The Stacks project

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

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

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

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

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

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

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. $\square$

Proof. Omitted. $\square$

Proof. With notation and assumptions as in (1). Let $(S^{-1}B)' \to S^{-1}B$ be the universal first order thickening of $S^{-1}B$ over $S^{-1}A$. Note that $S^{-1}B' \to S^{-1}B$ is a surjection of $S^{-1}A$-algebras whose kernel has square zero. Hence by definition we obtain a map $(S^{-1}B)' \to S^{-1}B'$ compatible with the maps towards $S^{-1}B$. Consider any commutative diagram

where $I \subset D$ is an ideal of square zero. Since $B'$ is the universal first order thickening of $B$ over $A$ we obtain an $A$-algebra map $B' \to D$. But it is clear that the image of $S$ in $D$ is mapped to invertible elements of $D$, and hence we obtain a compatible map $S^{-1}B' \to D$. Applying this to $D = (S^{-1}B)'$ we see that we get a map $S^{-1}B' \to (S^{-1}B)'$. We omit the verification that this map is inverse to the map described above.

With notation and assumptions as in (2). Let $(S^{-1}B)' \to S^{-1}B$ be the universal first order thickening of $S^{-1}B$ over $A$. Note that $(S')^{-1}B' \to S^{-1}B$ is a surjection of $A$-algebras whose kernel has square zero. Hence by definition we obtain a map $(S^{-1}B)' \to (S')^{-1}B'$ compatible with the maps towards $S^{-1}B$. Consider any commutative diagram

where $I \subset D$ is an ideal of square zero. Since $B'$ is the universal first order thickening of $B$ over $A$ we obtain an $A$-algebra map $B' \to D$. But it is clear that the image of $S'$ in $D$ is mapped to invertible elements of $D$, and hence we obtain a compatible map $(S')^{-1}B' \to D$. Applying this to $D = (S^{-1}B)'$ we see that we get a map $(S')^{-1}B' \to (S^{-1}B)'$. We omit the verification that this map is inverse to the map described above. $\square$

Proof. We are going to use the construction of $B'$ from the proof of Lemma 10.149.1 although in principle it should be possible to deduce these results formally from the definition. Namely, we choose a presentation $B = P/J$, where $P = A[x_ i]$ is a polynomial ring over $A$. Next, we choose elements $f_ i \in J$ such that $\text{d}f_ i = \text{d}x_ i \otimes 1$ in $\Omega _{P/A} \otimes _ P B$. Having made these choices we have $B' = P/J'$ with $J' = (f_ i) + J^2$, see proof of Lemma 10.149.1.

Consider the canonical exact sequence

see Lemma 10.131.9. By construction the classes of the $f_ i \in J'$ map to elements of the module $\Omega _{P/A} \otimes _ P B'$ which generate it modulo $J'/J^2$ by construction. Since $J'/J^2$ is a nilpotent ideal, we see that these elements generate the module altogether (by Nakayama's Lemma 10.20.1). This proves that $\Omega _{B'/A} = 0$ and hence that $B'$ is formally unramified over $A$, see Lemma 10.148.2.

Since $P$ is a polynomial ring over $A$ we have $\Omega _{P/R} = \Omega _{A/R} \otimes _ A P \oplus \bigoplus P\text{d}x_ i$. We are going to use this decomposition. Consider the following exact sequence

see Lemma 10.131.9. We may tensor this with $B$ and obtain the exact sequence

If we remember that $J' = (f_ i) + J^2$ then we see that the first arrow annihilates the submodule $J^2/(J')^2$. In terms of the direct sum decomposition $\Omega _{P/R} \otimes _ P B = \Omega _{A/R} \otimes _ A B \oplus \bigoplus B\text{d}x_ i $ given we see that the submodule $(f_ i)/(J')^2 \otimes _{B'} B$ maps isomorphically onto the summand $\bigoplus B\text{d}x_ i$. Hence what is left of this exact sequence is an isomorphism $\Omega _{A/R} \otimes _ A B \to \Omega _{B'/R} \otimes _{B'} B$ as desired. $\square$