Lemma 89.9.5. There exists an $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$ such that the equivalent conditions of Lemma 89.9.3 hold and moreover $H_1(L_{k/\Lambda }) = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{R/\Lambda } \otimes _ R k = \Omega _{k/\Lambda }$.

**Proof.**
In the classical case we choose $R = \Lambda $. More generally, if the residue field extension $k/k'$ is separable, then there exists a unique finite étale extension $\Lambda ^\wedge \to R$ (Algebra, Lemmas 10.153.9 and 10.153.7) of the completion $\Lambda ^\wedge $ of $\Lambda $ inducing the extension $k/k'$ on residue fields.

In the general case we proceed as follows. Choose a smooth $\Lambda $-algebra $P$ and a $\Lambda $-algebra surjection $P \to k$. (For example, let $P$ be a polynomial algebra.) Denote $\mathfrak m_ P$ the kernel of $P \to k$. The Jacobi-Zariski sequence, see (89.3.10.2) and Algebra, Lemma 10.134.4, is an exact sequence

We have the $0$ on the left because $P/k$ is smooth, hence $\mathop{N\! L}\nolimits _{P/\Lambda }$ is quasi-isomorphic to a finite projective module placed in degree $0$, hence $H_1(\mathop{N\! L}\nolimits _{P/\Lambda } \otimes _ P k) = 0$. Suppose $f \in \mathfrak m_ P$ maps to a nonzero element of $\Omega _{P/\Lambda } \otimes _ P k$. Setting $P' = P/(f)$ we have a $\Lambda $-algebra surjection $P' \to k$. Observe that $P'$ is smooth at $\mathfrak m_{P'}$: this follows from More on Morphisms, Lemma 37.38.1. Thus after replacing $P$ by a principal localization of $P'$, we see that $\dim (\mathfrak m_ P/\mathfrak m_ P^2)$ decreases. Repeating finitely many times, we may assume the map $\mathfrak m_ P/\mathfrak m_ P^2 \to \Omega _{P/\Lambda } \otimes _ P k$ is zero so that the exact sequence breaks into isomorphisms $H_1(L_{k/\Lambda }) = \mathfrak m_ P/\mathfrak m_ P^2$ and $\Omega _{P/\Lambda } \otimes _ P k = \Omega _{k/\Lambda }$.

Let $R$ be the $\mathfrak m_ P$-adic completion of $P$. Then $R$ is an object of $\widehat{\mathcal{C}}_\Lambda $. Namely, it is a complete local Noetherian ring (see Algebra, Lemma 10.97.6) and its residue field is identified with $k$. We claim that $R$ works.

First observe that the map $P \to R$ induces isomorphisms $\mathfrak m_ P/\mathfrak m_ P^2 = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{P/\Lambda } \otimes _ P k = \Omega _{R/\Lambda } \otimes _ R k$. This is true because both $\mathfrak m_ P/\mathfrak m_ P^2$ and $\Omega _{P/\Lambda } \otimes _ P k$ only depend on the $\Lambda $-algebra $P/\mathfrak m_ P^2$, see Algebra, Lemma 10.131.11, the same holds for $R$ and we have $P/\mathfrak m_ P^2 = R/\mathfrak m_ R^2$. Using the functoriality of the Jacobi-Zariski sequence (89.3.10.3) we deduce that $H_1(L_{k/\Lambda }) = \mathfrak m_ R/\mathfrak m_ R^2$ and $\Omega _{R/\Lambda } \otimes _ R k = \Omega _{k/\Lambda }$ as the same is true for $P$.

Finally, since $\Lambda \to P$ is smooth we see that $\Lambda \to P$ is formally smooth by Algebra, Proposition 10.138.13. Then $\Lambda \to P$ is formally smooth for the $\mathfrak m_ P$-adic topology by More on Algebra, Lemma 15.37.2. This property is inherited by the completion $R$ by More on Algebra, Lemma 15.37.4 and the proof is complete. In fact, it turns out that whenever $\underline{R}|_{\mathcal{C}_\Lambda }$ is smooth, then $R$ is isomorphic to a completion of a smooth algebra over $\Lambda $, but we won't use this. $\square$

## 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)