The Stacks project

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 10.139.4. 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 10.131.10 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 10.138.13) 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 $J^{n - 1}$. 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 (3)

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.

Comment #8338 by Et on

Typo in the last paragraph: "is equal to σ^n−1 modulo J^n" should be "is equal to σ^n−1 modulo J^n-1"


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05D5. Beware of the difference between the letter 'O' and the digit '0'.