I am trying to prove the general case of the incomparability theorem for integral extensions and I was stuck on one part, so I was hoping somebody could help me out. (It has been a while since I've looked at commutative algebra, so I may have just forgotten some basic facts.)
Let $A \subseteq B$ be an integral extension and let $q \subseteq b$ be ideals in $B$, with $q$ a prime ideal (and $b$ may not be prime). Suppose $q^c = b^c = p,$ a prime ideal in $A$. I want to show that $q = b.$
Localizing at $p$, we know that $B_p$ is integral over $A_p,$ and that the latter is a local ring. So $p_p,$ the extension of the prime ideal $p$ in $A_p$, is a maximal ideal. Let $q_p$ and $b_p$ be the extensions of $q$ and $b$ respectively in $B_p$. Now $q_p^c = q_p \cap A_p = (q \cap A)_p = p_p$ is maximal, so we have that $q_p$ is also a maximal ideal.
Because $b$ (and hence $b_p$) is not necessarily prime, we cannot do the same thing for $b$. Instead, let $q_p'$ be a maximal ideal of $B_p$ containing $b_p.$ Now here is my question: can we know that $q \subseteq b \subseteq q'$? We have that $q \subseteq b$ and $b_p \subseteq q_p',$ but because $b$ and $b_p$ are not prime, we cannot use the correspondence between prime ideals of $B$ and the ring of fractions $B_p$. Couldn't there be some weird way $b$ could act, so that the above inclusions hold, but $q'$ does not contain $b$?
The reason I want to answer this question is, if we know $q' \supseteq b \supset q,$ then $q'$ should also lie over $p$, and so by our usual incomparability theorem, $q' = q = b$.
Is this the right direction for the proof, and if so, how would I answer that question?