For fractional ideal, why $AB=R$ implies $B=A^{-1}$?

Question

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

Answer 1

Also for the other readers: $A^{-1}$ is the largest fractional ideal with the property that $A A^{-1} \subseteq R$. Explicitly, we have $A^{-1} = \{x \in K : xA \subseteq R\}$, where $K$ is the field of fractions of $A$.

Thus, $AB=R$ clearly implies $B \subseteq A^{-1}$. Conversely, we have

$$A^{-1} = A^{-1} (A B) = (A^{-1} A) B \subseteq R B = B.$$

This proves $A^{-1} = B$. All this works for any integral domain $R$.

Answer 2

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".