Proposition 10.38.7. Let $R \subset S$ be an inclusion of domains. Assume $R$ is normal and $S$ integral over $R$. Let $\mathfrak p \subset \mathfrak p' \subset R$ be primes. Let $\mathfrak q'$ be a prime of $S$ with $\mathfrak p' = R \cap \mathfrak q'$. Then there exists a prime $\mathfrak q$ with $\mathfrak q \subset \mathfrak q'$ such that $\mathfrak p = R \cap \mathfrak q$. In other words: the going down property holds for $R \to S$, see Definition 10.41.1.

Proof. Let $\mathfrak p$, $\mathfrak p'$ and $\mathfrak q'$ be as in the statement. We have to show there is a prime $\mathfrak q$, with $\mathfrak q \subset \mathfrak q'$ and $R \cap \mathfrak q = \mathfrak p$. This is the same as finding a prime of $S_{\mathfrak q'}$ mapping to $\mathfrak p$. According to Lemma 10.17.9 we have to show that $\mathfrak p S_{\mathfrak q'} \cap R = \mathfrak p$. Pick $z \in \mathfrak p S_{\mathfrak q'} \cap R$. We may write $z = y/g$ with $y \in \mathfrak pS$ and $g \in S$, $g \not\in \mathfrak q'$. Written differently we have $zg = y$.

By Lemma 10.38.4 there exists a monic polynomial $P = x^ m + b_{m-1} x^{m-1} + \ldots + b_0$ with $b_ i \in \mathfrak p$ such that $P(y) = 0$.

By Lemma 10.38.6 the minimal polynomial of $g$ over $K$ has coefficients in $R$. Write it as $Q = x^ n + a_{n-1} x^{n-1} + \ldots + a_0$. Note that not all $a_ i$, $i = n-1, \ldots , 0$ are in $\mathfrak p$ since that would imply $g^ n = \sum _{j < n} a_ j g^ j \in \mathfrak pS \subset \mathfrak p'S \subset \mathfrak q'$ which is a contradiction.

Since $y = zg$ we see immediately from the above that $Q' = x^ n + za_{n-1} x^{n-1} + \ldots + z^{n}a_0$ is the minimal polynomial for $y$. Hence $Q'$ divides $P$ and by Lemma 10.38.5 we see that $z^ ja_{n - j} \in \sqrt{(b_0, \ldots , b_{m-1})} \subset \mathfrak p$, $j = 1, \ldots , n$. Because not all $a_ i$, $i = n-1, \ldots , 0$ are in $\mathfrak p$ we conclude $z \in \mathfrak p$ as desired. $\square$

Comment #4230 by Daniel Gulotta on

I believe this argument still works if the assumpton "$S$ is a domain'' is replaced with the weaker assumption "$S$ is torsionfree as an $R$-module".

Comment #4410 by on

You could also say this: if $R \to S$ is a ring map and $R \to S/\mathfrak s$ has going down for every minimal prime ideal $\mathfrak s$ of $S$, then $R \to S$ has going down. Namely, every prime $\mathfrak q' \subset S$ is going to contain one of the minimal primes $\mathfrak s$ of $S$ and then you can work in $S/\mathfrak s$.

Thus if $R$ is a normal domain, $R \to S$ is integral, and every minimal prime of $S$ maps to the generic point $(0)$ of $\text{Spec}(R)$, then we have going down for $R \to S$ by the proposition and what I just said. Your statement is a special case of this.

If somebody needs this I will add it to the Stacks project, but there currently isn't a really good place to put it.

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