I was wondering if there is as much irreducible monic polynomial in a finite field than irreducible polynomial ? I would say yes since $a_nx^n+...+a_1x+a_0$ is irreducible if and only if $x^n+b_{n-1}x^{n-1}+...+b_1x+b_0$ is irreducible where $b_i=\frac{a_i}{a_n}\in \mathbb F_{p^r}$. If no, can is there a formula that join the number of monic irreducible polynomial and the number of irreducible of polynomial ?
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Well, how many possible values of $a_n$? (You have a typo in your quotient, the divisor was $a_n$.) – ancient mathematician May 08 '17 at 17:59
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If you choose to view scalar multiples of a monic irreducible polynomial as distinct irreducible polynomials, then their number is clearly multiplied by $q-1$. – Jyrki Lahtonen May 08 '17 at 18:39
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Mind you, IMHO this question was a duplicate already five years ago, so I won't give my usual upvote to the correct answer. – Jyrki Lahtonen May 08 '17 at 18:40
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The number of degree $n$ irreducible polynomials in a finite field $k$ of order $q$ is $$\frac1n\sum_{d\mid n}\mu(d)q^{n/d}$$ where $\mu$ is the Mobius function. This is obtained by considering the field of $q^n$ elements and considering which field each generates over $k$.
Angina Seng
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