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I'm a little bit confused that the number of irreducible polynomials of degree $n$ over $\mathbb{F}_{p}$.

In Dummit & Foote's textbook(Abstract algebra, 3rd), the number of $\textbf{irreducible}$ polynomials of degree $n$ over $\mathbb{F}_{p}$ is given by the following equation:$$\frac{1}{n}\cdot\sum_{d\mid n}\mu(d)\cdot p^{\tfrac{n}{d}}.$$

Is this only counting the number of $\textbf{monic irreducible}$ polynomials of degree $n$ over $\mathbb{F}_{p}$?

There is no constraint about $\textbf{monic}$ in the textbook. Can anyone help me? Thank you.

AnonyMath
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    Yes, "monic" , see the other posts about this topic, e.g., here, and in particualr here. – Dietrich Burde Apr 01 '20 at 09:16
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    Theorem: Let $\mu(n)$ denote the Möbius function. The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}q$ is the Necklace Polynomial $$M_n(q) = \frac{1}{n} \sum{d | n} \mu(d) q^{n/d}.$$

    To get the number of irreducible polynomials just multiply by $q - 1$

     – Dietrich Burde Apr 01 '20 at 09:21
  • @DietrichBurde Thank you. In Dummit & Foote's textbook, this theorem that 'the polynomial $x^{p^{n}}-x\in\mathbb{F}{p}[x]$ is precisely the product of all the distinct $\textbf{irreducible}$ polynomials in $\mathbb{F}{p}[x]$ of degree $d$ runs through all divisors of $n$' was used to derive that formula. So, I was confused for the monic condition. I think it needs to be fixed, isn't it? – AnonyMath Apr 01 '20 at 12:51

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