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. $\square$

Comment #2637 by Xiaowen Hu on

The left handside of the second equality in the proof should be $\mathfrak{m}_S/\mathfrak{m}_S^2$.

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