The Stacks project

Lemma 89.4.9. Let $R', R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$. Suppose that $R = R' \oplus I$ for some ideal $I$ of $R$. Let $x_1, \ldots , x_ r \in I$ map to a basis of $I/\mathfrak m_ R I$. Set $S = R'[[X_1, \ldots , X_ r]]$ and consider the $R'$-algebra map $S \to R$ mapping $X_ i$ to $x_ i$. Assume that for every $n \gg 0$ the map $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ has a left inverse in $\mathcal{C}_\Lambda $. Then $S \to R$ is an isomorphism.

Proof. As $R = R' \oplus I$ we have

\[ \mathfrak m_ R/\mathfrak m_ R^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus I/\mathfrak m_ RI \]

and similarly

\[ \mathfrak m_ S/\mathfrak m_ S^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus \bigoplus kX_ i \]

Hence for $n > 1$ the map $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ induces an isomorphism on cotangent spaces. Thus a left inverse $h_ n : R/\mathfrak m_ R^ n \to S/\mathfrak m_ S^ n$ is surjective by Lemma 89.4.2. Since $h_ n$ is injective as a left inverse it is an isomorphism. Thus the canonical surjections $S/\mathfrak m_ S^ n \to R/\mathfrak m_ R^ n$ are all isomorphisms and we win. $\square$

Comments (2)

Comment #2637 by Xiaowen Hu on

The left handside of the second equality in the proof should be .

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Your email address will not be published.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 06SG. Beware of the difference between the letter 'O' and the digit '0'.