For fractional ideal, why $AB=R$ implies $B=A^{-1}$? Let $A,B$ be two fractional ideals of $R$ (an integral domain). Could anyone tell me why $AB=R$ implies $B=A^{-1}$?

The inverse of an ideal $I\subseteq R$ is defined as $$I^{-1}:=\{x\in k(R) : xI\subseteq R\}$$ Then if $R$ is a Dedekind domain we have $II^{-1}=R$. Now in a Dedekind domain the set of fractional ideals forms a group under the following operation: $$(r^{-1}I)\cdot (s^{-1} J):=(rs)^{-1}IJ$$ where $r,s\in R$ and $I,J$ are ideals (note that any fractional ideal can be written as $r^{-1}I$). The inverse in this group is given by $$(r^{-1}I)^{-1}=rI^{-1}$$.
So you can not define inverse of a fractional ideal in any general domain, it only makes sense in a dedekind domain and your claim follows by the definition of "inverse".
