This question already has an answer here:

So $\alpha = \sqrt{2+\sqrt{2}}$, and I've already found the minimal polynomial of $\alpha$ over $\mathbb{Q}$ to be $p=x^4 - 4x^2 + 2$ and shown that $\mathbb{Q}(\alpha)$ is a normal extension. Now I want to describe the Galois group. We know $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 4$, so the Galois group has $4$ elements. The basis of the extension is $\{1, \alpha, \alpha^2, \alpha^3\}$, so the automorphism of $\mathbb{Q}(\alpha)$ leaving $\mathbb{Q}$ fixed need to act on $\sqrt{2}$ and $\sqrt{2+\sqrt{2}}$, I think. These would be: \begin{align*} \sigma_1 &: \sqrt{2}\rightarrow\sqrt{2}, \: \sqrt{2+\sqrt{2}} \rightarrow \sqrt{2+\sqrt{2}} \\ \sigma_2 &: \sqrt{2}\rightarrow -\sqrt{2}, \: \sqrt{2+\sqrt{2}} \rightarrow \sqrt{2+\sqrt{2}} \\ \sigma_3 &: \sqrt{2}\rightarrow\sqrt{2}, \: \sqrt{2+\sqrt{2}} \rightarrow -\sqrt{2+\sqrt{2}} \\ \sigma_4 &: \sqrt{2}\rightarrow -\sqrt{2}, \: \sqrt{2+\sqrt{2}} \rightarrow -\sqrt{2+\sqrt{2}} \end{align*}

Since all of these automorphisms are of order $2$, the Galois group of the extension is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$

Is it correct? Could the argument be simpler?


marked as duplicate by Watson, Behrouz Maleki, Claude Leibovici, Alex M., Namaste Dec 26 '16 at 15:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ the definition of the second automorphism is incompatible, because if it fixes the second element, it fixes its square, therefore it fixes the square root of 2.Observe more closely what the conjugates of alpha are. $\endgroup$ – Bogdan Simeonov May 26 '16 at 12:08

EDIT : I was wrong, the generators are actually not of order $2$ as I thought. The Galois group is actually cyclic.

I believe you have the wrong generators. $\sigma_2$, for example, cannot be an homomorphism as you should have $$2+\sqrt{2}=\left(\sqrt{2+\sqrt{2}}\right)^2=\left(\sigma_2\left(\sqrt{2+\sqrt{2}}\right)\right)^2 = \sigma_2(2+\sqrt{2})=2-\sqrt{2}.$$

The trick is that the roots of the minimal polynomial of $\alpha$ are $\pm \sqrt{2\pm \sqrt{2}}$, and the elements of the Galois group must be of the form $$\sqrt{2+\sqrt{2}}\mapsto \pm\sqrt{2\pm\sqrt{2}}.$$ You need to choose the sign twice, which gives you $4$ automorphisms.

Let us define $\varphi$ such that $\varphi\left(\sqrt{2+\sqrt{2}} \right)= \sqrt{2-\sqrt{2}}$; then squaring the two terms show that $\varphi(\sqrt{2}) = -\sqrt{2}$. Moreover, $$\left(\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2}}\right)=\sqrt{4-2}=\sqrt{2};$$ applying $\varphi$ on both sides yields $$\varphi\left(\sqrt{2+\sqrt{2}}\right)\varphi\left(\sqrt{2-\sqrt{2}}\right)=\varphi(\sqrt{2})$$and thus $$\varphi^2\left(\sqrt{2+\sqrt{2}}\right)=\varphi\left(\sqrt{2-\sqrt{2}}\right)=-\frac{\sqrt{2}}{\sqrt{2-\sqrt{2}}}=-\sqrt{2+\sqrt{2}}.$$ Thus this $\varphi$ is of order $4$, and thus the Galois group is $\mathbb{Z}/4\mathbb{Z}$.

  • $\begingroup$ Thank you. I realised after I posted it that $\sigma_2$ was a problem. Now, $\mathbb{Z}_2\times\mathbb{Z}_2$ has $3$ non-trivial proper subgroups, all of which are isomorphic to $\mathbb{Z}_2$. Should there then be only one intermediate field extension $\mathbb{Q} \subset K \subset\mathbb{Q}(\alpha)$? The field $\mathbb{Q}(\sqrt{2})$ should be one I think, are there supposed to be more? Do you have any hints on how to find them? $\endgroup$ – Auclair May 26 '16 at 12:18
  • $\begingroup$ Yes, that seems more reasonable than my suggestion. I'll have a closer look at the extensions you provided. Thank you. $\endgroup$ – Auclair May 26 '16 at 12:33
  • $\begingroup$ @Auclair I was wrong, sorry. The Galois group is cyclic, so in fact there are no other intermediate field extension. I've edited my answer. $\endgroup$ – Arnaud D. May 26 '16 at 12:52
  • $\begingroup$ In fact I just realized this has been asked before : math.stackexchange.com/questions/1301995/… $\endgroup$ – Arnaud D. May 26 '16 at 12:59
  • $\begingroup$ Cool. Thank you for your help, I'll check out the other thread as well. $\endgroup$ – Auclair May 26 '16 at 13:01

Not the answer you're looking for? Browse other questions tagged or ask your own question.