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Tag 06SG

Chapter 80: Formal Deformation Theory > Section 80.4: The completed base category

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}

    Comments (1)

    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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