# Invertibility of prime ideals in a number ring lying over prime numbers

I have trouble understanding an argument in the proof of the Kummer-Dedekind theorem. I am referring to a proof given in Peter Stevenhagen's notes. http://websites.math.leidenuniv.nl/algebra/ant.pdf

This is theorem 3.1 on page 27. For the proof of the second part, the first statement says that since $p_i$ contains $pR$, it is invertible. I am not sure what the author is invoking here. For every ideal contains a principal ideal, but that is not sufficient for invertibility. Also, the fact that $p_i$ contains $pR$ should follow from the first part of the theorem since $p_i=pR+g_i(\alpha)R$. So I don't see what fact is used in proving invertibility of $p_i$ based on its containment of $pR$. Any help would be very appreciated.

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Principal ideals are invertible, and so too are divisors of invertible ideals, so $\mathfrak{p}\:|\:pR\ \Rightarrow\ \mathfrak{p}\$ invertible.
Note that the ring in the theorem is not necessarily a a Dedekind domain so it need not be true that $\ I \supset J\ \Rightarrow\ I|J$
@Bill: I thought they were equivalent. That is the way the author defines division of ideals on page 14 of the above notes. How do you define $I|J$. I am sorry I am having a brain freeze here. –  Timothy Wagner Nov 15 '10 at 3:54
OK, I think I follow why $p_i$ is invertible, since $p_i J=pR$ so $p_i (Jp^{-1}R)=R$ where $J$ is the rest of the stuff in that product of $p_i$'s. I am guessing then, that $I|J$ means that there exists an ideal $A$ such that $IA=J$. –  Timothy Wagner Nov 15 '10 at 4:05
@Tim: It appears the author is inconsistent in his use of "divides". The standard definition for ideals in any ring is $\ I|J \iff I I' = J\$ for some $I'$. That's what is intended in the theorem. –  Bill Dubuque Nov 15 '10 at 4:08