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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!

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  • $\begingroup$ 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 $? $\endgroup$
    – Pedro
    Dec 23, 2021 at 16:39
  • $\begingroup$ @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 $(df)?$ $\endgroup$
    – Tengen
    Dec 23, 2021 at 16:58
  • $\begingroup$ 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 ? $\endgroup$
    – Tengen
    Dec 23, 2021 at 17:06
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    $\begingroup$ 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 $d\alpha = \frac{-1}{f'(\alpha)} \sum_{n=0}^{\deg(f)} \alpha^n dc_n$ $\endgroup$
    – reuns
    Dec 23, 2021 at 17:20
  • $\begingroup$ @reuns that's true, I added the hypothesis "separated". $\endgroup$
    – Tengen
    Dec 23, 2021 at 19:24

1 Answer 1

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Instead of the conormal exact sequence, you can use the cotangent exact sequence

$$L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0$$

Since everything is just a field, we have smoothness (every vector space is free), so this sequence is actually exact on the left as well:

$$0 \to L \otimes_K \Omega_{K/k}^1 \to \Omega^1_{L/k} \to \Omega^1_{L/K} \to 0.$$

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.

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