How to show that the trace maps a Galois extension to the base field Let $K$ be a finite extension of the finite field $F$, then the trace is defined as
$$\operatorname{Tr}(\alpha) = \sum_{\sigma \,\in\, \operatorname{Gal}(K/F)}\sigma(\alpha)$$
How can one show that $\operatorname{Tr}(\alpha) \in F, \forall \alpha \in K$?
 A: Observe that for any $\;\tau\in \text{ Gal}\,(K/F)\;$ we get
$$\tau(Tr(\alpha))=\sum_{\sigma\in Gal(K/F)}\tau\sigma\alpha=Tr(\alpha)$$
since as $\;\sigma\;$ runs over all the elements of the Galois group so does $\;\tau\sigma\;$ , and this means the trace is in the fixed field of the whole group, which means $\;Tr(\alpha)\in F$
A: But in his original question, Anfänger did not assume that the extension K/F is Galois, although he states that his morphisms s run through a certain Gal(K/F). Perhaps it's just a matter of notations, but I think that the correct definition should be: K/F is separable, and Tr(a) is defined as the sum of the s(a) when s runs through the [K : F] embeddings of K into an algebraic closure which fix F pointwise. If K = F(a), an easy computation shows that - Tr(a) is the second coefficient of the monic irreducible polynomial for a over F, so Tr(a) belongs to F. In the general case, with F < F(a) < K, the "transition formula" for the trace maps does the job. 
NB : if K/F admits a non trivial inseparable part, the trace is identically null.
