Lemma 110.43.1. There exists a formally étale surjective ring map $A \to B$ with $L_{B/A}$ not equal to zero.

## 110.43 A formally étale ring map with nontrivial cotangent complex

Let $k$ be a field. Consider the ring

Let $A$ be the localization at the maximal ideal generated by all $x_ n, y_ n$ and denote $J \subset A$ the maximal ideal. Set $B = A/J$. By construction $J^2 = J$ and hence $A \to B$ is formally étale (see Section 110.42). We claim that the element $x_1 \otimes y_1$ is a nonzero element in the kernel of

Namely, $(A, J)$ is the colimit of the localizations $(A_ n, J_ n)$ of the rings

at their corresponding maximal ideals. Then $x_1 \otimes y_1$ corresponds to the element $x_ n^ n \otimes y_ n^ n \in J_ n \otimes _{A_ n} J_ n$ and is nonzero (by an explicit computation which we omit). Since $\otimes $ commutes with colimits we conclude. By [III Section 3.3, cotangent] we see that $J$ is not weakly regular. Hence by [III Proposition 3.3.3, cotangent] we see that the cotangent complex $L_{B/A}$ is not zero. In fact, we can be more precise. We have $H_0(L_{B/A}) = \Omega _{B/A}$ and $H_1(L_{B/A}) = 0$ because $J/J^2 = 0$. But from the five-term exact sequence of Quillen's fundamental spectral sequence (see Cotangent, Remark 92.12.5 or [Corollary 8.2.6, Reinhard]) and the nonvanishing of $\text{Tor}_2^ A(B, B) = \mathop{\mathrm{Ker}}(J \otimes _ A J \to J)$ we conclude that $H_2(L_{B/A})$ is nonzero.

**Proof.**
See discussion above.
$\square$

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