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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