# The field trace is surjective over the fixed field

This problem it's from Stewart Galois Theory book. I want to solve it only using the theorems used on the book. Or at least simple tools.

Let $K$ be a field of characteristic $0$. And let $L/K$ be a finite and normal extension (thus in this case Galois extension , since is also separable). For any $a\in L$ define $T(a)= \sum_{\sigma\in G} \sigma (a)$ Where $G = Gal (L,K)$. Prove that $T(a)\in K$ and that $T$ is a surjective map $L\to K$.

Well it's clear that $T(a)\in K$ since is fixed by all the element of the galois group, and then it's just to consider the Galois theorem. But how I can prove that it's surjective?

I realized that $T$ is also a linear transformation, maybe I can use the dimension theorem is some way. But I don't know how to compare the dimension of the kernel, with the dimension of L over K. Or maybe that it's not a good way to solve the problem.

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Hint: What is the trace of an element of $K$?
Addendum: Also it is crucial here that the characteristic of your field is $0$ as Makoto points out below. Otherwise the proof has to use slightly more sophisticated machinery.
By the way, what if $char(K) = p > 0$ and $L/K$ is separable and $(L:K)$ is divisible by $p$? –  Makoto Kato Dec 4 '12 at 16:31
I think that I did it, but I'm not sure. Look this. Since $T:E\to F$ is a F-linear transformation, then $Im T \le F$ is a F-subspace of F. Since $dim_F F =1$ the image has dimension 1 or 0. But it's not zero , since the characteristic is not zero. Then por example 1 is not in the kernel. And Then $dim_F Im T = 1$ and since it's also contained in F , then $Im T = F$ –  Daniel Dec 4 '12 at 16:44
Yes, your solution is great. You should post it as an answer and accept it (posting answers to your own questions is encouraged on this site). Btw, I hope you are using characteristic $0$ (you said not $0$ in your comment above) to show that the trace of $1$ is not zero, or something like that. –  Rankeya Dec 4 '12 at 16:49