The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.145.1. Let $R \to S$ be a formally unramified ring map. There exists a surjection of $R$-algebras $S' \to S$ whose kernel is an ideal of square zero with the following universal property: Given any commutative diagram

\[ \xymatrix{ S \ar[r]_ a & A/I \\ R \ar[r]^ b \ar[u] & A \ar[u] } \]

where $I \subset A$ is an ideal of square zero, there is a unique $R$-algebra map $a' : S' \to A$ such that $S' \to A \to A/I$ is equal to $S' \to S \to A/I$.

Proof. Choose a set of generators $z_ i \in S$, $i \in I$ for $S$ as an $R$-algebra. Let $P = R[\{ x_ i\} _{i \in I}]$ denote the polynomial ring on generators $x_ i$, $i \in I$. Consider the $R$-algebra map $P \to S$ which maps $x_ i$ to $z_ i$. Let $J = \mathop{\mathrm{Ker}}(P \to S)$. Consider the map

\[ \text{d} : J/J^2 \longrightarrow \Omega _{P/R} \otimes _ P S \]

see Lemma 10.130.9. This is surjective since $\Omega _{S/R} = 0$ by assumption, see Lemma 10.144.2. Note that $\Omega _{P/R}$ is free on $\text{d}x_ i$, and hence the module $\Omega _{P/R} \otimes _ P S$ is free over $S$. Thus we may choose a splitting of the surjection above and write

\[ J/J^2 = K \oplus \Omega _{P/R} \otimes _ P S \]

Let $J^2 \subset J' \subset J$ be the ideal of $P$ such that $J'/J^2$ is the second summand in the decomposition above. Set $S' = P/J'$. We obtain a short exact sequence

\[ 0 \to J/J' \to S' \to S \to 0 \]

and we see that $J/J' \cong K$ is a square zero ideal in $S'$. Hence

\[ \xymatrix{ S \ar[r]_1 & S \\ R \ar[r] \ar[u] & S' \ar[u] } \]

is a diagram as above. In fact we claim that this is an initial object in the category of diagrams. Namely, let $(I \subset A, a, b)$ be an arbitrary diagram. We may choose an $R$-algebra map $\beta : P \to A$ such that

\[ \xymatrix{ S \ar[r]_1 & S \ar[r]_ a & A/I \\ R \ar[r] \ar@/_/[rr]_ b \ar[u] & P \ar[u] \ar[r]^\beta & A \ar[u] } \]

is commutative. Now it may not be the case that $\beta (J') = 0$, in other words it may not be true that $\beta $ factors through $S' = P/J'$. But what is clear is that $\beta (J') \subset I$ and since $\beta (J) \subset I$ and $I^2 = 0$ we have $\beta (J^2) = 0$. Thus the “obstruction” to finding a morphism from $(J/J' \subset S', 1, R \to S')$ to $(I \subset A, a, b)$ is the corresponding $S$-linear map $\overline{\beta } : J'/J^2 \to I$. The choice in picking $\beta $ lies in the choice of $\beta (x_ i)$. A different choice of $\beta $, say $\beta '$, is gotten by taking $\beta '(x_ i) = \beta (x_ i) + \delta _ i$ with $\delta _ i \in I$. In this case, for $g \in J'$, we obtain

\[ \beta '(g) = \beta (g) + \sum \nolimits _ i \delta _ i \frac{\partial g}{\partial x_ i}. \]

Since the map $\text{d}|_{J'/J^2} : J'/J^2 \to \Omega _{P/R} \otimes _ P S$ given by $g \mapsto \frac{\partial g}{\partial x_ i}\text{d}x_ i$ is an isomorphism by construction, we see that there is a unique choice of $\delta _ i \in I$ such that $\beta '(g) = 0$ for all $g \in J'$. (Namely, $\delta _ i$ is $-\overline{\beta }(g)$ where $g \in J'/J^2$ is the unique element with $\frac{\partial g}{\partial x_ j} = 1$ if $i = j$ and $0$ else.) The uniqueness of the solution implies the uniqueness required in the lemma. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 10.145: Conormal modules and universal thickenings

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04EB. Beware of the difference between the letter 'O' and the digit '0'.