Intermediate fields of $x^4+1$ Had this question on my test. We were asked to give irreducible polynomial for $e^\frac{2\pi i}{8}$ over $\mathbb{Q}$, which I said was $x^4+1$.
We were asked to find the splitting field of this polynomial, I said it was $K=\mathbb{Q}(e^\frac{2\pi i}{8})$, since the other roots of $x^4+1$ are powers of this root. [specifically third, fifth and seventh powers]
We were then asked to find the Galois group for this field, I said it was the Klein four group. Because once you permute one root into another, you find that the root you permuted to gets sent back. And then the other two roots get permuted. I.e. the Galois group is $\{e,(12)(34),(13)(24),(14)(23)\}$.
All the above I'm fairly confident about, although not completely certain, and I guess if I got any of that wrong my question is probably wrong as well.
We were asked to find the intermediate fields $\mathbb{Q}\subset L\subset K$. These correspond to subgroups of Klein four, of which there are three. Namely, the cyclic subgroup of order two generated by any non-identity element. I know that the intermediate field corresponding to a subgroup is the field fixed by that subgroup. But the only intermediate field I was able to find was $\mathbb{Q}(i)$, which is fixed by one out of three of those subgroups, while the other two double transpositions send $i \rightarrow -i$.
What are the other two intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}(e^\frac{2\pi i}{8})$?
 A: You already realized that "once you permute one root into another, you find that the root you permuted to gets sent back".
So, for example, lets take the element $\phi$ of the Galois group that sends $\zeta_8$ to $\zeta_8^3$ (and back). Then $\zeta_8 + \zeta_8^3$ is invariant under this map. This gives ${\mathbb Q}(\zeta_8 + \zeta_8^3)$ as the intermediate field that is invariant under $\langle \phi \rangle$.
If you want this field as something more recognisable, realise that it has degree $2$ over ${\mathbb Q}$. Writing $\alpha = \zeta_8 + \zeta_8^3$, you get $\alpha^2 = \zeta_8^2 + 2 \zeta_8^4 + \zeta_8^6 = -2$, so the field is ${\Bbb Q}(i\sqrt{2})$.
Doing the same with the element of the Galois group that sends $\zeta_8$ to $\zeta_8^7$, you find ${\Bbb Q}(\sqrt{2})$.
Luckily you've already found the intermediate field $\Bbb Q(i)$, because the same trick doesn't work for the element of the Galois group that sends $\zeta_8$ to $\zeta_8^5$. It is true that $\zeta_8 + \zeta_8^5$ is invariant under this map, but that doesn't get you anywhere because $\zeta_8^5 = - \zeta_8$, so $\zeta_8 + \zeta_8^5 = 0$. That does mean that $\zeta_8^2$ is invariant, though, which gives $\Bbb Q(\zeta_8^2) = \Bbb Q(i)$ as intermediate field.
Without the inspiration that such a sum of two roots is invariant, you can deduce this rather mechanically as well. The field ${\Bbb Q}(\zeta_8)$ has $1, \zeta_8, \zeta_8^2, \zeta_8^3$ as basis over ${\Bbb Q}$. Now $\phi$ maps the general element $a + b \zeta_8 + c \zeta_8^2 + d \zeta_8^3$ to $a + b \zeta_8^3 + c \zeta_8^6 + d \zeta_8^9 = a + d \zeta_8 - c \zeta_8^2 + b \zeta_8^3$. For this to be an invariant element under $\phi$, you need $b = d$ and $c = 0$, once again giving $\zeta_8 + \zeta_8^3$ as invariant under $\langle \phi \rangle$.
