Lemma 89.3.11. There is a canonical map

\[ \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \longrightarrow H_1(L_{k/\Lambda }). \]

If $k' \subset k$ is separable (for example if the characteristic of $k$ is zero), then this map induces an isomorphism $\mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \otimes _{k'} k = H_1(L_{k/\Lambda })$. If $k = k'$ (for example in the classical case), then $\mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 = H_1(L_{k/\Lambda })$. The composition

\[ \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \longrightarrow H_1(L_{k/\Lambda }) \longrightarrow \mathfrak m_ A/\mathfrak m_ A^2 \]

comes from the canonical map $\mathfrak m_\Lambda \to \mathfrak m_ A$.

**Proof.**
Note that $H_1(L_{k'/\Lambda }) = \mathfrak m_\Lambda /\mathfrak m_\Lambda ^2$ as $\Lambda \to k'$ is surjective with kernel $\mathfrak m_\Lambda $. The map arises from functoriality of the naive cotangent complex. If $k' \subset k$ is separable, then $k' \to k$ is an étale ring map, see Algebra, Lemma 10.143.4. Thus its naive cotangent complex has trivial homology groups, see Algebra, Definition 10.143.1. Then Algebra, Lemma 10.134.4 applied to the ring maps $\Lambda \to k' \to k$ implies that $\mathfrak m_\Lambda /\mathfrak m_\Lambda ^2 \otimes _{k'} k = H_1(L_{k/\Lambda })$. We omit the proof of the final statement.
$\square$

## Comments (0)

There are also: