Finding minimum polynomial of $e^{i\pi/6 }$

Question

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 $\Bbb Q$?

Answer 1

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 $\Bbb{Q}$.

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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.

Answer 2

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.

Answer 3

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.