Can the prime ideals of $A_a$ be identified with those prime ideals of $A$ which do not contain $a$?

Question

Let $A$ be a ring, and let $a \in A$. Can the prime ideals of $A_a$ be identified with those prime ideals of $A$ which do not contain $a$?

Answer 1

Can the prime ideals of $A_a$ be identified with those prime ideals of $A$ which do not contain $a$ ?

Yes. In general, if we consider the natural homomorphism $\phi \colon A\rightarrow S^{-1}A$ given by $a\mapsto \frac{a}{1}$, then for ideals $I$ of $A$ and $\mathscr{I}$ of $S^{-1}A$, $I^e=\langle \phi(I)\rangle$ is an ideal of $S^{-1}A$ called the extension of $I$, and $\mathscr{I}^c=\phi^{-1}(\mathscr{I})$ is an ideal of $A$ called the contraction of $\mathscr{I}$

We have the following:

Proposition: Let the situation be as in the previous paragraph. Then there exits a bijective map $\Phi$ between the sets $\mathscr{A}=\{P\in \text{Spec}(A): P\cap S=\emptyset\}$ and $\mathscr{B}=\text{Spec}(S^{-1}A)$ given by $$\Phi\colon \mathscr{A}\rightarrow \mathscr{B}$$ $$\;\;\;\;\;\;P\mapsto P^e.$$

Proof: This is the corollary of theorem 5.32 given in Sharp's "Steps in Commutative Algebra". $\blacksquare$

On the other hand the set $S=\{1, a, a^2,\ldots\}$ is a multiplicative closed subset of $A$. This means by the above proposition that the prime ideals of $S^{-1}A=A_a$ can be identified, by the bijection, with the prime ideals of $A$ such that are disjoint with $S$, so in particular those prime ideals do not contain $a$.