Intuitively, why $\operatorname{spec}(S^{-1}A) \cong \lbrace \mathfrak{p} \in \operatorname{spec} (A)| \mathfrak{p} \cap S = \emptyset \rbrace$?

Question

If $A$ is a commutative ring, $S$ is a multiplicative subset of $A$, I’d like to understand intuitively why is there a bijection between $\operatorname{spec}(S^{-1}A)$ and $\lbrace \mathfrak{p} \in \operatorname{spec}(A) | \mathfrak{p} \cap S = \emptyset \rbrace$.

I’ve read a proof of this fact on my lecture notes, but I still don’t feel like I understand it in depth as the proof was quite technical and didn’t seem to me very illuminating. I do understand intuitively why if $S=A \setminus \mathfrak{p}$ in particular, then $\operatorname{spec}(A_{\mathfrak{p}})$ and $\lbrace \mathfrak{q} \in \operatorname{spec}(A) | \mathfrak{p} \cap (A \setminus \mathfrak{p}) = \emptyset \rbrace = \lbrace \mathfrak{q} \in \operatorname{spec} (A)| \mathfrak{q} \subset \mathfrak{p} \rbrace$ are bijective (thanks to this very clear answer).

I feel like if I could read a similarly intuitive explanation for the more general result, it would help my understanding a lot.

Answer 1

You can think of the localisation $S^{-1}A$ of a ring $A$ at a multiplicatively closed set $S$ as some sort of a "nice union" (formally, this is called a colimit or more precisely, in this case, a direct limit) of $A_f$ where $f \in S$.

Fix $f \in S$. Observe that $A_f \cong A[x]/(xf-1)$ with the isomorphism induced by the usual map $A \hookrightarrow A[x] \to A[x]/(xf-1)$ (note that this mapping indeed sends $f$ to the unit $[f]$ in the quotient). Thus you can think of the localization map $A \to A_f$ as the composite map $A \to A[x]/(xf-1)$. The prime ideals of $A[x]/(fx-1)$ are in bijection with the prime ideals of $A[x]$ containing $xf-1$. It is easy to see that the map $Q \mapsto Q \cap A$ is a bijection of $\{\text{Prime Ideals of }A[x]\text{ containing }xf-1\}$ onto $\{\text{Prime Ideals of }A\text{ not containing }f\}$.

Now since $A_S$ is roughly a "union" of $A_f$s, prime ideals of $A_S$ are in bijection with the prime ideals of $A$ which don't contain $f$ for any $f \in S$. The point is that $Spec$ (considered as a functor (transformation)) changes "nice unions" into "nice intersections" and hence, $$Spec(S^{-1}A) = Spec(\cup A_f) = \cap Spec(A_f)$$ so that only those prime ideals "survive" in $S^{-1}A$ which are primes in every $A_f$, that is, which don't contain any $f \in S$.

A bit more precisely, the localization map $A \to S^{-1}A$ factors uniquely through each $A \to A_f$, so that for each $f \in S$, you have that the composite $A \to A_f \to S^{-1}A$ is equal to the localization $A \to S^{-1}A$. Thus, if $P$ is any prime ideal of $S^{-1}A$, then it's pullback under the second map should be a prime ideal of $A_f$, and hence pulling it back further under the first map should give you a prime ideal of $A$ not containing $f$. Hence, the pullback of $P$ under the composed map should be a prime ideal of $A$ not containing any $f \in S$, that is, one which is disjoint with $S$.

The essential idea in the above explanations is that a prime ideal of a ring, being proper, cannot contain any units. In particular, for a prime ideal $P$ of $S^{-1}A$, $P$ cannot contain any element of the form $s/1$ where $s \in A$. Thus, it's pullback under $A \xrightarrow{j} S^{-1}A$ (which is a prime ideal of $A$) should be disjoint from $S$.

This pullback map $P \mapsto j^{-1}(P)$ is the required bijective map with inverse mapping $Q \mapsto S^{-1}Q$.

Of course checking this rigorously is what the proof is all about and it is easy if you understand the definitions correctly.