Let $K|F$ be a finite separable extension (algebraic), then show that $\operatorname{Tr}_{K|F} : K \to F$ is surjective.

Question

Let $K|F$ be a finite separable extension (algebraic), then show that $\DeclareMathOperator{\Tr}{Tr}\Tr_{K|F} : K \to F$ is surjective.

Note: $F$ is not assumed to be finite like here , so not a duplicate!

My attempt:

The finite case is done here , for $F$ infinite , $K$ must also be infinite. Now, I am trying to apply the following theorem (citation : Theorem 12.2 Page 311, Algebra - Lang (2nd Edition) ):

Let $K$ be an infinite field and let $\sigma_1,\dots,\sigma_n$ be the distinct elements of a finite group of automorphisms of $K$ . Then $\sigma_1,\dots,\sigma_n$ are algebraically independent.

Since, $K|F$ is given to be a finite separable extension, $\exists \alpha \in K$ s.t. $K=F(\alpha)$ . Since, $[K :F]= \deg(\min_F (\alpha))=n$ (say) , Consider the Galois closure (which in this case is just the normal closure) say L, then $L|F$ is Galois and $\operatorname{Gal}(L|F)$ is isomorphic to a subgroup of $S_n$ (hence finite).

Then, using the above theorem, I could show that $ \Tr_{L|F}$ must be non-zero. Now since, $\Tr_{L|F} = \Tr_{K|F} \circ \Tr_{L|K}$ it follows that, $\Tr_{K|F}$ is non-zero. Since $\Tr_{K|F} : K \to F$ is an $F$-linear map, it follows that the map is surjective.

Please point out mistakes if any in the proof. Also I believe that my approach might be using too many results, can someone give a rather simple proof using relatively less machineries?

Thanks in advance for help!

Answer 1

Note that $Tr : K \rightarrow F$ is an $F$ linear functional . And hence it suffices to show that $Tr \neq 0$ .

Let $\sigma_i ; \ i=1, \dots, n$ denote the number of distinct $F$ embeddings of $K$ in $\bar F$.

Then $ Tr(\alpha)= \sum _{i=1} ^n \sigma_i (\alpha) $ (since $K|_{F} $ is separable)

By linear independence of characters $\exists \alpha_0 \in K $ such that $\sum _{i=1} ^n \sigma_i (\alpha_0) \neq 0$ that is $ Tr (\alpha_0)\neq 0$