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 10.118.12 (Krull-Akizuki). Let $R$ be a domain with fraction field $K$. Let $K \subset L$ be a finite extension of fields. Assume $R$ is Noetherian and $\dim (R) = 1$. In this case any ring $A$ with $R \subset A \subset L$ is Noetherian.

Proof. To begin we may assume that $L$ is the fraction field of $A$ by replacing $L$ by the fraction field of $A$ if necessary. Let $I \subset A$ be a nonzero ideal. Clearly $I$ generates $L$ as a $K$-vector space. Hence we see that $I \cap R \not= (0)$. Pick any nonzero $x \in I \cap R$. Then we get $I/xA \subset A/xA$. By Lemma 10.118.11 the $R$-module $A/xA$ has finite length as an $R$-module. Hence $I/xA$ has finite length as an $R$-module. Hence $I$ is finitely generated as an ideal in $A$. $\square$


Comments (2)

Comment #2596 by Aaron Landesman on

Here is a very minor comment: in the proof of Krull-Akizuki, you should assume is nonzero, or else the "Clearly ..." statement is false.


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