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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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You're right in looking for a reduction to a simpler case, but I think you've gone about it the wrong way. The question is local on $A$, so we may assume $A$ is a field and that $\mathfrak{p} = (0)$; to prove the claim it is enough to show that $\mathfrak{q}$ is maximal. –  Zhen Lin May 27 '12 at 19:52
    
What do you mean by saying "the question is local on A"? –  Manos May 27 '12 at 19:54
    
"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. –  Zhen Lin May 27 '12 at 21:27

1 Answer 1

up vote 3 down vote accepted

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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I understand the idea of your proof. However, $A/p$ is not a subring of $B/q$ and this confuses me: how can you say $A \subseteq B$? –  Manos May 28 '12 at 19:14
    
Since $p = A \cap q$, the monomorphism $A \to B$ extends to a monomorphism $A/p \to B/q$. –  Martin Brandenburg May 29 '12 at 9:26

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