbacks clearly give you a sequence of increasing ideals $(f^{-1}(S^{-1}I_k))_{k=1}^{\infty} \subset R$ which eventually stabilize since $R$ is Noetherian.

Hence there's some $t$ such that $f^{-1}(S^{-1}I_k) = f^{-1}(S^{-1}I_t)$ for all $k \geq t$.

Every answer I've looked at also happens to stop here and states the rest of the argument follows immediately. I don't see how it does however, since I don't see how this sequence of ideals stabilizing implies the sequence $(S^{-1}I_k)$ stabilizes. $f$ isn't onto its image (if for example $x/s \in S^{-1}I_k$ for $s \notin R^{\times}$, then there is no $y \in I_k$ with $\frac{ys}{s} = \frac{x}{s}$, since we can't choose $y = xs^{-1}$), so $f(f^{-1}(S^{-1}I_k)) \subset S^{-1}I_k$ so the image may not even be an ideal, and even if it was, we don't know that it equals $S^{-1}I_k$, so at best we've proven there exists a sequence of ideals $(f(f^{-1}(S^{-1}I_k))) \subset S^{-1}I_k$ that eventually stabilize, which isn't sufficient.

What am I missing here?

Answer 1

The following may help to clarify things: for an ideal $J\subseteq S^{-1}R$, write $J^{c}:=f^{-1}(J)\subseteq R$ for its contraction (an ideal of $R$), and for an ideal $I\subseteq R$ write $I^e:=(f(r)\mid r\in I)\subseteq S^{-1}R$ for its extension (i.e. the ideal of $S^{-1}R$ generated by the image of $I$). Then $J^{ce}=J$ for all $J\subseteq S^{-1}R$.

Indeed, it is always the case that $J^{ce}\subseteq J$ for set-theoratic reasons: we have that $f(J^c)=f(f^{-1}(J))\subseteq J$, and thus $J$ contains the ideal generated by $f(J^{c})$, i.e. $J^{ce}$.

For the reverse inclusion, which may fail if $f$ is just an arbitrary ring homomorphism, we have to use that $f$ is the localisation map. Let $x\in J$ be arbitrary and write it as $x=r/s$ for some $r\in R$ and $s\in S$. Then $r\in J^{c}$, because $f(r)=sx\in J$. So $f(r)=r/1\in J^{ce}$, but then we must also have $x=(1/s)f(r)\in J^{ce}$. Hence $J\subseteq J^{ce}$.

So we can conclude that $J=J^{ce}$. It should also be clear that both $J\mapsto J^{c}$ and $I\mapsto I^e$ respect inclusions. Therefore, if you have an increasing sequence of ideals $(J_n)_{n\geq 1}$ in $S^{-1}R$, then $(J_n^c)_{n\geq 1}$ is an increasing sequence of ideals in $R$ and thus stabilizes. Hence $(J_n^{ce})_{n\geq 1}$ stabilizes as well, and from we have seen above, we have $(J_n^{ce})_{n\geq 1}=(J_n)_{n\geq 1}$.

So the key point you were missing is that in the final step, you don't just conclude that the sequence of ideals $(f(J_n))_{n\geq 1}$ of $f(R)$ stabilizes, but you conlcude that the sequence of ideals $(J_n^{ce})_{n\geq 1}$ in $S^{-1}R$ generated by $f(J_n)$ stablizes, and that this is in fact $(J_n)_{n\geq 1}$.