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I have written a proof of the Going-down theorem that doesn't use some of the assumptions so it's false but I can't find the mistake. Can you tell where it's wrong?

*Going-down*$^\prime$: Let $R,S$ be rings such that $S \subset R$ and $R$ is integral over $S$. Let $q_1$ be a prime ideal in $R$ such that $p_1 = q_1 \cap S$ is a prime ideal in $S$. Let $p_2 \subset p_1$ be another prime ideal in $S$. Then there exists a prime ideal $q_2$ in $R$ s.t. $q_2 \cap S = p_2$. (Note: the missing assumptions are that $R,S$ are integral domains and $S$ is integrally closed)

Theorems we use for the false proof:

1.(5.10) Let $A \subset B$ be rings, $B$ integral over $A$, and let $p$ be a prime ideal of $A$. Then there exists a prime ideal $q$ of $B$ such that $q \cap A = p$.

2.(5.6.)If $S$ is a multiplicatively closed subset of $A$ and $B$ is integral over $A$ then $S^{-1}B$ is integral over $S^{-1}A$.

3.(3.11 iv)) The prime ideals of $S^{-1}A$ are in one-to-one correspondence ($p \leftrightarrow S^{-1}p$) with prime ideals of $A$ which don't meet $S$.

False proof:

By (5.6), $R_{q_1}$ is integral over $S_{p_1}$. By (3.11), $\overline{p_2} = {p_2}_{p_1}$ is prime in $S_{p_1}$ since $p_2 \subset p_1$ by assumption. By (5.10) there exists a prime ideal $\overline{q_2}$ in $R_{q_1}$ such that $\overline{q_2} \cap S_{p_1} = \overline{p_2}$. We know that the following diagram commutes: $$ \begin{matrix} S & \xrightarrow{i} & R \\ \left\downarrow{\psi}\vphantom{\int}\right. & & \left\downarrow{\varphi}\vphantom{\int}\right.\\ S_{p_1}& \xrightarrow{i_{p_1}} & R_{q_1} \end{matrix} $$

We claim that $q_2 = \varphi^{-1}\overline{q_2}$ is an ideal such that $q_2 \subset q_1$ and $q_2 \cap S = p_2$. The claim $q_2 \subset q_1$ follows by construction (or from (3.11)). We also have $$ \varphi^{-1} (\overline{q_2}) \cap S = i^{-1}\varphi^{-1}(\overline{q_2}) = \psi^{-1}i^{-1}_{p_1}(\overline{q_2}) = \psi^{-1}(S_{p_1} \cap \overline{q_2} ) = \psi^{-1}(\overline{q_2}) = p_2$$

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I like this idea of giving false proofs: it keeps one awake! +1 –  Georges Elencwajg Jul 17 '12 at 10:32
@GeorgesElencwajg Thank you! : ) –  Rudy the Reindeer Jul 17 '12 at 10:48

1 Answer 1

up vote 3 down vote accepted

You begin your proof with "By (5.6), $R_{q_1}$ is integral over $S_{p_1}$"
But this recourse to (5.6) is illegitimate: the multiplicative set here is $\Sigma=S\setminus \mathfrak p_1 $ .
However $R_{\mathfrak q_1}=T^{-1}R $ where $T=R\setminus \mathfrak {q}_1$ and we only have $\Sigma \subset T$, not $\Sigma=T$.
(Don't think, draw a Venn diagram!)

So $S_{\mathfrak p_1}\to \Sigma^{-1} R$ is indeed integral, but $S_{p_1}\to R_{q_1}$ has no reason to be integral.

In Exercise 4 of the same chapter 5 the authors give an example where indeed $R_{q_1}$ is not integral over $S_{p_1}$, so that it is not just that one cannot apply (5.6), but the result (unsurprisingly) is impossible to deduce from the too weak hypothesis.

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Thank you very much for finding my mistake. I'd spent quite some time trying to find it on my own. –  Rudy the Reindeer Jul 17 '12 at 11:39
You're welcome, Clark. As I commented above, it is a pleasure to think about that sort of questions ! –  Georges Elencwajg Jul 17 '12 at 11:50

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