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 10.35.2. Let $\varphi : R \to S$ be a ring map. Let $y \in S$. If there exists a finite $R$-submodule $M$ of $S$ such that $1 \in M$ and $yM \subset M$, then $y$ is integral over $R$.

Proof. Let $x_1 = 1 \in M$ and $x_ i \in M$, $i = 2, \ldots , n$ be a finite set of elements generating $M$ as an $R$-module. Write $yx_ i = \sum \varphi (a_{ij}) x_ j$ for some $a_{ij} \in R$. Let $P(T) \in R[T]$ be the characteristic polynomial of the $n \times n$ matrix $A = (a_{ij})$. By Lemma 10.15.1 we see $P(A) = 0$. By construction the map $\pi : R^ n \to M$, $(a_1, \ldots , a_ n) \mapsto \sum \varphi (a_ i) x_ i$ commutes with $A : R^ n \to R^ n$ and multiplication by $y$. In a formula $\pi (Av) = y\pi (v)$. Thus $P(y) = P(y) \cdot 1 = P(y) \cdot x_1 = P(y) \cdot \pi ((1, 0, \ldots , 0)) = \pi (P(A)(1, 0, \ldots , 0)) = 0$. $\square$


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