I know that there's a bijection between the prime ideals of $S^{-1}A$ and the prime ideals of $A$ not meeting $S$ -- some people (see here and here) have asked about whether this is true for all ideals. I see that this isn't true in general, but I'm wondering what conditions have to be met for it to be true.
Generally, I know that every ideal in $S^{-1}A$ is an extension of an ideal of $A$, but this "correspondence" is not injective. So maybe one way I could approach this problem is figuring out when two ideals have the same extension. On the flip side, knowing that every ideal of $S^{-1}A$ is equal to the extension of its contraction, maybe I could also try figuring out why the contraction of $I \subset S^{-1}A$ is "special" among the ideals of $A$ that extend to it.
Unfortunately, I'm having some difficulty with either of these approaches. Could someone possibly finish a classification of when this happens (maybe using some other idea)? One possible example is the obvious isomorphism when $S = \{1\}$; is this the only example?
Let $A$ be a nonzero commutative ring and $S\subset A$ a multiplicatively closed subset. If $s\in S$ is not invertible in $A$ then the ideal $(s)$ generated by $s$ is contained in a maximal ideal, so $(s)\neq A$. The canonical map $$A\ \longrightarrow\ S^{-1}A,$$ maps the principal ideal $(s)\subset A$ to the unit ideal $(1)\subset S^{-1}A$, and so this map does not induce a bijection between the ideals of the two rings.
On the other hand, if $S$ contains only units, then the canonical map is an isomorphism.