Product of monic polynomials in finite fields

Question

I am trying to show that the product of monic polynomials of degree $n$ in $\mathbb{F}_p[T]$ is given by $\prod_{i=0}^{n}(T^{p^n}-T^{p^i})$. I tried generating function but with no luck. Any hint?

Answer 1

Let $P_n(T)$ be the product of all monic polynomials of degree $n$ in $\mathbb{F}_p[T]$. We can show that $P_n(T)=\prod_{i=0}^{n-1}(T^{p^n}-T^{p^i})$ (the bounds on the product were slightly off as stated in the original question). Clearly $P_1(T)=T^p-T$, so for induction purposes, it suffices to show that $P_n(T)=(T^{p^n}-T)\cdot P_{n-1}(T)^p$.

Consider the unique factorization of $P_n(T)$ into irreducible factors. Let $g(T)\in \mathbb{F}_p[T]$ be irreducible and denote $m=\deg(g)$. Set $k=\lfloor n/m \rfloor$. The number of times $g(T)$ shows up in the factorization of $P_n(T)$ is given by \begin{align*} \text{ord}_g(P_n)=\sum_{i=1}^k f_i \end{align*} where $f_i$ is the number of distinct monic polynomials of degree $n$ in $\mathbb{F}_p[T]$ divisible by $g(t)^i$. In other words, $f_i$ is equal to the number of monic polynomials of degree $n-im$ in $\mathbb{F}_p[T]$, i.e. $f_i=p^{n-im}$.

This shows that \begin{align*} \text{ord}_g(P_n)= \begin{cases} p\cdot\text{ord}_g(P_{n-1}), &&\text{if}\; m\nmid n,\\ p\cdot\text{ord}_g(P_{n-1}) + 1, &&\text{if}\; m\mid n. \end{cases} \end{align*}

We arrive at \begin{align*} P_n(T)&=\bigg(\prod_{\substack{g \text{ irreducible, monic}\\\deg(g)\mid n}} g(T)\bigg)P_{n-1}(T)^p\\ &=(T^{p^n}-T)P_{n-1}(T)^p, \end{align*} which proves the claim.