# Minimal polynomial of $\omega:=\zeta_7+\overline{\zeta_7}$

Let $\omega:=\zeta_7+\overline{\zeta_7}$, where $\zeta_7$ is a primitive $7$th root of $1$. I want to find the minimal polynomial of $\omega$ over $\mathbb{Q}$. I've found $$\omega=\zeta_7+\overline{\zeta_7}\quad;\quad\omega^2=\zeta_7^2+\overline{\zeta_7}+2\quad;\quad\omega^3=\zeta_7^3+\overline{\zeta_7}^3+3\omega$$

Now how can I find the minimal polynomial?

• Related. Probably others too. – anon Oct 2 '13 at 5:42

You have :

$$\omega=\zeta+\zeta^6, \ \ \ \omega^2=\zeta^2+\zeta^5+2, \ \ \ \omega^3=\zeta^3+\zeta^4+3\omega$$

Adding all those three up, you obtain

$$\omega^3+\omega^2+\omega =\sum_{k=1}^{6} \zeta^k+(3\omega+2)= -1+(3\omega+2)=3\omega+1$$

So

$$\omega^3+\omega^2-2\omega-1=0$$

Since that polynomial has degree three and no rational root, it is irreducible. So it is the minimal polynomial.

You seek to solve

$$a + b \omega + c \omega^2 + d \omega^3 = 0$$

for $(a,b,c,d)$. That looks like a linear algebra problem.

The general purpose method is:

• Choose a basis for the 6-dimensional rational vector space $\mathbb{Q}(\zeta_7)$.
• Find the coefficients of $\omega^i$ with respect to this basis
• Arrange them into a suitable matrix equation
• Solve.