Lemma 10.149.4. Let $A \to B$ be a formally unramified ring map. Let $\varphi : B' \to B$ be the universal first order thickening of $B$ over $A$.

Let $S \subset A$ be a multiplicative subset. Then $S^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $S^{-1}A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/S^{-1}A}$.

Let $S \subset B$ be a multiplicative subset. Then $S' = \varphi ^{-1}(S)$ is a multiplicative subset in $B'$ and $(S')^{-1}B' \to S^{-1}B$ is the universal first order thickening of $S^{-1}B$ over $A$. In particular $S^{-1}C_{B/A} = C_{S^{-1}B/A}$.

Note that the lemma makes sense by Lemma 10.148.4.

**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

\[ \xymatrix{ B \ar[r] & S^{-1}B \ar[r] & D/I \\ A \ar[r] \ar[u] & S^{-1}A \ar[r] \ar[u] & D \ar[u] } \]

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

\[ \xymatrix{ B \ar[r] & S^{-1}B \ar[r] & D/I \\ A \ar[r] \ar[u] & A \ar[r] \ar[u] & D \ar[u] } \]

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.
$\square$

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