Question about localization

Question

If $A \not= \{0 \}$ is a commutative ring and $P \subset Q$ are prime ideals of $A$ then of course $P \cap (A \setminus Q) = \varnothing$ so that $PA_Q = S^{-1}P$ is a prime ideal of $A_Q$ where $S=A \setminus Q$.

My book says that it is easily seen that $A_P$ then is the localization of $A_Q$ at $PA_Q$ but I cant see this easily, can anyone clarify this to me?

Answer 1

Let $A$ be commutative ring with identity $1 \neq 0$ and let $S, T$ be two multiplicatively closed subset of $A.$ Le $U$ be the image of $T$ in $S^{-1}A.$ Then $(ST)^{-1}A \cong U^{-1}(S^{-1}A).$

Consider the natural ring homomorphism $A \rightarrow (ST)^{-1}A, a \mapsto \frac{a}{1}.$ Note that, for each $s \in S, \frac{s}{1}$ is a unit in $(ST)^{-1}A.$ So this will induce a ring homomorphism $\phi': S^{-1}A \rightarrow (ST)^{-1}A.$ Also $\phi'(t/s)$ is a unit in $(ST)^{-1}A, \forall t \in T.$ We conclude that there is a ring homomorphism $\phi: U^{-1}(S^{-1}A) \rightarrow (ST)^{-1}A$ defined by $\phi \left(\dfrac{(a/s)}{(t/1)} \right) = \dfrac{a}{ts}.$

Now consider the ring homomorphism $A \rightarrow U^{-1}(S^{-1}A), a \mapsto \frac{(a/1)}{(1/1)}.$ Under this map image of $st$ has an inverse, for each $s \in S, t \in T,$ namely, $\frac{(1/s)}{(t/1)}.$ So this will induce a ring homomorphism $\psi: (ST)^{-1}A \rightarrow U^{-1}(S^{-1}A)$ defined by $\psi \left(\dfrac{a}{st} \right) = \dfrac{(a/s)}{(t/1)}.$

Now check that $\phi$ and $\psi$ are inverse to each other.

In this particular case, choose $S := A \setminus Q, T := A \setminus P.$ Then $U = A_Q \setminus PA_Q.$

Answer 2

your question is Expressive and comments almost covers the answer. I only give the isomorphism:
$$f: A_P \to (A_Q)_{PA_Q}$$

$$f(r/t)=(r/1)/(t/1)$$
for all $r\in R$ and $t \in R- P$

Answer 3

This could be easily stated with complements, $S=A\setminus Q$, $T=A\setminus P$, then $S\subset T$ and they are multiplicative subsets. Then prove that given any multiplicative subsets $S,T$ s.t. $S\subset T$ then $A[T^{-1}]=A[S^{-1}][T^{-1}]$ by the universal property of the ring of fractions. Never try to define a concrete morphism and check that it is bijective by hand, this is the kind of stuff that made many proofs in old books unreadable.