Lemma 10.140.4. Let k be a field. Let R be a Noetherian local ring containing k. Assume that the residue field \kappa = R/\mathfrak m is a finitely generated separable extension of k. Then the map
is injective.
Lemma 10.140.4. Let k be a field. Let R be a Noetherian local ring containing k. Assume that the residue field \kappa = R/\mathfrak m is a finitely generated separable extension of k. Then the map
is injective.
Proof. We may replace R by R/\mathfrak m^2. Hence we may assume that \mathfrak m^2 = 0. By assumption we may write \kappa = k(\overline{x}_1, \ldots , \overline{x}_ r, \overline{y}) where \overline{x}_1, \ldots , \overline{x}_ r is a transcendence basis of \kappa over k and \overline{y} is separable algebraic over k(\overline{x}_1, \ldots , \overline{x}_ r). Say its minimal equation is P(\overline{y}) = 0 with P(T) = T^ d + \sum _{i < d} a_ iT^ i, with a_ i \in k(\overline{x}_1, \ldots , \overline{x}_ r) and P'(\overline{y}) \not= 0. Choose any lifts x_ i \in R of the elements \overline{x}_ i \in \kappa . This gives a commutative diagram
of k-algebras. We want to extend the left upwards arrow \varphi to a k-algebra map from \kappa to R. To do this choose any y \in R lifting \overline{y}. To see that it defines a k-algebra map defined on \kappa \cong k(\overline{x}_1, \ldots , \overline{x}_ r)[T]/(P) all we have to show is that we may choose y such that P^\varphi (y) = 0. If not then we compute for \delta \in \mathfrak m that
because \mathfrak m^2 = 0. Since P'(y)\delta = P'(\overline{y})\delta we see that we can adjust our choice as desired. This shows that R \cong \kappa \oplus \mathfrak m as k-algebras! Now either a direct computation of \Omega _{\kappa \oplus \mathfrak m/k} or an application of Lemma 10.131.10 finishes the proof. \square
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Comment #8331 by Et on
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