Lemma 10.134.11. Let $S \subset A$ is a multiplicative subset of $A$. Let $S^{-1}A \to B$ be a ring map. Then $\mathop{N\! L}\nolimits _{B/A} \to \mathop{N\! L}\nolimits _{B/S^{-1}A}$ is a homotopy equivalence.

Proof. Choose a presentation $\alpha : P \to B$ of $B$ over $A$. Then $\beta : S^{-1}P \to B$ is a presentation of $B$ over $S^{-1}A$. A direct computation shows that we have $\mathop{N\! L}\nolimits (\alpha ) = \mathop{N\! L}\nolimits (\beta )$ which proves the lemma as the naive cotangent complex is well defined up to homotopy by Lemma 10.134.2. $\square$

Comment #2797 by Dario Weißmann on

The second sentence of the statement should be: ... is a homotopy equivalence.

Comment #8130 by Et on

It may be worth saying a word on why the two complexes are equal. The equality in degree 0 is clear, but in degree 1 it relies on the fact we are quotienting by I^2 and the surjectivity of the presentation.

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