Let $R$ be a left-Noetherian ring, and suppose $S \subset R$ is left-localizable ($S$ is assumed to be a multiplicative closed subset and $1 \in S$).
Let $\phi: R \to S^{-1}R$ be the canonical homomorphism; $S^{-1}R = \{\phi(s)^{-1}\phi(a): s \in S, a\ \in R\}$.
Let $I \subset R$ be a two-sided ideal. Why does it follow that $S^{-1}R\phi(I)$ is a two sided ideal?
Where I'm at:
Let $\phi(s_0)^{-1}\phi(a_0)\phi(i) \in S^{-1}R\phi(I)$ and let $\phi(s)^{-1}\phi(a) \in S^{-1}R$. Using the left Ore condition it's clear that $[\phi(s)^{-1}\phi(a)][\phi(s_0)^{-1}\phi(a_0)\phi(i)] \in S^{-1}R\phi(I)$.
However I can't find a reason why multiplcation on the other side will yield in element in $S^{-1}R\phi(I)$. One can write $Sa_0i = As$ to get $\hat{s}a_0i = \hat{a}s \in I$ and so $[\phi(s_0)^{-1}\phi(a_0)\phi(i)][\phi(s)^{-1}\phi(a)] = \phi(s_0)^{-1}\phi(\hat{s})^{-1}\phi(\hat{a})\phi(a)$. But I see no reason why this element will be in the needed set.
I've tried looking at the annihilator of $s$ in $A/I$ as it is left-Noetherian too, but to no avail.
Hints?
I'll assume $\phi$ is injective which means no element of $S$ is a zero divisor. Let $Q:=S^{-1}R$ and let $I^e= QI$.
We know $I^e$ is a left ideal. Let $s^{-1}r$ be a typical element of $Q$; we want to show $I^e(s^{-1}r)\subseteq I^e$. Since $I$ is a right ideal, we get $Is\subseteq I$, multiplying on the right by $s^{-1}$ in $Q$, this gives $I\subseteq Is^{-1}$. Applying this repeatedly, we get a chain of left ideals in $Q$: $$I\subseteq Is^{-1} \subseteq Is^{-2} \subseteq \cdots$$ Since $Q$ is left noetherian, this chain terminates, say $$Is^{-n} = Is^{-(n+1)}$$ for some $n$. Cancelling $s^n$ we get $I^e = I^es^{-1}$. Thus $I^er = I^e(s^{-1}r)$, as required.