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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?

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

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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.
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