If $A \subseteq B$, suppose that there is a subset $S$ of $A$ that is closed under multiplication, every element of $S$ is a unit in $B$, and $B = \left \{ a/s : a \in A, s \in S\right \}$. If $a\in A—S$, show that $aB\cap A = aA$. ($A$ and $B$ are commutative rings)
I have many doubts with this demonstration, here is my attempt, I would appreciate any suggestion:
This exercise is from the book Field and Galois Theory by Patrick Morandi, And he on his website says that there is an error with this problem but I do not understand very well, someone could explain, the author says: " Page $239$, Exercise $18$. Part (b) is wrong as stated. If $A$ and $B$ are UFDs and a is assumed irreducible in $B$, then I think it is correct (and is what is needed in the proof of Theorem $4.5$)."
Take $x\in aB\cap A$ then $x = ab_1$ and $x = a_1$ for some $b_1 \in B$ and $a_1 \in A$ (I do not know what else I can do here). For the other contention take $x \in aA$, then $x = aa_1$, but since $A \subseteq B$ then $a_1\in B$ and thus $x\in aB$ so $x\in aB\cap A$. what do you think? why $a\in A—S$?