# Product of all irreducibles with degree divisible by $n$ in $\mathbb{F}_{q^n}$?

In the finite field of $q^n$ elements, the product of all monic irreducible polynomials with degree dividing $n$ is known to simply be $X^{q^n}-X$. Why is this?

I understand that $q^n=\sum_{d\mid n}dm_d(q)$, where $m_d(q)$ is the number of irreducible monic polynomials with degree $d$, and that each element of the field satisfies $X^{q^n}-X$.

How does the conclusion follow from this? I tried substituting the exponent to $X^{\sum_{d\mid n}dm_d(q)}-X$ but this doesn't seem to do much good.

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Your first $q^n$ should be $q$. The product of all monic irreducible polynomials of degree dividing $n$ over the field of $q^n$ elements is $X^{(q^n)^n}-X$: there are at least $q^n$ of degree $1$, and you also have irreducibles of degree $n$, among others. – Arturo Magidin Feb 7 '12 at 17:07

Let $\mathbf{F}$ be the field of $q$ elements, and fix $n\geq 1$. The extension of degree $n$ over $\mathbf{F}$ is the field with $q^n$ elements. Call that $\mathbf{K}$.

If $a\in\mathbf{K}$, then $\mathbf{F}(a)$ is an intermediate field, so $[\mathbf{F}(a):\mathbf{F}]$ divides $n$; that is, the monic irreducible polynomial of $a$ is of degree dividing $n$ over $\mathbf{F}$. So every element corresponds to a monic irreducible.

Moreover, every monic irreducible of degree dividing $n$ corresponds to an element in $\mathbf{K}$: if $f(x)$ is such an irreducible, and $a$ is a root, then $\mathbf{F}(a)$ has degree $\deg(f)$, which divides $n$, and so is contained in the field extension of degree $n$ (remember that $\mathbf{F}_{q^r}\subseteq \mathbf{F}_{q^s}$ if and only if $r|s$).

That means that if you let $P(X)$ be the product of all monic irreducible polynomials over $\mathbf{F}$ that have degree dividing $n$, then its roots are precisely the elements of $\mathbf{K}$.

We also know that $\mathbf{K}$ is the splitting field of $X^{q^n}-X$: every element of $\mathbf{K}$ satisfies this polynomial (by Lagrange's Theorem, every nonzero element satisfies $a^{q^n-1}=1$, and then there's $0$), and no field strictly smaller than $\mathbf{K}$ can be the splitting field (not enough roots). So now we have two polynomials that are satisfied by every element of $\mathbf{K}$ and only by all the elements of $\mathbf{K}$: $X^{q^n}-X$ and our $P(X)$. So $X^{q^n}-X$ certainly must divide $P(X)$, and $P(X)$ must be a product of linear factors over $\mathbf{K}$.

So the only question that remains is: does $P(X)$ have any repeated roots?

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As noted by Arturo, the problem should be stated about $\mathbb F_q$, not $\mathbb F_{q^n}$.

There are two steps here.

1. Prove that a prime $\pi(x)$ in $\mathbb F_q[x]$ divides $x^{q^n}-x$ if and only if $deg(\pi(x))|n$
2. Prove that there are no repeated prime factors of $x^{q^n}-x$.

First prove that for integers $d,n$, $x^{q^d}-x|x^{q^n}-x$ if and only if $d|n$.

I'll prove the first part of (1) for you. If $deg(\pi(x))=d$, and $d|n$ then consider the field $F=\mathbb F_q[x]/\left<\pi(x)\right>$. It is of order $q^d$, so we know that for every element $y\in F$, $y^{q^d}=y$. In particular, $x$ is an element of $\mathbb F_q[x]$, so the image $\bar x$ of $x$ in $F$ has the property that ${\bar x}^{q^d}-\bar{x}=0$. But that means that the image of $x^{q^d}-x$ is in the kernel of the map from $\mathbb F_p[x]$ to $F$. So $x^{q^d}-x$ is divisible by $\pi(x)$. So, by our first step above, $x^{q^n}-x$ is divisible by $\pi(x)$ when $d|n$.

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