Consider the conormal sequence proposition from Eisenbud's Commutative Algebra:
If $\pi:S\to T$ is a surjective morphism of $R$-algebras, and $I=\ker(\pi)$ we have an exact sequence $$I/I^2\to T\otimes_S\Omega_{S/R}\to \Omega_{T/R}\to 0.$$
If we consider $(R,m,K=R/m)$ a local ring such that $\exists K_0\subseteq R$ a subfield such that $\pi_{|K_0}:K_0\to K$ is an isomorphism. Note that $R=K_0\oplus m$. We have that $R,K$ are $K_0$-algebras and $\pi:R\to K$ an epimorphism, so by the proposition we have $$m/m^2\to K\otimes_R\Omega_{R/K_0}\to \Omega_{K/K_0}\to 0$$
with the left hand map $\delta(\overline{x})=1\otimes dx$.
We have that $\delta$ is surjective and to show that it's injective, we show $\delta^{*}:\text{Hom}_K(\Omega_{R/K_0},K)\to \text{Hom} _K(m/m^2,K)$ is surjective. For $\phi\in \text{Hom}_K(m/m^2,K)$ I want to define a $R$-bilinear map $f:\Omega_{R/K_0}\oplus K\to K$ to use the universal property of the tensor product to get something in $\text{Hom}_K(\Omega_{R/K_0},K)$. But my attempts don't work. For example I find $f(d(k_0+u),\lambda)=\lambda k_0\phi(u)$ is not bilinear (we identify $k_0$ with it's image in $K$).
Is this a good idea ?
Best Answer
It is probably simpler to use that for an $R$-module $N$, and $I$ an ideal, $N \otimes_{R/I} R/I \cong N/IN$ rather than using universal property of tensor products. (See https://amathew.wordpress.com/2010/02/15/kahler-differentials-ii for a write up.)
However, if you are set on using universal property of tensor products, you can define
$$f(rd(k_0 + u), \lambda) := r\lambda \phi(u)$$