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I am attempting to find the minimal polynomial for $\omega=\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}+i\frac{1}{2}$ over $\mathbb{Q}$. I'm doing this in the context of cyclotomic field extensions. The book gives $x^2-x+1$, the 6th cyclotomic polynomial, as an answer. Is this correct? When I do the calculation, it appears that $\omega$ isn't even a root.

I believe the answer is the 12th cyclotomic polynomial, $f(x)=x^4-x^2+1$. We have $f(\omega)=0$, it is monic, and by the mod $p$ irreducibility test (or the irreducibility test of your choice), it is irreducible over $\mathbb{Q}$. It also seems nicely intuitive that the minimal polynomial of a primitive $n$th root of unity would be the $n$th cyclotomic polynomial, since all the cyclotomic polynomials (I believe) are irreducible over $\mathbb{Q}$. So is this the minimal polynomial, or am I missing something?

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What do you mean by "mod $p$ irreducibility test" ? as far as I remember it can only show that a polynomial is reducible but not thew other way around –  Belgi May 4 '12 at 13:13
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Actually $\omega$ has two square roots. Sure, you are fine, $x^2-x+1$ is certainly not it, nothing quadratic can work. –  André Nicolas May 4 '12 at 13:20
    
The field extension generated by $\omega$ contains $$\omega+\omega^{-1}=\omega+\overline{\omega}=\sqrt3.$$ Furthermore, the field $\mathbb{Q}(\omega)$ is not real. Putting these two bits together tells us the minimal polynomial must be at least quartic. Also any book on cyclotomic polynomials will tell you that $$\phi_{12}(x)=x^4-x^2+1.$$ –  Jyrki Lahtonen May 4 '12 at 13:26
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What book has this mistake? What page? –  Gerry Myerson May 4 '12 at 13:33
    
@Belgi: Take a prime $p$ and let $f(x) \in \mathbb{Z}[x]$ with $\mbox{deg}(f(x)) \ge 1$. If $\bar{f}(x) \in \mathbb{Z}_p[x]$ obtained by reducing all coefficients of $f(x)$ mod $p$ is irreducible over $\mathbb{Z}_p$ and $\mbox{deg}\bar{f}(x)) = \mbox{deg}(f(x))$, then $f(x)$ is irreducible over $\mathbb{Q}$. –  Alex Petzke May 4 '12 at 15:33

2 Answers 2

up vote 5 down vote accepted

Here are three ways to prove that the degree of $\zeta_{12}$ over $\Bbb{Q}$ is not 2. Two of them will actually prove to you that the degree of this over the rationals is actually 4.

$\textbf{Approach 1:}$ Let's do this in a very simple way and try to determine the degree of $\zeta_{12}$ over $\Bbb{Q}$. Notice that

$$\Bbb{Q}(\zeta_{12}) \cong \Bbb{Q}(\sqrt{3},i) \cong \left(\Bbb{Q}(\sqrt{3})\right)(i). $$

Now the degree of $i$ over $\Bbb{Q}$ is 2 because the minimal polynomial of $i$ over $\Bbb{Q}$ is $x^2 +1$. Now we show that the degree of $\sqrt{3}$ over $\Bbb{Q}(i)$ is two as well. We already have a polynomial with coefficients in $\Bbb{Q}(i)$ with $\sqrt{3}$ as a root, namely the polynomial $x^2 - 3$. Now suppose that this polynomial is reducible, meaning that $\sqrt{3} \in \Bbb{Q}(i)$. Then we get that since as a vector space over $\Bbb{Q}$, $\Bbb{Q}(i)$ is two dimensional we can write

$$\sqrt{3} = a + bi$$

for some $a,b \in \Bbb{Q}$. Then squaring both sides we get that $3 = a^2 - b^2 + 2abi$ and hence that $$\frac{3 - a^2 + b^2}{2ab} = i$$

in particular that $i$ is real. This is a contradiction so that the polynomial $x^2 - 3$ is irreducible, and hence that $[\Bbb{Q}(\sqrt{3},i) : \Bbb{Q}(i)] = 2$. It follows by the dimension counting formula that $[\Bbb{Q}(\sqrt{3},i):\Bbb{Q}] = 2 \times 2 = 4$ and hence the degree of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$ must be 4 and you are right. Therefore if you have found a monic degree polynomial of degree 4 with $\zeta_{12}$ as a root, it must be the case that that polynomial is irreducible otherwise this would contradict our result that $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}] = 4$.

$\textbf{Add-on to approach 1:}$ Suppose you did not know about cyclotomic polynomials (like me) and wanted to compute the minimal polynomial of $\zeta_12$ over $\Bbb{Q}$. You already know that it must be of degree 4 by the argument above. Once you find a monic one of degree 4, it must be unique by uniqueness of the minimal polynomial (exercise). Write

$$x = \frac{\sqrt{3} + i}{2}.$$

Then $2x = \sqrt{3} + i$ so that squaring both sides gives that $4x^2 = 3 - 1 + 2\sqrt{i}$ which implies that $4x^2 - 2 = 2\sqrt{3}i$. Hence

$$\begin{eqnarray*} 2x^2 - 1 &=& \sqrt{3}i\\ \implies 4x^4 - 4x^2 + 1 &=& -3 \\ \implies 4x^4 - 4x^2 + 4 &=& 0 \\ \implies x^4 - x^2 +1 &=& 0 \\ \end{eqnarray*}$$

and voilà! This is exactly the cyclotomic polynomial of degree 4 that you found.

$\textbf{Approach 2: Galois Theory}$ Suppose that $[\Bbb{Q}(\zeta_{12}) : \Bbb{Q}] = 2$. Then this is a degree 2 extension of a field of characteristic zero and hence is a Galois Extension. It follows that the Galois Group $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$, the cyclic group of order 2.

Now take some $\sigma \in G$. Then $\sigma$ must induce a permutation on the roots of the minimal polynomial of $\zeta_{12}$ over $\Bbb{Q}$. But then since $\sigma(\sqrt{3})^2 = \sigma(3) = 3$ which implies that $\sigma$ fixes $\sqrt{3}$ (similarly it fixes $i$) it follows that

$$\sigma(\zeta_{12}) = \zeta_{12}.$$

But then since $\sigma \in \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ was arbitrary this means that every $ \sigma \in \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the identity permutation so that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q})$ is the trivial group. But then this contradicts the fact that $ \operatorname{Gal}(\Bbb{Q}(\zeta_{12})/\Bbb{Q}) \cong C_2$ so it is not possible for $[\Bbb{Q}(\zeta_{12}):\Bbb{Q}]$ to be equal to 2.

$\textbf{Edit:}$ The proof above on Galois Theory is incorrect. Namely because it is not true that $\sigma(\sqrt{3}) = \sqrt{3}$ alone, but rather $\sigma(\sqrt{3}) = \pm \sqrt{3}$ and similarly for $\sigma(i)$.

$\textbf{Approach 3:}$ There is this result that is useful:

$$[\Bbb{Q}(\zeta_n) : \Bbb{Q}] = \varphi(n)$$

where $\varphi(n)$ is the Euler totient function. It counts the number of integers $k$ such that $1 \leq k \leq n$ with $k$ relatively prime to $n$. In your case, the integers relatively prime to 12 are $1,5,7,11$ which is $4$, exactly what you need.

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It's a lot easier to show that $i$ has degree 2 over ${\bf Q}(\sqrt3)$ than to show that $\sqrt3$ has degree 2 over ${\bf Q}(i)$. –  Gerry Myerson May 4 '12 at 13:31
    
@GerryMyerson Yeah probably I guess both worked for me anyway :D –  user38268 May 4 '12 at 13:35
    
Excellent, thank you. –  Alex Petzke May 4 '12 at 18:01

If you know that $\prod_{d\mid n}\Phi_d(X) = X^n - 1$ (which is simple to prove), then it is easy to find $\Phi_{12}$, which has degree $\phi(12)=4$. One difficulty is proving that $\Phi_n$ is indeed irreducible, a non-trivial result due to Gauss for general $n$, but it should be easy for $\Phi_{12}$.

Wikipedia lists $\Phi_{12}(x)=x^4-x^2+1$.

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