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
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