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

$0 \to H_1(\mathop{N\! L}\nolimits _{k/\Lambda }) \to \mathfrak m_ P/\mathfrak m_ P^2 \to \Omega _{P/\Lambda } \otimes _ P k \to \Omega _{k/\Lambda } \to 0$

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. $\square$

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).