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 10.36.7. Let $R$ be a domain with fraction field $K$. Suppose $f = \sum \alpha _ i x^ i$ is an element of $K[x]$.

  1. If $f$ is integral over $R[x]$ then all $\alpha _ i$ are integral over $R$, and

  2. If $f$ is almost integral over $R[x]$ then all $\alpha _ i$ are almost integral over $R$.

Proof. We first prove the second statement. Write $f = \alpha _0 + \alpha _1 x + \ldots + \alpha _ r x^ r$ with $\alpha _ r \not= 0$. By assumption there exists $h = b_0 + b_1 x + \ldots + b_ s x^ s \in R[x]$, $b_ s \not= 0$ such that $f^ n h \in R[x]$ for all $n \geq 0$. This implies that $b_ s \alpha _ r^ n \in R$ for all $n \geq 0$. Hence $\alpha _ r$ is almost integral over $R$. Since the set of almost integral elements form a subring (Lemma 10.36.4) we deduce that $f - \alpha _ r x^ r = \alpha _0 + \alpha _1 x + \ldots + \alpha _{r - 1} x^{r - 1}$ is almost integral over $R[x]$. By induction on $r$ we win.

In order to prove the first statement we will use absolute Noetherian reduction. Namely, write $\alpha _ i = a_ i / b_ i$ and let $P(t) = t^ d + \sum _{j < d} f_ j t^ j$ be a polynomial with coefficients $f_ j \in R[x]$ such that $P(f) = 0$. Let $f_ j = \sum f_{ji}x^ i$. Consider the subring $R_0 \subset R$ generated by the finite list of elements $a_ i, b_ i, f_{ji}$ of $R$. It is a domain; let $K_0$ be its field of fractions. Since $R_0$ is a finite type $\mathbf{Z}$-algebra it is Noetherian, see Lemma 10.30.3. It is still the case that $f \in K_0[x]$ is integral over $R_0[x]$, because all the identities in $R$ among the elements $a_ i, b_ i, f_{ji}$ also hold in $R_0$. By Lemma 10.36.4 the element $f$ is almost integral over $R_0[x]$. By the second statement of the lemma, the elements $\alpha _ i$ are almost integral over $R_0$. And since $R_0$ is Noetherian, they are integral over $R_0$, see Lemma 10.36.4. Of course, then they are integral over $R$. $\square$


Comments (1)

Comment #744 by Wei Xu on

In the statement, I think it would be better if "integral over ", "almost integral over " are replaced by "integral over ", "almost integral over ".

And some typos, "such that " should be "such that ". "" in "It is still the case that is integral over " should be "".


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