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I am trying to solve an exercise, where i have to determine the number of the primitive polynomials of degree $10$ over the finite field $F_{2}$.

My approach was using the formula $\frac{\varphi (p^{d}-1)}{d}$, where $p$ is the number of the elements of the finite field, $d$ the degree and $\varphi$ is the Euler totient function. So, i get $\frac{\varphi (2^{10}-1)}{10}=\frac{600}{10}=60$.

Do you agree with this solution? I would be glad to read your comments and remarks.

Thanks in advance!

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    $\begingroup$ I'd say this is correct, but it needs more explanation. $\endgroup$ – Magdiragdag Nov 8 '13 at 12:59
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    $\begingroup$ Looks good to me. Presumably some facts (e.g. that formula) have been covered in your course/book. If not, then it is a good exercise to prove that formula! $\endgroup$ – Jyrki Lahtonen Nov 8 '13 at 20:22
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If $q$ is a prime power, then the number of primitive monic degree $d$ polynomials in $\mathbb{F}_q[x]$ is indeed $$ \frac{φ(q^d − 1)}{d}. $$ And primitive polynomials are necessarily irreducible. We have $99$ monic irreducible polynomials of degree $10$ over $\mathbb{F}_2[x]$, and $60$ of them are primitive.

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  • $\begingroup$ Yes, i used the same formulas. Thank you! $\endgroup$ – Lullaby Nov 8 '13 at 20:34
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    $\begingroup$ You are welcome. The formula for the number of monic irreducible polynomials, which I have used is $\frac{1}{n}\sum_{d\mid n}\mu(d)q^{n/d}$. $\endgroup$ – Dietrich Burde Nov 8 '13 at 21:19

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