# Degree of the minimal polynomial of $\sum\sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$.

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.

• 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. Sep 24, 2022 at 6:23
• @JyrkiLahtonen, so can I say that the splitting field is precisely $\mathbb{Q}(2 \cos(\pi/16)$? Sep 24, 2022 at 6:59
• 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 $8$? Sep 24, 2022 at 7:02
• 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. Sep 24, 2022 at 7:31

Preamble:

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, $$\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})\cong\text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})/\{\text{id},\tau\}\cong C_8$$.

Derivations:

Now, let $$f(x)$$ be your polynomial, and let $$E$$ be the splitting field of $$f(x)$$. Note that $$$$\sqrt{2+\sqrt{2+\sqrt{2}}}=2\cos\left(\frac{\pi}{16}\right)$$$$ This means that $$E/\mathbb{Q}(\cos(\pi/16))$$ is a possibly trivial field extension. In fact, we have that $$$$8=|\text{Gal}(\mathbb{Q}(\cos(\pi/16))/\mathbb{Q})|=[\mathbb{Q}(\cos(\pi/16)):\mathbb{Q}]\leq[E:\mathbb{Q}]\leq\deg(f)=8$$$$ so $$[\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 $$$$\alpha=2\left[\cos\left(\frac{\pi}{16}\right)+\cos\left(\frac{3\pi}{16}\right)+\cos\left(\frac{5\pi}{16}\right)+\cos\left(\frac{7\pi}{16}\right) \right]$$$$ which you can get by noting that $$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 $$[\mathbb{Q}(\alpha):\mathbb{Q}]=8$$.

• what about the degree of the minimal polynomial of $\sum \sqrt{2\pm\sqrt{2\pm\sqrt{2}}}$? Sep 24, 2022 at 12:49
• @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. Sep 24, 2022 at 23:11
• 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. Sep 25, 2022 at 5:36
• I will work upon all these in details, and will get back in next few days. @JyrkiLahtonen Sir Sep 25, 2022 at 6:30
• A typo two comments up. Should be $\dfrac12x^8f(2/x)$, of course. Potentially confusing, sorry about that. Oct 2, 2022 at 4:06