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Can someone please help with this question.

Prove that degree of minimal polynomial over $\mathbb{Q} $ of $\zeta_{7} $ , a primitive 7th root of unity is not a prime number.

I thought as $\zeta_{7} $ =$1^{1/7} $ so I can write $1^{1/7} $ =x which implies $x^{7} $ =1 . So, it's prime.

But answer is not prime. So, I am missing some concept.

Can anyone please tell what mistake I am doing.

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The mistake is, $x^7-1$ is not the minimal polynomial of $\zeta_7$, as it is not irreducible over $\Bbb{Q}$. The minimal polynomial of $\zeta_7$ is the irreducible factor $x^6+x^5+x^4+x^3+x^2+x+1$ of $x^7-1$ over $\Bbb{Q}$.

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  • $\begingroup$ how is $x^{6}$ + ....+x +1 is minimal polynomial , you need to prove it!! $\endgroup$ – Ben Jul 1 at 17:43
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    $\begingroup$ Yes. Recall the definition of minimal polynomial $\alpha$ over $F$. It is the irreducible monic polynomial over $F$ with $\alpha$ as a root. Now that polynomial is a cyclotomic polynomial which is irreducible over $\Bbb{Q}$ (can be shown from Eisenstein's criteria). Now $\zeta_7$ is a root of $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$ but not a root of $x-1$, so it must be root of the other factor. $\endgroup$ – user598858 Jul 1 at 17:50
  • $\begingroup$ Eisenstein crieria can't prove $x^{6} $ + $x^{5} $ +... +1 to be irreducible . So, how can I prove $ x^{6} $ + $x^{5} $ +... +1 to be irreducible. $\endgroup$ – Ben Jul 1 at 18:05
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    $\begingroup$ Although I am giving a rough idea of the proof. Let denote that polynomial by $p(x)=\frac{x^7-1}{x-1}$ then you can show $p(x+1)$ is irreducible polynomial over $\Bbb{Q}$ by Eisenstein's criteria using the prime 7. Now $p(x+1)$ is irreducible implies $p(x)$ is so. $\endgroup$ – user598858 Jul 1 at 18:13
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    $\begingroup$ Well, you try to prove if $p(x)$ is reducible then for any $a$, $p(x+a)$ is also reducible. It is easy. $\endgroup$ – user598858 Jul 1 at 18:33
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A minimal polynomial should be irreducible. But all its roots are non-real roots of 1. Since coefficients are real (even rational) all roots of the minimal polynomial must be decomosed into pairs $(a, \bar a)$ , so the number of roots and the degree must be even. The only even prime is 2 but a quadratic rational polynomial cannot have a non-real root which is the seventh root of 1. So, indeed, the degree of minimal polynomial is not prime.

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