I'm to look at $\mathbb{Q}(\zeta_7)$ and $\mathbb{Q}(\zeta_{10})$, and they both have a common thing I can't see how to work around.
For $\mathbb{Q}(\zeta_7)$, skipping some details I doubt matter too much, I get that its Galois group has two subgroups, one of which corresponds to $\mathbb{Q}(i\sqrt{7})$ (which I got to calculating a trace), but the other subgroup, using the trace, gives me the element $\zeta + \zeta^6$; the problem is it turned out that $\mathbb{Q}(\zeta +\zeta^6)= \mathbb{Q}(\zeta_7)$. I tried the norm but that was worse; came out as 1.
Same issue with $\mathbb{Q}(\zeta_{10})$. This has a Galois group with only one subgroup, which using a trace to get a fixed element gave $\zeta+\zeta^9$ which has identical problems to the case above.
Given a group of automorphisms, are there any other methods of finding fixed elements to get the required subfield? I only know of the trace and norm, if there isn't some candidate suspected by inspection.
One way of finding an element fixed by automorphisms is to take the sum of an orbit: say $\sigma$ is an automorphism with $\sigma^4 = \mathrm{id}$, then $x + \sigma(x) + \sigma^2(x) + \sigma^3(x)$ is always fixed, regardless of $x$. Of course, whether or not this is an interesting example depends on what you choose for $x$. But, for example, you can immediately find $\zeta + \zeta^6$ this way, because that's just the sum of the orbit of $\zeta$ under complex conjugation.