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 10.143.1. Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $R \to B$ be a ring map. If $h \in B[x]/(f)$ is integral over $R$, then the element $f' h$ can be written as $f'h = \sum _ i b_ i x^ i$ with $b_ i \in B$ integral over $R$.

Proof. Say $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ with $r_ i \in R$. There exists a finite free ring extension $B \subset B'$ such that $f = (x - \alpha _1) \ldots (x - \alpha _ d)$ for some $\alpha _ i \in B'$, see Lemma 10.134.9. Note that each $\alpha _ i$ is integral over $R$. We may represent $h = h_0 + h_1 x + \ldots + h_{d - 1} x^{d - 1}$ with $h_ i \in B$. Then it is a universal fact that

\[ f' h \equiv \sum \nolimits _{i = 1, \ldots , d} h(\alpha _ i) (x - \alpha _1) \ldots \widehat{(x - \alpha _ i)} \ldots (x - \alpha _ d) \]

as elements of $B[x]/(f)$. You prove this by evaluating both sides at the points $\alpha _ i$ over the ring $B_{univ} = \mathbf{Z}[\alpha _ i, h_ j]$ (some details omitted). By our assumption that $h$ satisfies $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ we see that

\[ h(\alpha _ i)^ e + r_1 h(\alpha _ i)^{e - 1} + \ldots + r_ e = 0 \]

in $B'$. Hence $h(\alpha _ i)$ is integral over $R$. Using the formula above we see that $f'h \equiv \sum _{j = 0, \ldots , d - 1} b'_ j x^ j$ in $B'[x]/(f)$ with $b'_ j \in B'$ integral over $R$. However, since $f' h \in B[x]/(f)$ and since $1, x, \ldots , x^{d - 1}$ is a $B'$-basis for $B'[x]/(f)$ we see that $b'_ j \in B$ as desired. $\square$


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