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 10.36.4. Let $R$ be a domain with fraction field $K$. If $u, v \in K$ are almost integral over $R$, then so are $u + v$ and $uv$. Any element $g \in K$ which is integral over $R$ is almost integral over $R$. If $R$ is Noetherian then the converse holds as well.

Proof. If $ru^ n \in R$ for all $n \geq 0$ and $v^ nr' \in R$ for all $n \geq 0$, then $(uv)^ nrr'$ and $(u + v)^ nrr'$ are in $R$ for all $n \geq 0$. Hence the first assertion. Suppose $g \in K$ is integral over $R$. In this case there exists an $d > 0$ such that the ring $R[g]$ is generated by $1, g, \ldots , g^ d$ as an $R$-module. Let $r \in R$ be a common denominator of the elements $1, g, \ldots , g^ d \in K$. It is follows that $rR[g] \subset R$, and hence $g$ is almost integral over $R$.

Suppose $R$ is Noetherian and $g \in K$ is almost integral over $R$. Let $r \in R$, $r\not= 0$ be as in the definition. Then $R[g] \subset \frac{1}{r}R$ as an $R$-module. Since $R$ is Noetherian this implies that $R[g]$ is finite over $R$. Hence $g$ is integral over $R$, see Lemma 10.35.3. $\square$


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