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I was given a question about finite field. Let $F$ be a finite field with $q$ elements, say with characteristic $p$ and $x_k$ be the number of monic irreducible polynomials with degree $k$ in $F[x]$. Why we always have the formal power series$1+qt+...+q^nt^n+...=\prod_{i=1}^{\infty}\frac{1}{{(1-t^i)}^{x_i}}$. I recognize that the left hand side is just 1 but I don't see why the right hand side is 1.

singlovehk
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  • It is a FORMAL series the letter $t$ is indeterminated. If your left hand side were equal to 1 you always have $qt+....+q^nt^n+...=0$ – Piquito Oct 17 '15 at 12:48
  • I am not sure if the left hand side is equal to 1. However, I thought that the field has characteristic p which divides q, hence all the non-constant term will be equal to 0. – singlovehk Oct 17 '15 at 14:31
  • In what course were you given that formula? – Georges Elencwajg Oct 17 '15 at 21:00
  • In a first course of graduate algebra, which talks about caregory theory, module and galois theory – singlovehk Oct 18 '15 at 00:42

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The equality $$ 1+qt+q^2t^2+\ldots+q^nt^n+\ldots=\prod_{i=1}^{\infty}\dfrac{1}{(1-t^i)^{x_i}}$$ holds in $\mathbb Z[[t]]$. The left hand side is the generating function associated to the sequence $$l_k=|\{p(x)\in \mathbb F_q[x]| p \mbox{ is monic }, \deg p=k\}|=q^k.$$ If you expand the right hand side you can see that is equal to $$ (1+t+t^2+\ldots)^{x_1}(1+t^2+t^4+\ldots)^{x_2}\cdot \ldots \cdot(1+t^k+t^{2k}+\ldots)^{x_k}\cdot\ldots$$ Now you should observe that the coefficients $r_k$ of $t^k$ in this series represent the number of ways to multiply irreducible monic polynomials in order to have a monic polynomial of degree $k$, that is $$ r_k=\sum_{a_1+2a_2+\ldots +ka_k=k}\prod_{i=1}^k {x_i}^{a_i}$$

Sfarla
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