Finding minimum polynomial of $e^{i\pi/6 }$ Finding minimum polynomial of $e^{i\pi/6 }$:
I know it satisfies $t^6 +1 = 0$. I factorized $t^6+1 = (t^2+1)(t^4-t^2+1)$
obviously it does not satisfy $t^2 +1$ so it must satisfy $t^4-t^2 +1$. How do I show that this is indeed the minimum?
Is it enough to say: in modulo $2$ we have that $t^4 - t^2 + 1 \equiv t^4 +t^2 + 1$. In mod $2$ we only have 2 elements $0,1$. Obviously none of these satisfy $t^4+t^2+1 = 0$ hence it is irreducible in mod $2$ therefore in $\Bbb Q$?
 A: Let $\zeta:=e^{i\pi/6}$. As you note, $\zeta$ is a root of $t^4-t^2+1$, because it is clearly a root of 
$$t^6+1=(t^2+1)(t^4-t^2+1),$$
and it is not hard to check that it is not a root of $t^2+1$. For the same reason, also $\zeta^5$, $\zeta^7$ and $\zeta^{11}$ must be roots of $t^4-t^2+1$. So apparently
$$t^4-t^2+1=(t-\zeta)(t-\zeta^5)(t-\zeta^7)(t-\zeta^{11}).$$
If $t^4-t^2+1$ were reducible, then either $t-\zeta^m\in\Bbb{Q}[t]$ for some $m\in\{1,5,7,11\}$, which is clearly false, or otherwse $(t-\zeta)(t-\zeta^m)\in\Bbb{Q}[t]$ for distinct $m\in\{5,7,11\}$. Trying all three options shows that this is not the case, so $t^4-t^2+1$ is irreducible over $\Bbb{Q}$.

Alternatively, by the rational root test it is clear that $t^4-t^2+1$ has no roots in $\Bbb{Q}$. We can check by brute force that it is not a product of two quadratics by solving
$$(t^2+at+b)(t^2+ct+d)=t^4-t^2+1,$$
for $a,b,c,d\in\Bbb{Q}$, which comes down to solving the system of equations
\begin{eqnarray*}
a+c&=&0,\\
ac+b+d&=&-1,\\
ad+bc&=&0,\\
bd&=&1,
\end{eqnarray*}
First it is immediate that $c=-a$ and $d=b^{-1}$, so the third equation becomes
$$ab^{-1}-ba=0\qquad\text{ and so }\qquad ab^2=a,$$
that is, either $a=c=0$ or $b=d=\pm1$. In the first case the second equation becomes
$$b+b^{-1}=-1\qquad\text{ and so }\qquad b^2-b+1=0,$$
which is impossible. In the second case the second equation becomes
$$-a^2\pm2=-1,$$
which is also impossible. So there is no such factorization, so $t^4-t^2+1$ is irreducible.
A: Here's a slick way of showing irreducibility of $x^4- x^2+1$, using uniqueness of factorization in both $\Bbb Q[x]$ and $K[x]$, where $K$ is the field of Gaussian numbers, $\Bbb Q(i)$.
Note that $x^4-x^2+1=(x^2+ix-1)(x^2-ix-1)$, and that this is a factorization into irreducibles over $K$.  If the original quartic had any factorization over $\Bbb Q$, it would also be a factorization over $K$, but this product of quadratics is the only $K$-factorization that there is. So the original quartic is irreducible over the rationals. 
A: Be careful. Whilst it is true that if a polynomial $f$ is irreducible modulo some prime then it is irreducible over $\mathbb Q$, just because a quartic equation doesn't have a root, doesn't make it irreducible. For example, $X^4 + 4$ has no rational roots, but $$X^4 + 4 = (X^2 + 2X+2)(X^2 - 2X+2).$$
Indeed in this case, $$(t^2 + t + 1)^2 \equiv t^4 + t^2 + 1 \pmod 2$$so your polynomial is reducible modulo $2$.
To show that it's irreducible, you can work manually. The fact that $f(t) = t^4 - t^2 + 1$ has no roots (as it has no roots modulo $2$) means that if it is reducible, then it must factorise as a pair of quadratics. Gauss's Lemma tells us that we only need to consider integer coefficients.
Suppose $a,b,c,d\in \mathbb Z$ satisfy$$f(t) = (t^2 + at + b)(t^2+ct+d)=t^4 + (a+c)t^3+(ac+b+d)t^2+(ad+bc)t + (bd).$$ Then, for example, we have $bd = 1$, so $b = d= \pm1$. And $a+c = 0$, so $a=-c$. Finally, looking at the $t^2$ coefficient, $$ac+b+d = \pm2-a^2 = -1$$and some quick checking shows that $a$ cannot be an integer.
