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*}

If $R$ is a summand of $S$ and $S$ is smooth over $R$, then the $I$-adic completion of $S$ is often a power series over $R$ where $I$ is the kernel of the projection map from $S$ to $R$.

Lemma 10.137.4. Let $\varphi : R \to S$ be a smooth ring map. Let $\sigma : S \to R$ be a left inverse to $\varphi $. Set $I = \mathop{\mathrm{Ker}}(\sigma )$. Then

  1. $I/I^2$ is a finite locally free $R$-module, and

  2. if $I/I^2$ is free, then $S^\wedge \cong R[[t_1, \ldots , t_ d]]$ as $R$-algebras, where $S^\wedge $ is the $I$-adic completion of $S$.

Proof. By Lemma 10.130.10 applied to $R \to S \to R$ we see that $I/I^2 = \Omega _{S/R} \otimes _{S, \sigma } R$. Since by definition of a smooth morphism the module $\Omega _{S/R}$ is finite locally free over $S$ we deduce that (1) holds. If $I/I^2$ is free, then choose $f_1, \ldots , f_ d \in I$ whose images in $I/I^2$ form an $R$-basis. Consider the $R$-algebra map defined by

\[ \Psi : R[[x_1, \ldots , x_ d]] \longrightarrow S^\wedge , \quad x_ i \longmapsto f_ i. \]

Denote $P = R[[x_1, \ldots , x_ d]]$ and $J = (x_1, \ldots , x_ d) \subset P$. We write $\Psi _ n : P/J^ n \to S/I^ n$ for the induced map of quotient rings. Note that $S/I^2 = \varphi (R) \oplus I/I^2$. Thus $\Psi _2$ is an isomorphism. Denote $\sigma _2 : S/I^2 \to P/J^2$ the inverse of $\Psi _2$. We will prove by induction on $n$ that for all $n > 2$ there exists an inverse $\sigma _ n : S/I^ n \to P/J^ n$ of $\Psi _ n$. Namely, as $S$ is formally smooth over $R$ (by Proposition 10.136.13) we see that in the solid diagram

\[ \xymatrix{ S \ar@{..>}[r] \ar[rd]_{\sigma _{n - 1}} & P/J^ n \ar[d] \\ & P/J^{n - 1} } \]

of $R$-algebras we can fill in the dotted arrow by some $R$-algebra map $\tau : S \to P/J^ n$ making the diagram commute. This induces an $R$-algebra map $\overline{\tau } : S/I^ n \to P/J^ n$ which is equal to $\sigma _{n - 1}$ modulo $J^ n$. By construction the map $\Psi _ n$ is surjective and now $\overline{\tau } \circ \Psi _ n$ is an $R$-algebra endomorphism of $P/J^ n$ which maps $x_ i$ to $x_ i + \delta _{i, n}$ with $\delta _{i, n} \in J^{n -1}/J^ n$. It follows that $\Psi _ n$ is an isomorphism and hence it has an inverse $\sigma _ n$. This proves the lemma. $\square$


Comments (1)

Comment #2161 by Daniel Smolkin on

Suggested slogan: If R is a summand of S and S is smooth over R, then we can determine whether or not the I-adic completion of S is a power series over R by looking at I/I^2, where I is the kernel of the projection map from S to R.


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