Invariant fields of the Galois group of $x^4 + 1$ Let $f(x) = x^4 + 1 \in \mathbb{Q}[x]$. We can show that if $\alpha$ is a zero of $f(x)$, then the full set of zeros is given by $\{\alpha, -\alpha, i\alpha, -i\alpha\}$. Since $\alpha^2 = \pm i$ we can easily see that $L = \mathbb{Q}(\alpha)$ is the splitting field of $f$ over $\mathbb{Q}$. Since $L$ is the splitting field of a seperable polynomial, we see that $|\operatorname{Gal}(L/\mathbb{Q})|$ = $[L : \mathbb{Q}] = 4$. Define two automorphisms $\sigma$ and $\tau$ by $\sigma(\alpha) = -\alpha$ and $\tau(\alpha) = \alpha^3$. We can check that $\sigma^2 = \tau^2 = id$ and that $\sigma \tau = \tau \sigma$ to conclude that $|\operatorname{Gal}(L/\mathbb{Q})|$ is Klein's viergroup. Now I want to determine the invariant fields of this Galois group. I've already found that $\sigma \tau$ fixes $\alpha^2$ (since $\sigma \tau(\alpha^2) = \sigma(\alpha^6) = \sigma(-\alpha^2) = \alpha^2$ which shows that the invariant field of the subgroup $\{id, \sigma \tau\} = \mathbb{Q}(i)$. However, I can't seem to find the fixed points of $\sigma$ and $\tau$. From the way $\sigma$ is defined I doubt that $\sigma$ has any fixed points. For $\tau$ I thought it was logical to try $\tau(\alpha + \alpha^3)$ and got $\tau(\alpha + \alpha^3) = -2(\alpha + \alpha^3)$ but I don't see how this helps. How do I continue?
 A: Expanding on my comment, here’s my way of looking at things. Let $\zeta$ be a root of your polynomial $f(x)=x^4+1$. You know, of course, by looking at $f(x+1)$ and applying Eissenstein, that $f$ is irreducible. (In fact, it’s one of the cyclotomic polynomials, all of which are irreducible over $\mathbb Q$.) Now look at $\zeta^2$, which you’ve already noticed is a square root of $-1$. Also look at $\zeta+\zeta^{-1}$, which you check is a square root of $2$. Since you’re only asking about what the properly intermediate fields of $\mathbb Q(\zeta)$ are, even without bringing in the Galois group you see what they are. They’re $\mathbb Q(\sqrt2)$, $\mathbb Q(i)$, and $\mathbb Q(\sqrt{-2})$. Now it’s easy to see which elements of the Galois group, described as $\zeta\mapsto\zeta^m$ for various odd $m$, fix which of these intermediate fields.
I should add that the moral of the story is that you get farther with explicit computations, and loads of explicit information gained by working out examples.
A: Trying to apply the nice comments below the post:
$$\Bbb Q(\alpha)=\text{Span}_{\Bbb Q}\{\,1\,,\,\alpha\,,\,\alpha^2=i\,,\,\alpha^3=i\alpha\,\}$$
where in fact
$$\alpha=\frac1{\sqrt 2}(1+i)\;,\;\alpha^2=i\;,\;\;\alpha^3=\frac1{\sqrt 2}(-1+i)$$
Let $\,x:=a+b\alpha+ci+d\alpha i\in\Bbb Q(\alpha)\,,\;\;a,b,c,d\in\Bbb Q$ .  We define
$$\sigma\alpha:=-\alpha\implies \sigma(\alpha^2)=\sigma(i)=(-\alpha)^2=\alpha^2=i\;,\;\sigma(\alpha^3)=\sigma\alpha\cdot\sigma(\alpha^2)=-\alpha i\;,\;\text{and thus:}\;$$
$$x\in\Bbb Q(\alpha)^\sigma\iff a+b\alpha+ci+d\alpha i=x=\sigma x=a-b\alpha+ci-d\alpha i\iff$$
$$\iff b=d=0\implies x=a+c\alpha^2=a+ci\in\Bbb Q(\alpha^2)=\Bbb Q(i)$$
Also, defining
$$\tau\alpha:=\alpha i\;,\;\;\tau i=\tau(\alpha^2)=-i\;,\;\;\tau\alpha i=\tau\alpha\cdot\tau i=(\alpha i)(-i)=\alpha\,,\,\,\text{thus:}$$
$$x\in\Bbb Q(\alpha)^\tau\iff a+b\alpha+ci+d\alpha i=x=\tau x=a+b\alpha i-c i+d\alpha\iff$$
$$\iff \;b=d\;,\;c=0\implies x= a+b\alpha+b\alpha i=a+b\alpha(1+i)=a+b\sqrt 2\,i\in\Bbb Q(\sqrt 2\, i)$$
