Hi,
I'm trying to find the minimal polynomial of the 12th primitive root of unity over $\displaystyle \mathbb{Q}$
Is there a best way of going about this?
Thanks,
I would start by finding the primitive roots then constructing the minimal polynomial from that.
The twelfth roots of unity, of course, satisfy the equation $\displaystyle x^{12}= 1$. They are of the form $\displaystyle cos(n\pi/6)+ i sin(n\pi/6)$ with n from 0 to 11: 1, $\displaystyle \sqrt{3}/2+ i/2$, $\displaystyle 1/2+ i\sqrt{3}/2$, $\displaystyle i$, $\displaystyle -1/2+ i\sqrt{3}{2}$, $\displaystyle -\sqrt{3}/2+ i/2$, $\displaystyle -1$, $\displaystyle -\sqrt{3}/2- i/2$, $\displaystyle -0.5- i\sqrt{3}/2$, $\displaystyle -i$, $\displaystyle 1/2- i\sqrt{3}/2$, and $\displaystyle \sqrt{3}/2- i/2$.
A primitive root is one that does NOT satisfy $\displaystyle x^n- 1= 0$ for any smaller value of n which removes 1, -1, i, and -i. What about cube roots?
It might help to note that $\displaystyle x^{12}= 1$ is equivalent to
$\displaystyle x^{12}- 1= (x^3)^2- 1= (x^3- 1)(x^3+ 1)= (x- 1)(x^2+ x+ 1)(x+ 1)(x^2- x+ 1)= 0$.
Perhaps my terminology isn't quite right, by the "12th primitive root of unity" I meant:
$\displaystyle e^{i\pi/6} =: \omega $
So I am trying to find $\displaystyle [\mathbb{Q}(\omega):\mathbb{Q}] $
As $\displaystyle \omega$ is a root of $\displaystyle x^{12}-1$ we have that the min poly divides $\displaystyle x^{12}-1$.
I think perhaps the factorisation is off by power 2?
$\displaystyle x^{12} -1 = (x^6)^2-1 = (x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1) $
Hence, the min polynomial is one of the above bracketed terms:
$\displaystyle \omega^4 = \dfrac{1}{2}(-1+\sqrt3 i) \ \omega^2=\dfrac{1}{2}(1+ \sqrt3 i) $
Hence $\displaystyle x^4-x^2+1 $ is the min poly.
Thanks!