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Let $p$ be a prime number and $n \in \mathbb{N}^+$. Let $H_n$ be the product of all monic irreducible polynomials in $\mathbb{F}_p[T]$ whose degree is equal to $n$. What is known about $H_n$? Is there an explicit formula? Where does it appear in the literature?

It is well-known that the product of all monic irreducible polynomials whose degree divides $n$ is equal to $T^{\large p^n}-T$. It follows that, if $\ell$ is a prime number, then $$H_{\ell} = (T^{\large p^\ell} - T)/(T^p - T).$$ More generally, if $n = \ell^s$ is a power of a prime number, then $$H_{\large \ell^s} = (T^{\large p^{\ell^s}} - T)/(T^{\large p^{\ell^{s-1}}} - T).$$ For the product of two primes it is also easy, but for $n = 12 = 2^2 \cdot 3$ we need the principle of inclusion-exclusion to find (if I am not mistaken) $$H_{12} = \dfrac{(T^{\large p^{12}} - T) (T^{\large p^2}-T)}{(T^{\large p^4}-T)(T^{\large p^6}-T)}.$$ I assume that a similar approach works in general. I am looking in particular for a good reference.

Martin Brandenburg
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2 Answers2

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We have the formula, valid for all $n$: $$T^{p^n}-T=\prod_{d\mid n}H_d(T).$$ The general (multiplicative) version of Möbius inversion then gives the answer $$H_n(T)=\prod_{d\mid n}(T^{p^d}-T)^{\mu(n/d)}.$$

This formula is also the content of Theorem 3.29 in Lidl & Niederreiter. The variant formulation in Qiaochu Yuan's answer is Theorem 3.31 (loc.cit.).

Jyrki Lahtonen
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We can write down something with no divisions as follows. $T^{p^n} - T$ has the factorization (over $\mathbb{Z}$!)

$$T^{p^n} - T = T \prod_{d \mid p^n - 1} \Phi_d(T)$$

where $\Phi_d$ are the cyclotomic polynomials. The roots of $\Phi_d(T)$ over $\overline{\mathbb{F}_p}$ are the primitive $d^{th}$ roots of unity; on the other hand, if $\alpha \in \overline{\mathbb{F}_p}$ then the condition that its minimal polynomial have degree exactly $n$ is the condition that $\alpha$ has order exactly $n$ under the Frobenius map, meaning $n$ is the smallest positive integer $k$ such that $\alpha^{p^k} = \alpha$, or equivalently, if $\Phi_d(\alpha) = 0$, that $n$ is the smallest positive integer $k$ such that $p^k \equiv 1 \bmod d$. This gives

$$\boxed{ H_n(T) = \prod_{\text{ord}_d(p) = n} \Phi_d(T) }.$$

Qiaochu Yuan
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