Inspired by a previous question what let $x = \sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot (7.5^\circ)$. What is the minimal polynomial of $x$ ?

The theory of algebraic extensions says the degree is $4$ since we have the degree of the field extension $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{6}): \mathbb{Q}] = [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] =4$

Does trigonometry help us find the other three conjugate roots?

  • $+\sqrt{2}-\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_1$
  • $-\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6} = \cot \theta_2$
  • $-\sqrt{2}-\sqrt{3}+\sqrt{4}+\sqrt{6} = \cot \theta_3$

This problem would be easier if we used $\cos$ instead of $\cot$. If I remember the half-angle identity or... double-angle identity:

$$ \cot \theta = \frac{\cos \theta}{\sin \theta} = \sqrt{\frac{1 - \sin \frac{\theta}{2}}{1 + \sin \frac{\theta}{2}}}$$

Sorry I am forgetting, but I am asking about the relationship between trigonometry and the Galois theory of this number.


By Galois theory the intermediate fields are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. Your number is not an element of any of those, so it generates the whole 4-d extension. In particular, we know that the minimal polynomial will be a quartic. Therefore any quartic polynomial with integer coefficients with this number as a root is the minimal polynomial.

Let $x=2+\sqrt2+\sqrt3+\sqrt6$. Then $$ 0=(x-2-\sqrt3)^2-(\sqrt2+\sqrt6)^2=x^2-(4+2\sqrt3)x-1. $$ The idea here is that squaring removes "the $\sqrt2$ content" from $\sqrt2+\sqrt6$ leaving only irrationalities coming from $\sqrt3$.

Using the obvious algebraic conjugate as an extra factor we see that $$ m(x)=(x^2-(4+2\sqrt3)x-1)(x^2-(4-2\sqrt3)x-1)=x^4-8x^3+2x^2+8x+1 $$ fits the bill.

The algebraic conjugates of values of trig functions (at rational multiples of $\pi$) are of the same type: $$ \begin{aligned} 2+\sqrt2+\sqrt3+\sqrt6&=\cot\frac{\pi}{24},\\ 2+\sqrt2-\sqrt3-\sqrt6&=\cot\frac{17\pi}{24},\\ 2-\sqrt2+\sqrt3-\sqrt6&=\cot\frac{13\pi}{24},\\ 2-\sqrt2-\sqrt3+\sqrt6&=\cot\frac{5\pi}{24}. \end{aligned} $$ I haven't checked the details, but I'm fairly sure that when you dig for the integer multiple angle formulas for cotangent, the polynomial $m(x)$ pops out. After all, we can write $\cot 6x$ as a rational function of $\cot x$ with polynomials of degrees $\le6$ as numerators and denominators. When you use that formula in the l.h.s. of the equation $$ \cot 6x=1 $$ and clear the denominator, the resulting degree $6$ polynomial equation in $\cot x$ factors as a product of a quartic (obviously $m(\cot x)$) and a quadratic - the latter accounting for solutions $x=3\pi/8$ and $x=7\pi/8$.

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All you need to do is write your number and its square, cube and fourth powers in the form $z^j = a_j + b_j \sqrt{2} + c_j \sqrt{3} + d_j \sqrt{6}$ with $a_j,b_j,c_j,d_j$ rational and solve a system of four equations in four unknowns

$$ \eqalign{a_4 + a_3 x_3 + a_2 x_2 + a_1 x_1 + a_0 x_0 &= 0\cr b_4 + b_3 x_3 + b_2 x_2 + b_1 x_1 + b_0 x_0 &= 0\cr c_4 + c_3 x_3 + c_2 x_2 + c_1 x_1 + c_0 x_0 &= 0\cr d_4 + d_3 x_3 + d_2 x_2 + d_1 x_1 + d_0 x_0 &= 0\cr }$$

to determine a polynomial $Z^4 + x_3 Z^3 + x_2 Z^2 + x_1 Z + x_0$ of which your number is a root.

Maple says the answer is $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$.

EDIT: Here is another way, maybe a bit deeper. Your number is $z = 1+(1+\sqrt{2})(1+\sqrt{3})$. $u = 1+\sqrt{2}$ is an eigenvalue of $A = \pmatrix{0 & 1\cr 1 & 2\cr}$, and $v = 1+\sqrt{3}$ is an eigenvalue of $B = \pmatrix{0 & 2\cr 1 & 2\cr}$.
So $uv = (1+\sqrt{2})(1+\sqrt{3})$ is an eigenvalue of $$ A \otimes B = \pmatrix{0 & 0 & 0 & 2\cr 0 & 0 & 1 & 2\cr 0 & 2 & 0 & 4\cr 1 & 2 & 2 & 4}$$ which has characteristic polynomial $X^4 - 4 X^3 - 16 X^2 - 8 X + 4$. Substitute $X = Z - 1$ to obtain $Z^4 - 8 Z^3 + 2 Z^2 + 8 Z + 1$.

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  • $\begingroup$ the computation of the minimal polynomial is pretty mechanical... but you can still get it wrong :-) I am asking if we pick out some deeper geometric meaning or symmetric to help us understand better. In the same way roots of unity correspond to regular polygons. $\endgroup$ – cactus314 Jan 4 '16 at 1:47

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