Showing $E^H = K(\operatorname{Tr}_{E/{E^H}}(a))$ Suppose $E = K(a)$ and $E/K$ is a Galois extension. Show that if $H \leq \operatorname{Gal}(E/K)$ then the fixed field $E^H = K(\operatorname{Tr}_{E/{E^H}}(a))$.
$K(\operatorname{Tr}_{E/{E^H}}(a)) \subseteq E^H$  is trivial, but what about the other containment?
 A: Let $K=\mathbb{Q}(i)$, $a=2^{1/4}$. Then $\text{Gal}(E/K)$ is the group of $4$-th roots of $1$, acting on $a$ by ($u^4=1$) $a\mapsto u a$. Take $H=\{1,-1\}$, then $\text{Tr}_{E/E^H}(a)=2^{1/4}-2^{1/4}=0$. The claim is therefore false.
A: Another counterexample - one that I should have seen right away - is the following:
Let $K$ be the field of two elements, and let $a$ be a root of $x^4+x+1=0$. In this case $E=K(a)$ is the field of 16 elements, so $Gal(E/K)$ is cyclic of order four. The Frobenius
mapping $F:E\rightarrow E, x\mapsto x^2$ is a generator of the Galois group. If $H$ is the subgroup generated by $F^2$, then we run into a similar problem, because
$$
Tr_{E/E^H}(a)=\sum_{\sigma\in H}\sigma(a)=a+a^4=1
$$
is in the prime field.
The most obvious positive result related to this question is the following. Let's make a stronger assumption that $a$ generates a normal basis of $E/K$, in other words, the set of conjugates of $a$ is a $K$-basis of $E$. In that case it is clear that
$$
s:=Tr_{E/E^H}(a)=\sum_{\sigma\in H}\sigma(a)
$$
is a fixed point of an automorphism $\tau\in Gal(E/K)$, iff $\tau\in H$, because otherwise we violate the linear independence (over $K$) of the conjugates of $a$. Thus, by Galois correspondence, the smallest extension field of $K$ containing $s$ is $K(s)=E^H.$
