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Prop. 3.16 tells us that if $f: A \to B$ is a ring homomorphism and $\mathfrak p$ is a prime ideal of $A$ then $\mathfrak p$ is the contraction of a prime ideal of $B$ if and only if $\mathfrak p^{ec} = \mathfrak p$.

This is used in the proof of the Going-down theorem: If I understand correctly, AM claim it's enough to show that $B_{\mathfrak q_1} \mathfrak p_2 \cap A = \mathfrak p_2$.

Let $i: A \to B$ be the inclusion. Then $B_{\mathfrak q_1} \mathfrak p_2 \cap A = i^{-1} B_{\mathfrak q_1} \mathfrak p_2 = (B_{\mathfrak q_1} \mathfrak p_2 )^c$.

Let $f:B \to B_{\mathfrak q_1}$ be the (inclusion) map $b \mapsto \frac{b}{1}$.

Then $(B_{\mathfrak q_1} \mathfrak p_2 )^c = \mathfrak p_2^{ec}$.

Here is my question: In prop. 3.16, $()^e$ and $()^c$ are both with respect to the same homomorphism. What's the justification that we may use prop. 3.16 in the proof of the Going-down theorem where $()^e$ and $()^c$ are with respect to two different homomorphisms? Or am I misreading the proof of the Going-down theorem?

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Both the extension and contraction are along the composite map $A \to B_{\mathfrak q_1}$, I think. – Dylan Moreland Jul 10 '12 at 15:23
@DylanMoreland Then perhaps this is a typo in the book. For the extension along $f \circ i (p_1)$ I get $$ f \circ i (p_1) = f(p_1) = (p_1)_{q_1} B_{q_1}$$ whereas in the book it's $$ (p_1) B_{q_1}$$ – Rudy the Reindeer Jul 11 '12 at 6:00
@DylanMoreland And then I need to apply the contraction. I see. Thank you, by now I'm starting to think that it works! – Rudy the Reindeer Jul 11 '12 at 6:03

I'm posting Dylan Moreland's comment above as a community-wiki answer in order to remove this from the unanswered queue.

Both the extension and contraction are along the composite map $A\rightarrow B_{q_1}$, I think.

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Would the downvoter care to explain? This comment is correct and was helpful to the asker (see comments above). – Potato Jul 9 '13 at 21:53

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