## 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)

## Add a comment on tag `06SG`

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 lower-right corner).

All contributions are licensed under the GNU Free Documentation License.