Lemma 90.4.9. 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 90.4.2. 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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #2637 by Xiaowen Hu on
Comment #2660 by Johan on