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*}

Lemma 10.112.2. Let $A \to B$ be a ring map. Assume

  1. $A \subset B$ is an extension of domains,

  2. the induced extension of fraction fields is finite,

  3. $A$ is Noetherian, and

  4. $A \to B$ is of finite type.

Let $\mathfrak p \subset A$ be a prime of height $1$. Then there are at most finitely many primes of $B$ lying over $\mathfrak p$ and they all have height $1$.

Proof. By the dimension formula (Lemma 10.112.1) for any prime $\mathfrak q$ lying over $\mathfrak p$ we have

\[ \dim (B_{\mathfrak q}) \leq \dim (A_{\mathfrak p}) - \text{trdeg}_{\kappa (\mathfrak p)} \kappa (\mathfrak q). \]

As the domain $B_\mathfrak q$ has at least $2$ prime ideals we see that $\dim (B_{\mathfrak q}) \geq 1$. We conclude that $\dim (B_{\mathfrak q}) = 1$ and that the extension $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is algebraic. Hence $\mathfrak q$ defines a closed point of its fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$, see Lemma 10.34.9. Since $B \otimes _ A \kappa (\mathfrak p)$ is a Noetherian ring the fibre $\mathop{\mathrm{Spec}}(B \otimes _ A \kappa (\mathfrak p))$ is a Noetherian topological space, see Lemma 10.30.5. A Noetherian topological space consisting of closed points is finite, see for example Topology, Lemma 5.9.2. $\square$


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