Let I be a principal ideal of a domain R, say I = aR. Show that $\displaystyle I^{-1} = a^{-1}R$ and that I is invertible.
The inverse of I is the set of $\displaystyle I^{-1} = \{ u \in K : uI \subseteq R \} $, where K is the quotient field of R.
I think I have the solution:
pick j to be an element of $\displaystyle I^{-1}$, then $\displaystyle jI \subset R $, so we have $\displaystyle ji=r$ for some r in R, and i in I. Let i = am, for some elements m in R.
So we have j(am) = r
$\displaystyle j(am)=r$
$\displaystyle j=rm^{-1}a^{-1}$
$\displaystyle j=a^{-1}(rm^{-1})$
$\displaystyle rm^{-1} $ is an element of R, so $\displaystyle I^{-1} \subseteq a^{-1}R$
Well, now I have to show that $\displaystyle a^{-1}R$ is a subseteq of of $\displaystyle I^{-1}$...