Lemma 109.42.1. There exists a formally étale surjective ring map $A \to B$ with $L_{B/A}$ not equal to zero.

## 109.42 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 109.41). 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 91.12.5 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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)