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slijmigrd 2022-07-14 Answered
Let A B be rings, B integral over A; let q , q be prime ideals of B such that q q and q c = q c = p say. Then q = q .
Question 1. Why n c = n c = m?
My attempt: Since p q, we have m = S 1 p S 1 q = n B p . But m is maximal in A p , which is not necessarily maximal in B p . I can't get m = n by this.
Question 2. When we use that notation A p , which means the localization S 1 A of A at the prime ideal p of A. But in this corollary, p doesn't necessarily be a prime ideal of B. Why can he write B p ? Should we write S 1 B rigorously?
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Answers (1)

bap1287dg
Answered 2022-07-15 Author has 13 answers
To answer your first question, look at the commutative diagram
A B A p B p
Going the right way, n is contracted to p by assumption. Hence the same holds for the left way. But the map Spec A p Spec A is well known to be injective and the sole pre-image of p is m. Thus n is contracted to m by the map A p B p . Of course the same arguments works for n .
For your second question, note that B is an A-module, so this is just the usual notation M p , when M is an A-module and p a prime of A.
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