Let $A$ be a left and right-Noetherian ring and let $S \subset A$ be left-localisable.
Let $P \subset A$ be a prime ideal such that $P \cap S = \emptyset$. I wish to show $S^{-1}AP = \{\phi(s)^{-1}\phi(p): s \in S, p \in P\} := P^e$ is a prime ideal.
Where I'm at:
Let $I, J$ be ideals in $S^{-1}A$ such that $IJ \subset P^e$.
Define $I^c = \{i \in A : \phi(i) \in I \}$, $J^c = \{j \in A : \phi(j) \in J \}$. It's easily seen these are ideals in $A$.
If we can show $I^cJ^c \subset P$ it will follow w.l.o.g $I^c \subset P$. From here given $\phi(s)^{-1}\phi(a) \in I$ yields $\phi(a) \in I$ and so $a \in I^c$. Therefore $\phi(a) \in \phi(P)$ and $\phi(s)^{-1}\phi(a) \in P^e$. So it suffices to prove $I^cJ^c \subset P$.
From here I'm less sure: $S \cap P = \emptyset$ implies that $S/P$ is left-localisable in $A/P$ (it is clearly left-Ore, and because $A/P$ is left-Noetherian it is left-localisable). Now if $\overline{\phi}: A/P \to (S/P)^{-1}A/P$ is the canonical map we have $t_{S/P}(A/P) = \text{Ker}\overline{\phi}$. If I can show that $S/P$ consists of regular elements in $A/P$ then $\overline{\phi}$ will be injective (this is Ore's theorem) and we'd be done as $\text{Ker}\overline{\phi} = P$ then.
Questions:
Can we use the right-Noetherian condition to show $S/P$ consists of regular elements?
If not, a hint please?
We have a multiplicatively closed subset $S\subset A$ and a ring homomorphism $\phi\colon A\to B:=S^{-1}A$ such that
(1) $\phi(s)$ is invertible for all $s\in S$.
(2) every element in $B$ is of the form $\phi(s)^{-1}\phi(a)$ for some $s\in S$ and $a\in A$.
(3) $\ker(\phi) = \{a\in A : sa=0\textrm{ for some }s\in S\}$.
We have maps between left ideals of $A$ and left ideals of $B$, sending $M\leq A$ to $M^e=B\phi(M)$ and sending $I\leq B$ to $I^c=\phi^{-1}(I)$. Clearly $I^{ce}=I$. On the other hand, every element in $M^e$ is of the form $\phi(s)^{-1}\phi(m)$ for some $s\in S$ and $m\in M$, so $M^{ec}$ consists of those $a\in A$ such that $sa\in M$ for some $s\in S$.
The claim is that when $A$ is Noetherian, these constructions yield mutually inverse bijections between primes $P\lhd A$ with $P\cap S=\emptyset$ and primes $Q\lhd B$.
First some results on two-sided ideals.
If $I\lhd B$ is a two-sided ideal, then $I^c\lhd A$ is a two-sided ideal. Conversely, if $M\lhd A$ is a two-sided ideal, then $M^e\lhd B$ is a two-sided ideal. For, fix $s\in S$ and consider $$ M_n := \{a\in A:as^n\in M^{ec}\} = \big(M^e\phi(s)^{-n}\big)^c. $$ These form an ascending chain of left ideals $M^{ec}=M_0\subset M_1\subset\cdots$, which stabilises since $A$ is left Noetherian, so $M_n=M_{n+1}$ for some $n$. Thus $M_n^e=M_{n+1}^e$, so $M^e\phi(s)^{-n}=M^e\phi(s)^{-n-1}$, so $M^e\phi(s)^{-1}=M^e$.
Now let $Q\lhd B$ be prime. We know that $Q^{ce}=Q$. If $M,N\lhd A$ with $MN\subset Q^c$, then $M^e,N^e\lhd B$ are two-sided ideals, so $M^eN^e=(MN)^e\subset Q^{ce}=Q$. Without loss of generality, $M^e\subset Q$, so $M\subset M^{ec}\subset Q^c$. Hence $Q^c$ is prime. Also, $Q$ contains no units of $B$, so $Q^c\cap S=\emptyset$.
Conversely, let $P\lhd A$ be prime with $P\cap S=\emptyset$. Then $P^{ec}$ is finitely generated as a right ideal, and hence there exists some $s\in S$ such that $sP^{ec}\subset P$. So $P\cap S=\emptyset$ implies $P^{ec}=P$. Finally, if $I,J\leq B$ satisfy $IJ\subset P^e$, then clearly $I^cJ^c\subset P^{ec}=P$, so without loss of generality $I^c\subset P$, and hence $I=I^{ce}\subset P^e$. Thus $P^e$ is prime.
A good reference is Chapter 2 of Noncommutative Noetherian Rings by McConnell and Robson.