# The Stacks Project

## Tag 06SG

Lemma 80.4.9. Let $R', R \in \mathop{\rm Ob}\nolimits(\widehat{\mathcal{C}}_\Lambda)$. 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 $$\mathfrak m_R/\mathfrak m_R^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus \bigoplus kX_i$$ 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 80.4.2. 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. $\square$

The code snippet corresponding to this tag is a part of the file formal-defos.tex and is located in lines 1097–1106 (see updates for more information).

\begin{lemma}
\label{lemma-power-series}
Let $R', R \in \Ob(\widehat{\mathcal{C}}_\Lambda)$. 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.
\end{lemma}

\begin{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
$$\mathfrak m_R/\mathfrak m_R^2 = \mathfrak m_{R'}/\mathfrak m_{R'}^2 \oplus \bigoplus kX_i$$
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 \ref{lemma-surjective-cotangent-space}.
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.
\end{proof}

Comment #2637 by Xiaowen Hu on July 9, 2017 a 3:38 am UTC

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

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