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*}

The henselization of a Noetherian pair is a Noetherian pair with the same completion

Lemma 15.12.4. Let $(A, I)$ be a pair with $A$ Noetherian. Let $(A^ h, I^ h)$ be as in Lemma 15.12.1. Then the map of $I$-adic completions

\[ A^\wedge \to (A^ h)^\wedge \]

is an isomorphism. Moreover, $A^ h$ is Noetherian, the maps $A \to A^ h \to A^\wedge $ are flat, and $A^ h \to A^\wedge $ is faithfully flat.

Proof. The first statement is an immediate consequence of Lemma 15.12.2 and in fact holds without assuming $A$ is Noetherian. In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of ├ętale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism. For each such $A \to B$ the induced map $A^\wedge \to B^\wedge $ is an isomorphism (see proof of Lemma 15.12.2). By Algebra, Lemma 10.96.2 the ring map $B \to A^\wedge = B^\wedge = (A^ h)^\wedge $ is flat for each $B$. Thus $A^ h \to A^\wedge = (A^ h)^\wedge $ is flat by Algebra, Lemma 10.38.6. Since $I^ h = IA^ h$ is contained in the Jacobson radical of $A^ h$ and since $A^ h \to A^\wedge $ induces an isomorphism $A^ h/I^ h \to A/I$ we see that $A^ h \to A^\wedge $ is faithfully flat by Algebra, Lemma 10.38.15. By Algebra, Lemma 10.96.6 the ring $A^\wedge $ is Noetherian. Hence we conclude that $A^ h$ is Noetherian by Algebra, Lemma 10.158.1. $\square$


Comments (1)

Comment #3830 by slogan_bot on

Suggested slogan: "Henselization of a Noetherian pair is Noetherian and does not affect completions"


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