The polynomial $$f(x)= x^8-8x^6+20x^4-16x^2+2=((x^2-2)^2-2)^2-2=0$$ has $$y=\sqrt{2+\sqrt{2+\sqrt{2}}}$$ one of its roots.

How do I determine the degree of its spliting field and how do I determine its Galois Group?

What I know is the other roots are $$\pm \sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$$ and these $8$ elements of $\mathbb{R}$ are the only roots of this polynomial. But I am not sure how do I proceed further.

I am computing this because I was trying to find the degree of the minimal polynomial of the element

$$ \sum \sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$$ and this element is contained in the splitting field of $f(x)$ and these are fixed by conjugation(inaccurate). I need to work a bit more on this. I would appreciate if I can get an idea about it galois groups.

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    $\begingroup$ Finding the splitting field and the Galois group is greatly facilitated by the following list of observations (proof by induction using the half-angle formula): $2\cos(\pi/4)=\sqrt2$, $2\cos(\pi/8)=\sqrt{2+\sqrt2}$, $2\cos(\pi/16)=\sqrt{2+\sqrt{2+\sqrt2}}$, $\ldots$. The field generated by $2\cos(\pi/16)$ is a subfield of a cyclotomic field, hence already Galois itself, and with an abelian Galois group. $\endgroup$ Sep 24, 2022 at 6:23
  • $\begingroup$ @JyrkiLahtonen, so can I say that the splitting field is precisely $\mathbb{Q}(2 \cos(\pi/16)$? $\endgroup$ Sep 24, 2022 at 6:59
  • $\begingroup$ and hence the splitting field has degree $8$ because minimal polynomial of $\cos(\pi/16$ is of degree $8$(yet to show that it is irreducible) and hence the order of GG is $8$? $\endgroup$ Sep 24, 2022 at 7:02
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    $\begingroup$ Yes. Surely $f(x)$ is irreducible. Either by Eisenstein or by cyclotomic theory. The cyclotomic field $\Bbb{Q}(e^{\pi i/16})$ is a degree sixteen extension, and you are looking at its real subfield. Mind you, numerical testing suggests that the sum of positive zeros of $f(x)$ has a trivial stabilizer within the Galois group. So the minimal polynomial of the sum has degree eight as well. Mathematica crunched it out to $$m(x)=x^8-32x^6+160x^4-256x^2+128.$$ That can be seen to be irreducible by its 2-adic Newton's polygon. I would prefer a pencil & paper argument. $\endgroup$ Sep 24, 2022 at 7:31

1 Answer 1



Let $\zeta$ be a primitive $32$nd root of unity. First, note that $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})\cong(\mathbb{Z}/32\mathbb{Z})^\times\cong C_2\times C_8$. In fact, every element in $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ can be writen as $\sigma^i\tau^j$ for some unique $i\in\mathbb{Z}_8,j\in\mathbb{Z}_2$, where $\sigma:\zeta\rightarrow\zeta^3$ and $\tau:\zeta\rightarrow\zeta^{-1}$ is the complex conjugate. Furthermore, note that $\mathbb{Q}(\cos(\pi/16))$ is a subfield of $\mathbb{Q}(\zeta)$ that is invariant under the automorphism subgroup $\{\text{id},\tau\}$; therefore by the Fundemental Theorem of Galois Theory, $\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})/\{\text{id},\tau\}\cong C_8$.


Now, let $f(x)$ be your polynomial, and let $E$ be the splitting field of $f(x)$. Note that \begin{equation} \sqrt{2+\sqrt{2+\sqrt{2}}}=2\cos\left(\frac{\pi}{16}\right) \end{equation} This means that $E/\mathbb{Q}(\cos(\pi/16))$ is a possibly trivial field extension. In fact, we have that \begin{equation} 8=|\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})|=[\mathbb{Q}(\cos(\pi/16)):\mathbb{Q}]\leq[E:\mathbb{Q}]\leq\deg(f)=8 \end{equation} so $[\mathbb{Q}(\cos(\pi/16)):\mathbb{Q}]=[E:\mathbb{Q}]$ which implies that $E=\mathbb{Q}(\cos(\pi/16))$. As discussed before, this means that $\text{Gal}(E/\mathbb{Q})\cong C_8$, and furthermore the degree of the splitting field of $f(x)$ is $8$.

Let $\alpha=\sum_\pm\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$ be your sum. There are multiple ways to go about calculating $\text{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]$ (which is the degree of the minimal polynomial of $\alpha$).

One way is to notice that \begin{equation} \alpha=2\left[\cos\left(\frac{\pi}{16}\right)+\cos\left(\frac{3\pi}{16}\right)+\cos\left(\frac{5\pi}{16}\right)+\cos\left(\frac{7\pi}{16}\right) \right] \end{equation} which you can get by noting that $2\cos(k\pi/{16})$ for $k\in\{1,3,5,7\}$ are precisely the positive numbers which are the images of $2\cos(\pi/16)$ under the automorphisms in $\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$, meaning they must be some permutation of the numbers $\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$ which are the positive roots of $f(x)$. Now, we can note that $\alpha$ is only fixed by id and $\tau$, which by the same logic as in the above proof, tells us that $\text{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})\cong C_8$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]=8$.

Alternatively, one can further simplify to see that $\frac{1}{\alpha}=\cos\left(\frac{7\pi}{16}\right)$, which means that $\mathbb{Q}(\cos(\pi/16))=\mathbb{Q}(\cos(7\pi/16))\subseteq\mathbb{Q}(\alpha)\subseteq\mathbb{Q}(\cos(\pi/16))$ which again confirms that $\text{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})\cong C_8$ and $[\mathbb{Q}(\alpha):\mathbb{Q}]=8$.

  • $\begingroup$ what about the degree of the minimal polynomial of $\sum \sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$? $\endgroup$ Sep 24, 2022 at 12:49
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    $\begingroup$ @permutation_matrix I have expanded on your additional question. Please feel free to ask if you have any additional questions, or concerns. This was an interesting problem. $\endgroup$ Sep 24, 2022 at 23:11
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    $\begingroup$ It was criminal negligence of me to miss the fact that this sum simplifies to $1/\cos(7\pi/16)$ :-). This also immediately gives the minimal polynomial of the sum. After all, $2\cos(7\pi/16)$ is a zero of the degree $8$ polynomial $f(x)$ @permutation_matrix started with. It follows that the minimal polynomial of the sum of positive roots of $f$ is (go reciprocal) $$\frac12 x^2f(\frac2x)=x^8-32x^6+160x^4-256x^2+128$$ confirming what I got numerically. $\endgroup$ Sep 25, 2022 at 5:36
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    $\begingroup$ I will work upon all these in details, and will get back in next few days. @JyrkiLahtonen Sir $\endgroup$ Sep 25, 2022 at 6:30
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    $\begingroup$ A typo two comments up. Should be $\dfrac12x^8f(2/x)$, of course. Potentially confusing, sorry about that. $\endgroup$ Oct 2, 2022 at 4:06

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