# Does the natural bijection between the set of prime ideals in A disjoint from S and Spec$(S^{-1}A)$ restrict to maximals?

I was studying rings of fractions, and I was wondering about the problem of restricting the canonical bijection (induced by retraction and extension of ideals) $\{p\in \text{Spec}(A) \mid p\cap S=\emptyset \} \to \text{Spec}(S^{-1}A)$ to a map $\{\mathfrak{M} \in \text{M-Spec}(A) \mid \mathfrak{M}\cap S=\emptyset \} \to \text{M-Spec}(S^{-1}A)$. I proved that the extension map can be restricted to a map between these sets (which is therefore injective), but the problem is that there is no reason why the retraction of a maximal ideal in the ring of fraction should be maximal (I was trying to argue by contradiction, using the bijectivity of the map between the prime spectra, but I think the problem is that taking by Zorn's lemma a maximal element in the set of ideals disjoint from a multiplicative subset yields only a prime ideal, not a maximal one).

What do you know or think about this? Are there counterexamples? Are there interesting cases in which the restricted map gives actually a bijection between these two sets?

• I found a counterexample. If we take $A=\mathbb{Z}, S=A\setminus\{0\}$ then $S^{-1}A=\mathbb{Q}$ and $(0)$ is a maximal ideal in $\mathbb{Q}$ whose restriction is $(0)$ which is a prime ideal but not a maximal ideal in $\mathbb{Z}$. So in general the restricted map is not surjective... but are there cases in which this holds? – Diogenes Feb 15 '14 at 19:59
• @Georges Elencwajg: Ok thanks, so what I wrote is right. But are there cases in which the restricted map is also surjective? – Diogenes Feb 15 '14 at 20:25
• I have added an Edit addressing your question. – Georges Elencwajg Feb 15 '14 at 23:15
• @GeorgesElencwajg: Thanks a lot for your answer! I think I have understood what you're saying: you are taking that multiplicative subset because we know that in that case we have $A_f \cong A[Y]/(Yf-1)$ so if $A$ is finitely generated over $k$ also $A_f$ is (with $k$-algebra structure inherited by making the canonical homomorphism $A \to A_f$ a $k$-algebra homomorphism), and so we can use the fact that the retraction of a maximal ideal through a homomorphism of finitely generated $k$-algebras is not only prime, but actually maximal. Is it right or was there an easier way to see this? – Diogenes Feb 16 '14 at 11:57
• Dear Diogenes, your reasoning is quite right. – Georges Elencwajg Feb 16 '14 at 14:15

a) Maximal ideals of $S^{-1}A$ needn't restrict to maximal ideals of $A$:

Take $A=\mathbb Z, S=\mathbb Z\setminus \{0\}, S^{-1}A=\mathbb Q$.
The maximal ideal $(0)\subset \mathbb Q$ restricts to $(0)\subset \mathbb Z$ which is not maximal.

The explanation is that every maximal ideal $(0)\subsetneq (p)\subset \mathbb Z$ containing $(0)$ in $\mathbb Z$ will intersect $S$ and thus disappear from the correspondence.
Hence that small guy $(0)$ in $\mathbb Z$ will see all his competitors killed in $\mathbb Q$ and he will become the big boss there...

b) However the correspondence you mention between prime ideals in $A$ disjoint from $S$ and primes of $S^{-1}A$ immediately implies that a maximal ideal $\mathfrak m\subset A$ disjoint from $S$ extends to a maximal ideal $\mathfrak m^e= \mathfrak m.S^{-1}A\subset S^{-1}A$ of $S^{-1}A$.

Edit
Diogenes asks about cases where all maximal ideals of $S^{-1}A$ are extended from maximal ideals of $A$.
Here algebraic geometry comes to the rescue since maximal ideals in a ring $A$ exactly correspond to closed points in the corresponding affine scheme $\text{Spec} (A)$.
The simplest example is obtained by taking for $A$ a finitely generated algebra over a field $k$ and for $S$ the multiplicative set $S=\{1,f,f^2,\cdots \}$ generated by a non nilpotent element $f\in A$.
Every maximal ideal of $S^{-1}A=A_f$ is then extended from a maximal ideal of $A$.
The geometric interpretation is that a point $P\in D(f)\subset \text{Spec} (A)$ is closed in $D(f)$ if and only if it is closed in $\text{Spec} (A)$.
[As remarked by Diogenes this is due to $A_f$ being finitely generated over $A$, hence over $k$ too; see also here.]