# Going down theorem: Atiyah and Macdonald

This is the proof of the going down theorem from Atiyah and Macdonald. While the rest of the proof I can understand, what i dont is the sentence "we have to show that $$p_2$$ is the contraction of the prime ideal in the ring $$B_{q_1}$$ or equivalently that $$B_{q_1} p_2 \cap A = p_2$$."

Which is the prime ideal in $$B_{q_1}$$ ? Is it $$B_{q_1} p_2$$ ? If so, why is it a prime ideal? Does it mean I have to show $$B_{p_2}$$ is a prime ideal of $$B$$ ? If so, does that mean $$q_2 = B_{p_2}$$ ?

Then how do I show $$B_{p_2}$$ intersect $$A = p_2$$?

Also how do I show $$q_1$$ contains $$q_2$$?

Fabio's answer has everything you need but I am going to elaborate with extreme verbosity. It is not to patronize, but simply because I studied this theorem for my senior project in undergrad and it was confusing and not intuitive to me, and hard to organize the argument for some reason. There are a lot of hypotheses in this result and we also use many small results and identities to prove it without stating them. I hope this helps, but if not, hopefully it helps someone like me who really needed it!

After justifying the reduction to the case where $$m = 1, n = 2$$ as in (5.11) here are our assumptions:

1. $$A$$ is a subring of $$B$$, both are integral domains
2. $$A$$ is integrally closed
3. $$B$$ is integral over $$A$$
4. $$\mathfrak{p}_1 \supseteq \mathfrak{p}_2$$ is a chain of prime ideals in $$A$$
5. $$\mathfrak{q}_1$$ is a prime ideal of $$B$$ such that $$q_1 \cap A = \mathfrak{p}_1$$

and to show the result we need to find a prime ideal $$\mathfrak{q}_2$$ of $$B$$ such that $$\mathfrak{q}_1 \supseteq \mathfrak{q}_2$$ and $$\mathfrak{q}_2 \cap A = \mathfrak{p}_2$$.

'what i dont is the sentence "we have to show that $$\mathfrak{p}_2$$ is the contraction of the prime ideal in the ring $$B_{\mathfrak{q}_1}$$..'

We will show that to prove the result it suffices to show that $$\mathfrak{p}_2$$ is the contraction of a prime ideal in the ring $$B_{\mathfrak{q}_1}$$. Let's review what that means and see why.

Since $$A$$ is a subring of $$B$$ we have a natural embedding $$f \colon A \to B$$ and we also have the natural map $$g \colon B \to B_{\mathfrak{q}_1}$$, then the composition $$g \circ f$$ is a map from $$A$$ into the localization such that for $$a \in A$$, $$g \circ f(a) = a/1$$ in $$B_{\mathfrak{q}_1}$$. Then by definition (page 9) $$\mathfrak{p}_2$$ is the contraction of a prime ideal of $$B_{\mathfrak{q}_1}$$ if there is a prime ideal, say $$\mathfrak{q}'$$, such that $$(g \circ f)^{-1}(\mathfrak{q}') = \mathfrak{p}_2$$. So assume this is true, that we found the ideal $$\mathfrak{q}'$$.

Now, as Fabio mentions, we want to look at Proposition 3.11 part iv) (page 41) which tells us the prime ideals of $$B_{\mathfrak{q}_1}$$ are in bijection with the prime ideals of $$B$$ which don't meet $$B - \mathfrak{q}_1$$. Looking at the proof of part iv) tells us how to find the bijective image of the ideal $$\mathfrak{q}'$$; we simply take the contraction under the natural map from $$B$$ into the localization, this is the map $$g$$ from the last paragraph. Remember that for $$b \in B$$ we have $$g(b) = b/1$$ in $$B_{\mathfrak{q}_1}$$. So, we have a prime ideal of $$\mathfrak{q}$$ of $$B$$ given by $$\mathfrak{q} = (\mathfrak{q}')^{c} = g^{-1}(\mathfrak{q}')$$. We need to check two things, that $$\mathfrak{q}_1 \supseteq \mathfrak{q}$$ and $$\mathfrak{q} \cap A = \mathfrak{p}_2$$.

For the first part, remember that by the result 3.11 part iv) we used we know that the ideal $$\mathfrak{q}$$ does not meet $$B - \mathfrak{q}_1$$, but this is equivalent to saying $$\mathfrak{q}_1 \supseteq \mathfrak{q}$$. Also, recall that $$(g \circ f)^{-1}(\mathfrak{q}') = \mathfrak{p}_2$$, but $$(g \circ f)^{-1}(\mathfrak{q}') = f^{-1}(g^{-1}(\mathfrak{q}')) = f^{-1}(\mathfrak{q}) = \mathfrak{q} \cap A,$$

where the last equality is by definition of that map $$f$$ and the second to last equality is because we already established that $$g^{-1}(\mathfrak{q}') = \mathfrak{q}$$. This means we have shown that the ideal $$\mathfrak{q}$$ satisfies the properties we needed, so $$\mathfrak{q}$$ is the ideal $$\mathfrak{q}_2$$ for the theorem - thus to prove the theorem we know it is sufficient to show $$\mathfrak{p}_2$$ is the contraction of a prime ideal from $$B_{\mathfrak{q}_1}$$. But instead of proving this, we will prove an equivalent condition.

"or equivalently that $$B_{\mathfrak{q}_1} \mathfrak{p}_2 ∩ A = \mathfrak{p}_2$$."

As both Fabio and the text mention, by Proposition 3.16 we know that $$\mathfrak{p}_2$$ is the contraction of a prime ideal of $$B_{\mathfrak{q}_1}$$ if and only if $${\mathfrak{p}_{2}}^{ec} = \mathfrak{p}_2$$. However, considering we are using the map $$g \circ f$$ from above, the extension is $${\mathfrak{p}_2}^{e} = B_{\mathfrak{q}_1} \mathfrak{p}_2$$ and then the contraction of that is just $${\mathfrak{p}_2}^{ec} = B_{\mathfrak{q}_1} \mathfrak{p}_2 \cap A$$. These can easily be worked out using the definition of the maps $$f,g$$ and the definitions of extensions / contractions.

Thus at this point, to prove the theorem it suffices to show $${\mathfrak{p}_2} = B_{\mathfrak{q}_1} \mathfrak{p}_2 \cap A$$ which is what Atiyah & MacDonald give a proof of.

So the logic of the whole argument is

\begin{align*}{\mathfrak{p}_2} = B_{\mathfrak{q}_1} \mathfrak{p}_2 \cap A &\iff \mathfrak{p}_2 \text{ is the contraction of a prime ideal in } B_{\mathfrak{q}_1}\\ &\implies \text{ there exists a prime ideal } \mathfrak{q}_2 \text{with } \mathfrak{q}_2 \cap A = \mathfrak{p}_2 \text{ and } \mathfrak{q}_1 \supseteq \mathfrak{q}_2. \end{align*}

So if you understand the entirety of the proof that follows in the text that $${\mathfrak{p}_2} = B_{\mathfrak{q}_1} \mathfrak{p}_2 \cap A$$ then you are done (at least logically), because the arguments I give above justify that the result is satisfied. However, judging from your question

"Which is the prime ideal in Bq1?"

it seems like you want to know what the actual ideal $$\mathfrak{q}_2$$ is. Understand first that the way we are proving this theorem we don't need to explicitly find the ideal. But, for the sake of curiosity, to get a constructive idea you will want to chase through the proofs of all the theorems we are applying starting with the proof of Proposition 3.16 on page 43. It shows that there must be some maximal ideal $$\mathfrak{m}$$ in $$((A - \mathfrak{p}_2)^{e})^{-1}B_{\mathfrak{q}_1}$$ and defines $$\mathfrak{q}$$ to be the contraction of $$\mathfrak{m}$$ back to $$B_{\mathfrak{q}_1}$$ and further shows that it satisfies $$\mathfrak{q}^c = \mathfrak{p}_2$$. Then combining this with our work from above we know that the real ideal we are looking for is $$\mathfrak{q}_2 = \mathfrak{q}^c$$ where we are contracting $$q$$ back to $$B$$ from the localization. Unfortunately, I think the proof of the existence of this maximal ideal $$\mathfrak{m}$$ uses Zorn's lemma and is non-constructive so in practice you may not be able to explicitly find the ideal $$\mathfrak{q}$$.

No, you don't have to do any of this. At least the way Atiyah & Macdonald have set it up, we just need to show $${\mathfrak{p}_2} = B_{\mathfrak{q}_1} \mathfrak{p}_2 \cap A$$.
• Nice answer! Isn't ${\mathfrak{p}_2}^{ec} = (B_{\mathfrak{q}_1} \mathfrak{p}_2)^c \cap A$? – Shivering Soldier Nov 20 '19 at 2:55
• Technically yes. The way I have it written is an abuse of notation, but in what you have written the contraction on the left is already going to be a subset of $A$ so there is no reason to intersect with $A$ at that point. I wrote it the way I did since that is how it is in the picture OP posted :) – Prince M Nov 20 '19 at 3:57
• Unless the way you have it written means to contract the left back to $B$, then intersect with $A$ – Prince M Nov 20 '19 at 4:03
By Proposition 3.11(iv) there exists a bijection between primes of $$B$$ contained in $$\mathfrak q_1$$ and primes of $$B_{\mathfrak q_1}$$. By Proposition 3.16, $$\mathfrak p_2$$ is the contraction of a prime ideal of $$B_{\mathfrak q_1}$$ if and only if $$\mathfrak p_2=\mathfrak p_2B_{\mathfrak q_1}\cap A$$.