Let $F=\mathbb{F}_{q}$ be a finite field (so $q=p^k$ for some prime $p$ and positive integer $k$), and let $\varphi(d)$ denote the number of monic irreducible polynomials of degree $d$ in $F[X]$. I'm supposed to show that $\displaystyle{\sum_{d \mid n} d \varphi(d) = q^n}$.
I see there are previous questions about this topic and even a paper, but all (save one) seem to employ the use of the Möbius function and Möbius inversion - both topics I have not covered yet in class. There is also this answer, but it appears to hinge upon the extension having prime degree. Is there some way to show this without explicitly coming up with a formula for the number of irreducible monic polynomials of a given degree in $F[X]$?
Any help would be greatly appreciated.
Answer
The splitting field of $X^{q^n}-X$ is ${\bf F}_{q^n}$. Every irreducible $\pi(X)$ of degree $d$ splits in ${\bf F}_{q^n}$ and every element of ${\bf F}_{q^n}$ is a root of $X^{q^n}-X$, and thus $\pi(X)\mid(X^{q^n}-X)$. Furthermore $X^{q^n}-X$ has no repeated roots so each irreducible $\pi(X)$ of degree $d\mid n$ must appear in its factorization precisely once. Therefore we have the conclusion
$$X^{q^n}-X=\prod_{d\mid n}\prod_{\deg\pi=d}\pi(X).$$
Taking degrees yields $\displaystyle q^n=\sum_{d\mid n}d\varphi(d)$ (and from here Möbius inversion yields $\varphi(d)$).
No comments:
Post a Comment