# Thread: Principal ideal problem

1. ## Principal ideal problem

Let I be a principal ideal of a domain R, say I = aR. Show that $I^{-1} = a^{-1}R$ and that I is invertible.

2. Originally Posted by tttcomrader
Let I be a principal ideal of a domain R, say I = aR. Show that $I^{-1} = a^{-1}R$ and that I is invertible.
Define what it means $I^{-1}$.

3. The inverse of I is the set of $I^{-1} = \{ u \in K : uI \subseteq R \}$, where K is the quotient field of R.

I think I have the solution:

pick j to be an element of $I^{-1}$, then $jI \subset R$, so we have $ji=r$ for some r in R, and i in I. Let i = am, for some elements m in R.

So we have j(am) = r

$j(am)=r$
$j=rm^{-1}a^{-1}$
$j=a^{-1}(rm^{-1})$

$rm^{-1}$ is an element of R, so $I^{-1} \subseteq a^{-1}R$

Well, now I have to show that $a^{-1}R$ is a subseteq of of $I^{-1}$...