The Stacks project

Lemma 10.149.1. 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 10.131.9. This is surjective since $\Omega _{S/R} = 0$ by assumption, see Lemma 10.148.2. 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. $\square$

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