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?

  • $\begingroup$ Hmm. Isn't $$\sigma(-\alpha^2)=\sigma(-1)\sigma(\alpha)^2=-\alpha^2?$$ It seems to me that $\sigma\tau(\alpha)=\sigma(\alpha^3)=(-\alpha)^3=\alpha^7=\overline{\alpha}$, so $\sigma\tau$ is the usual complex conjugation. So its fixed field should be $L\cap\mathbb{R}$. $\endgroup$ – Jyrki Lahtonen Jun 3 '13 at 19:03
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    $\begingroup$ I think you need to recompute $\tau(\alpha+\alpha^3)$. $\endgroup$ – Thomas Andrews Jun 3 '13 at 19:03
  • $\begingroup$ @ThomasAndrews $\tau(\alpha + \alpha^3)$ = $(\alpha + \alpha^3)^3$ = $(\alpha + \alpha^3)^2(\alpha + \alpha^3)$ = $(\alpha^2 + 2\alpha^4 + \alpha^6)(\alpha + \alpha^3)$ = $-2(\alpha + \alpha^3)$? Or did I make some stupid mistakes here? $\endgroup$ – user50945 Jun 3 '13 at 19:19
  • $\begingroup$ $\tau(\alpha)=\alpha^3$ doesn't mean $\tau(x)=x^3$ for all $x$. $\tau(\alpha+\alpha^3)=\tau(\alpha)+\tau(\alpha)^3 = \alpha^3+\alpha^9 = \alpha^3+\alpha$. $\endgroup$ – Thomas Andrews Jun 3 '13 at 19:33
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    $\begingroup$ Ah ofcourse, then $\tau(\alpha + \alpha^3)$ = $\tau(\alpha) + \tau(\alpha^3)$ = $\alpha^3 + \alpha^9$ = $\alpha^3 + \alpha$, correct? $\endgroup$ – user50945 Jun 3 '13 at 19:37

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.


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

$$\tau\alpha:=\alpha i\;,\;\;\tau i=\tau(\alpha^2)=-i\;,\;\;\tau\alpha i=\tau\alpha\cdot\tau i=(\alpha i)(-i)=\alpha\,,\,\,\text{thus:}$$

$$x\in\Bbb Q(\alpha)^\tau\iff a+b\alpha+ci+d\alpha i=x=\tau x=a+b\alpha i-c i+d\alpha\iff$$

$$\iff \;b=d\;,\;c=0\implies x= a+b\alpha+b\alpha i=a+b\alpha(1+i)=a+b\sqrt 2\,i\in\Bbb Q(\sqrt 2\, i)$$

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    $\begingroup$ Great answer! I'm doing the same exercise but I'm stuck with the computation of the invariant field associated to $\sigma \tau$. According to my reason, let $\beta := \sigma \tau$ then $\beta (x) = a - b \alpha^3 + c \alpha^2 - d \alpha$ so $x$ must be an element of $\mathbb{Q}(\alpha^2, \alpha - \alpha^3)$ but this "field" would have a dimension over Q strictly greater than 2 $\Rightarrow 4$ but again, this field is not $\mathbb{Q}(\alpha)$ and so a contradiction $\endgroup$ – Riccardo Dec 11 '13 at 0:29
  • $\begingroup$ I know this is an old answer, but if someone can clarify this to me I'd be grateful :) $\endgroup$ – Riccardo Dec 11 '13 at 0:30
  • $\begingroup$ Hi @RicPed . I don't even remember this but I'll try later to read all about it and address your doubt. $\endgroup$ – DonAntonio Dec 11 '13 at 4:53
  • $\begingroup$ Thanks for the help! Maybe i found the error after several tries. If $\beta \alpha = -\alpha^3 $ then it is not true that $\beta \alpha^2 =-(\alpha^2)^3 =\alpha^2$ but it is true that $\beta \alpha^2 =( \beta \alpha)^2= (-\alpha^3)^2=\alpha^6=-\alpha^2 $. The problem is that i used the former valutation ibstead the latter. I think this is the fault in my previous reasoning :) $\endgroup$ – Riccardo Dec 11 '13 at 8:12

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