# $\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}$ is an isomorphism when $K\subset L$ is finite

Let $$k\subset K\subset L$$ be field extensions. Then we have the following exact sequence of $$L$$-vector spaces: $$\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}\rightarrow \Omega^1_{L|K}\rightarrow 0.$$ If $$K\subset L$$ is algebraic then $$\Omega^1_{L|K}=0$$ and so we have a surjective $$L$$-linear map $$\Omega^1_{K|k}\otimes_KL\rightarrow\Omega^1_{L|k}.$$ I want to prove that if $$K\subset L$$ is finite and separated then this map is an isomorphism.

If $$K\subset L$$ is finite and separable then $$L\cong K[x]/(P)$$ for some polynomial $$P.$$ I thought of using the conormal exact sequence $$(P)/(P^2)\rightarrow \Omega^1_{K[x]|k}\otimes_{K[x]}L\rightarrow \Omega^1_{L|k}\rightarrow 0$$ but that doesn't seem to yield anything interesting.
Any help would be appreciated!

• If $L = K[x]/(f)$, then cannot you prove by hand that $\Omega_{L\mid k}^1$ is obtained by taking the quotient of forms on $K[x]$ by $df = 0$?
– Pedro
Dec 23, 2021 at 16:39
• @PedroTamaroff If I denote by $d:K[x]\rightarrow \Omega^1_{K[x]|k}$ and $d':L\rightarrow \Omega^1_{L|k}$ the canonical derivations then composing with the projection we get a drivation $d'\pi$ from $K[x]$ to $\Omega^1_{L|k}$ and so a unique $K[x]$-linear map $\varphi:\Omega^1_{K[x]|k}\rightarrow \Omega^1_{L|k}$ such that $\varphi d=d'\pi$ Clearly $\varphi$ is surjective and $\varphi (df)=0.$ But why does the kernel contain only $(df)?$ Dec 23, 2021 at 16:58
• So if $\varphi(g)=0$ then $d'\overline{g}=\overline{g'}d'\overline{x}=0$ so $g'\in (f)$ and so $dg=g'dx\in (df),$ right ? Dec 23, 2021 at 17:06
• You need $L/K$ to be separable, if $f(\alpha)=0$ with $f=\sum_{n=0}^d c_n x^n\in K[x]$ separable then $d\alpha = \frac{-1}{f'(\alpha)} \sum_{n=0}^{\deg(f)} \alpha^n dc_n$ Dec 23, 2021 at 17:20
• @reuns that's true, I added the hypothesis "separated". Dec 23, 2021 at 19:24

$$L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0$$
$$0 \to L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0.$$
Since $$L/K$$ is separable, $$\Omega_{L/K}^1 = 0$$ (easy exercise using explicit realization of differentials and separable polynomial), from which you obtain your desired isomorphism.