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I'm trying to understand the proof of the going down theorem in Introduction to Commutative Algebra by Atiyah and Macdonald. My main confusion is when they say it suffices to show that $B_{\mathfrak q_1}\mathfrak p_2\cap A = \mathfrak p_2$. Isn't it true that $x\in B_{\mathfrak q_1}\mathfrak p_2\cap A$ iff $x \in A$ and $x/1 \in B_{\mathfrak q_1}\mathfrak p_2$? I don't understand this notation.

user26857
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kfriend
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  • What is it that you don't understand in the notation? If you don't understand it, you might want to trace back and read the definitions. – Pedro Aug 10 '14 at 23:22
  • When we say that (B_1)2 intersect A = 2, do we mean that the contraction of (B_1)2 into B (under the standard map from B to B_1, given by f(x)=x/1), contracted again under the inclusion mapping into A, then intersected with A, is equal to 2? – kfriend Aug 11 '14 at 03:13
  • Bunch of symbols missing there. – Pedro Aug 11 '14 at 04:08
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    Well, if we write it carefully, $x \in \mathfrak{p}2 B{\mathfrak{q}_1} \cap A$ means that $x$ can be written as $b/s$ with $b \in \mathfrak{p}_2$ and $x$ can also be written as $c/1$, with $c \in A$. Note that the way to write a quotient is not uniquely determined.

    Could you write properly what's the relation between $\mathfrak{q}_1$ and $\mathfrak{p}_2$?

     – Pedro A. Castillejo Aug 11 '14 at 13:06
  • That makes sense. And for the going down theorem, we are given A $\subset$ B, A is integrally closed (in its field of fractions), B is integral over A, and given $p_1$ $\supset$ $p_2$, both prime ideals in A, and $q_1$ prime in B st $q_1 \cap A = p_1$ and we want to show there exists some $q_2 \in B$, prime, st $q_1 \supset q_2$ and $q_2 \cap A = p_2$ – kfriend Aug 11 '14 at 16:30

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