The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Proposition 10.37.7. 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 10.40.1.

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 10.16.9 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 10.37.4 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 10.37.6 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 10.37.5 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. $\square$


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