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

$\text{d} : \mathfrak m/\mathfrak m^2 \longrightarrow \Omega _{R/k} \otimes _ R \kappa (\mathfrak m)$

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

$\xymatrix{ R \ar[r] & \kappa \\ & k(\overline{x}_1, \ldots , \overline{x}_ r) \ar[lu]^\varphi \ar[u] }$

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

$P(y + \delta ) = P(y) + P'(y)\delta$

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! From a direct computation of $\Omega _{\kappa \oplus \mathfrak m/k}$ the lemma follows. $\square$

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).