The following proofs was left as an exercise in my class notes of commutative algebra and I am not able to prove it despite trying it a lot. I have no option rather than posting it here.
(a) Let B be an ideal of $S^{-1} R$ where R is a ring and $S= A/P$ where P is a prime ideal. Define $A = i^{-1} B$ where $i: A \to S^{-1} A$. Then prove that $S^{-1} A = B$.
(b) If $P \in Spec A$ , $P\cap S=\phi$ then show that $ S^{-1} P$ is a prime ideal of $S^{-1} A$.
(c) If $P\in Spec A$ and $P\cap S =\phi$ Then show that $i^{-1} (S^{-1} P) =P$.
Attempt:(a) Let $a/s \in S^{-1} / A$ . But now, how can I prove that this element also is in B? I am not able to think of it. Also, I wasn't able to prove the reverse inclusion.
(b) Let a/s and b/t be elements of $S^{-1} A$ such that $(a/s) (b/t) \in S^{-1} P$. To show that $(a/s) \in S^{-1} P$ or $(b/t) \in S^{-1} P$ . Given condition implies that ab/st = p/s' => there exists $s'' \in S/P $ such that s' ab = st p . But st p lies in P as $p \in P$ and $s'' \notin P$ => $ab \in P$ . But P is prime , so either $a\in P$ or $b \in P$. But how to prove from here that either $a/s\in P$ or $b/t \in P$.
(c) Why does $i^{-1}$ exists? It should be 1-1 and onto. I have proved it 1-1 and onto. I think then $ i^{-1} (S^{-1} P) =P$ is equivalent to showing that $i(P)= S^{-1} P$ which holds by definition. I hope my proof is fine? OR Why is $P\cap S=\phi$ important?
Kindly help me with these.
(a) We know that $a/s \in S^{-1}A$ which means that $a \in A$. So, we have $a/1 \in B$ and $a/s = (1/s)(a/1) \in B$ because B is an ideal.
(b) $ab/st = p/s'$ means that there exists a $u \in S$ such that $u(s'ab - stp) = 0$. So, $us'ab = ustp \in P$ and $ab \in P$.
(c) You have to show that $i^{-1}(S^{-1}P) \subset P$. Take $a \in i^{-1}(S^{-1}P)$, we have $a/1 = p/s$ where $p \in P$. This means that there exists a $u \in S$ such that $uas = up$. So, $uas \in P$ and it follows that $a \in P$ because $P \bigcap S = \emptyset$