# For $K$ the splitting field of $x^8+1$ over $\mathbb{Q}$, determine $Gal(K/\mathbb{Q})$.

Let $f(x) = x^8+1$. To determine the Galois group $G$, we first need the splitting field and before that we need to find the zeroes of $f$. So, $\left(re^{i\theta}\right)^8 = 0$ implies $r=1, \theta=\frac{\pi}{8}, \frac{3\pi}{8},\ldots, \frac{15\pi}{8}$ where by half angle formulas all of the roots are \begin{align*} \pm \left( \frac{\sqrt{2+\sqrt{2}}}{2} + i\frac{\sqrt{2-\sqrt{2}}}{2} \right) & &\pm \left(\frac{\sqrt{2-\sqrt{2}}}{2} + i\frac{\sqrt{2+\sqrt{2}}}{2}\right) \\ \pm \left( - \frac{\sqrt{2+\sqrt{2}}}{2} + i\frac{\sqrt{2-\sqrt{2}}}{2} \right)& & \pm \left( -\frac{\sqrt{2-\sqrt{2}}}{2} + i\frac{\sqrt{2+\sqrt{2}}}{2}\right) \end{align*} Adjoin these to $\mathbb{Q}$ to get the splitting field $K$. It's not hard to see that actually $$K= \mathbb{Q}\left(\frac{\sqrt{2+\sqrt{2}}}{2},\frac{\sqrt{2-\sqrt{2}}}{2},i\right) =: \mathbb{Q}(\alpha,\beta,i).$$ Assuming I've done all this right, we can now find the Galois group $Gal(K/\mathbb{Q})$. This is where I start running into trouble. In particular, the Galois permutes roots of the respective minimal polynomials of $\alpha,\beta$ and $i$. But $\alpha$ and $\beta$ are conjugates so $\alpha \mapsto \pm \alpha, \pm \beta$. Then $i \mapsto \pm i$ so we have $4\cdot 2 = \fbox{8}$ elements of $G$. This number should equal the index $[\mathbb{Q}(\alpha ,\beta,i):\mathbb{Q}]$. But I calculate $$[\mathbb{Q}(\alpha,\beta,i):\mathbb{Q}] = [\mathbb{Q}(\alpha,\beta,i):\mathbb{Q}(\alpha,\beta)][\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}] = 2\cdot 2\cdot 4 = \fbox{16}.$$ I'm certain $i$ is of degree two over $\mathbb{Q}(\alpha,\beta)$, and I'm also certain that $\alpha$ is of degree 4 over $\mathbb{Q}$, since the minimal polynomial can be found pretty easily to be of degree 4. What I'm iffy on is the middle part. My work shows that the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$ is found by $$\frac{\sqrt{2+\sqrt{2}}}{2} = x \Rightarrow x^2 = \frac{2+\sqrt{2}}{2} \Rightarrow 2x^2-(2+\sqrt{2}) = 0$$ so it's of degree 2. Is this where I am making a mistake? If not, then indeed the index is 16 so the order of $G$ must be 16 as well. Now if this is the case, what are the other 8 automorphisms I'm not seeing? Of course, there's also the possibility I messed up much earlier on, and that's what's making all this so difficult.

• Maybe I'm not seeing the relevant part, but at first glance it doesn't seem all that helpful since I'm asking about $x^8+1$ not $x^8-1$. Do some of the results carry over? – Derek Allums May 26 '12 at 1:00
• $x^{16}-1=(x^8+1)(x^8-1)$. – Gerry Myerson May 26 '12 at 1:02
• Thank you for showing your work so well. If only all posters were this thorough! – rschwieb May 26 '12 at 1:14
• @rschwieb No problem at all. Thank you for the quick and helpful answer. – Derek Allums May 26 '12 at 1:31

When you adjoin your $\alpha$, you already get your $\beta$ for free, since: $\beta=\frac{\beta\alpha}{\alpha}=\frac{\sqrt{2}}{\alpha}$
Also note that $\sqrt{2}$ was already adjoined when $\alpha$ was, since $\sqrt{2}=4\alpha^2 -2$
• You're absolutely right. I had a feeling there was some really simple relationship between $\alpha$ and $\beta$ I just wasn't seeing. Thanks. – Derek Allums May 26 '12 at 1:10
I think you are going about it the hard way. You know the roots are the odd powers of $\zeta=e^{\pi i/8}$, and you may find that a far more useful form for the roots than what you've done with all those square root of two thingies.
• Ah, so actually $K = \mathbb{Q}(\zeta)$, where of course $\zeta$ is of degree 8 over $\mathbb{Q}$, so $[\mathbb{Q}(\zeta):\mathbb{Q}]=8$, which is what we want. Then the automorphisms are $\zeta \mapsto \zeta^i$ for $i=1,\ldots,8$. This is a much cleaner approach, thank you. – Derek Allums May 26 '12 at 1:15