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 17.27.7. 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). $\square$

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).