1. ## Min polynomial

Hi,

I'm trying to find the minimal polynomial of the 12th primitive root of unity over $\mathbb{Q}$

Thanks,

2. ## Re: Min polynomial

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 $x^{12}= 1$. They are of the form $cos(n\pi/6)+ i sin(n\pi/6)$ with n from 0 to 11: 1, $\sqrt{3}/2+ i/2$, $1/2+ i\sqrt{3}/2$, $i$, $-1/2+ i\sqrt{3}{2}$, $-\sqrt{3}/2+ i/2$, $-1$, $-\sqrt{3}/2- i/2$, $-0.5- i\sqrt{3}/2$, $-i$, $1/2- i\sqrt{3}/2$, and $\sqrt{3}/2- i/2$.
A primitive root is one that does NOT satisfy $x^n- 1= 0$ for any smaller value of n which removes 1, -1, i, and -i. What about cube roots?

It might help to note that $x^{12}= 1$ is equivalent to
$x^{12}- 1= (x^3)^2- 1= (x^3- 1)(x^3+ 1)= (x- 1)(x^2+ x+ 1)(x+ 1)(x^2- x+ 1)= 0$.

3. ## Re: Min polynomial

Perhaps my terminology isn't quite right, by the "12th primitive root of unity" I meant:

$e^{i\pi/6} =: \omega$

So I am trying to find $[\mathbb{Q}(\omega):\mathbb{Q}]$

As $\omega$ is a root of $x^{12}-1$ we have that the min poly divides $x^{12}-1$.

I think perhaps the factorisation is off by power 2?

$x^{12} -1 = (x^6)^2-1 = (x^6-1)(x^6+1)=(x^2-1)(x^4+x^2+1)(x^2+1)(x^4-x^2+1)$

Hence, the min polynomial is one of the above bracketed terms:

$\omega^4 = \dfrac{1}{2}(-1+\sqrt3 i) \ \omega^2=\dfrac{1}{2}(1+ \sqrt3 i)$

Hence $x^4-x^2+1$ is the min poly.

Thanks!

4. ## Re: Min polynomial

Originally Posted by HallsofIvy
It might help to note that $x^{12}= 1$ is equivalent to
$x^{12}- 1= (x^3)^2- 1$
Little mistake here but you can still go through the whole difference of 2 squares process with the higher index.