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 10.37.2. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $A = \sum I^ nt^ n \subset R[t]$ be the subring of the polynomial ring generated by $R \oplus It \subset R[t]$. An element $s \in S$ is integral over $I$ if and only if the element $st \in S[t]$ is integral over $A$.

Proof. Suppose $st$ is integral over $A$. Let $P = x^ d + \sum _{j < d} a_ j x^ j$ be a monic polynomial with coefficients in $A$ such that $P^\varphi (st) = 0$. Let $a_ j' \in A$ be the degree $d-j$ part of $a_ i$, in other words $a_ j' = a_ j'' t^{d-j}$ with $a_ j'' \in I^{d-j}$. For degree reasons we still have $(st)^ d + \sum _{j < d} \varphi (a_ j'') t^{d-j} (st)^ j = 0$. Hence we see that $s$ is integral over $I$.

Suppose that $s$ is integral over $I$. Say $P = x^ d + \sum _{j < d} a_ j x^ j$ with $a_ j \in I^{d-j}$. Then we immediately find a polynomial $Q = x^ d + \sum _{j < d} (a_ j t^{d-j}) x^ j$ with coefficients in $A$ which proves that $st$ is integral over $A$. $\square$


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