Lemma 33.16.5. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point and let $s = f(x) \in S$. Assume that $\kappa (x) = \kappa (s)$. Then there are canonical isomorphisms
\[ \mathfrak m_ x/(\mathfrak m_ x^2 + \mathfrak m_ s\mathcal{O}_{X, x}) = \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x) \]
and
\[ T_{X/S, x} = \mathop{\mathrm{Hom}}\nolimits _{\kappa (x)}( \mathfrak m_ x/(\mathfrak m_ x^2 + \mathfrak m_ s\mathcal{O}_{X, x}), \kappa (x)) \]
This works more generally if $\kappa (x)/\kappa (s)$ is a separable algebraic extension.
Proof.
The second isomorphism follows from the first by Lemma 33.16.4. For the first, we can replace $S$ by $s$ and $X$ by $X_ s$, see Morphisms, Lemma 29.32.10. We may also replace $X$ by the spectrum of $\mathcal{O}_{X, x}$, see Modules, Lemma 17.28.7. Thus we have to show the following algebra fact: let $(A, \mathfrak m, \kappa )$ be a local ring over a field $k$ such that $\kappa /k$ is separable algebraic. Then the canonical map
\[ \mathfrak m/\mathfrak m^2 \longrightarrow \Omega _{A/k} \otimes \kappa \]
is an isomorphism. Observe that $\mathfrak m/\mathfrak m^2 = H_1(\mathop{N\! L}\nolimits _{\kappa /A})$. By Algebra, Lemma 10.134.4 it suffices to show that $\Omega _{\kappa /k} = 0$ and $H_1(\mathop{N\! L}\nolimits _{\kappa /k}) = 0$. Since $\kappa $ is the union of its finite separable extensions in $k$ it suffices to prove this when $\kappa $ is a finite separable extension of $k$ (Algebra, Lemma 10.134.9). In this case the ring map $k \to \kappa $ is étale and hence $\mathop{N\! L}\nolimits _{\kappa /k} = 0$ (more or less by definition, see Algebra, Section 10.143).
$\square$
Comments (0)