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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?

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  • $\begingroup$ 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
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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.

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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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