I am working on an alternative proof of Corollary 5.9, p.61 in Atiyah - MacDonald, "Introduction to Commutative Algebra". The Corollary reads as follows:
"If $A \subseteq B$ are rings, $B$ is integral over $A$ and $q \subseteq q'$ are prime ideals of $B$ such that their contractions in $A$ are equal to the ideal $p$, then $q=q'$. "
My idea is to try to show that for any element $x \in B$, $[x]_q=[x]_{q'}$, where $[x]_q$ is the image of $x$ in $B/q$. Towards this end, we know (by Proposition 5.6) that both $B/q, B/q'$ are integral over $A/p$. Thus there is a polynomial $\xi(t) \in {A/p}[t]$ that has roots both $[x]_q,[x]_{q'}$ and its degree is minimal with respect to this property. Now we can write $\xi(t)=(t-[x]_q)\xi'(t)$, where $deg(\xi'(t)) < deg(\xi(t))$. Substituting $t=[x]_{q'}$ we get $([x]_{q'}-[x]_q)\xi'([x]_{q'})=0$ and since all pertaining factor rings are integral domains (by the primality of the ideals) we must have $[x]_q=[x]_{q'}$.
Now, i realize that the above argument is not rigorous enough, since e.g. the quantity $[x]_{q'}-[x]_q$ does not make sense, since $[x]_{q'} \in B/q'$ and $[x]_q \in B/q$. Also it does not make sense to evaluate a polynomial with coefficients over $A/p$ to a point of e.g. $B/q$ (i realize though that e.g. $(A+q)/q \cong A/p$).
My question is: how can i make the above argument rigorous?
Thanks :-)