Let $B\subset A$ be commutative rings with identity. Furthermore $B$ is a domain. We are given the set $$l(B,A) = \{ b\in B\setminus \{0\}: B[b^{-1}] = A[b^{-1}]\} \cup \{0\},$$ where $B[b^{-1}]$ means the $B$-algebra generated by $b^{-1}$. It can be proved that $l(B,A)$ is an ideal. The problem is:
Show that if $l(B,A)$ is a prime ideal then $B$ is maximal in $A$.
Does anyone have any ideas?
Just to give this an answer: it seems that the problem was incorrectly stated in the source. As written the claim is not true. E.g. if $B= \mathbb C$ and $A=\mathbb C[x]$, then l(B,A)=0, which is prime, but there are many rings intermediate between A and B, e.g. $\mathbb C[x^2]$. Note that there is nothing special about the choice of $\mathbb C$; it could be any field, or even any integral domain. Another counterexample is given by $\mathbb Z \subset \mathbb Q$.