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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 :-)

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    "Local" means that the properties in question are stable under localisation, _and_ that proving the localised case is enough to prove the general case. Here it is even stable under quotients by prime ideals.2012-05-27

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Your proof may be fixed by working in just one quotient ring, namely $A/q$. But I would prove it as follows:

We may replace $A \to B$ by $A/p \to B/q$ and rename $q'$ by $q$. Then, $A \subseteq B$ are integral domains and $q \subseteq B$ is a prime ideal with $q \cap A = \{0\}$; the goal is to prove that $q=0$.

Well, choose any element $b \in q$ and an integral equation $b^n + a_{n-1} b^{n-1} + \dotsc + a_1 b + a_0 = 0$ with $n$ minimal and assume $b \neq 0$. We have $a_0 \in A \cap q = \{0\}$. Since $B$ is an integral domain, we may cancel $b$. But this contradicts the minimality. Done.

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    Since $p = A \cap q$, the monomorphism $A \to B$ extends to a monomorphism $A/p \to B/q$.2012-05-29