I have shown that if $f$ is irreducible with $deg(f) = d$ then $\mathbb{F}_p [t] / \langle f\rangle$ is a field with $p^d$ elements. I've also shown that $\langle f\rangle$ is precisely those polynomials $g$ where $g(\alpha) = 0$ where $\alpha$ is the image of $t$ in the quotient map.
Now I need to show that this implies $f$ divides $t^{p^d-1}-1$. I have worked out some facts but I'm having trouble putting them together:
- $t^{p^d} - t$ is the zero polynomial as it is zero on every element of the field.
- I want to some how show that then this polynomial is 0 and so it is in the ideal generated by $\langle f\rangle$ and then some how use the fact $\langle f\rangle$ is maximal to see that $f$ then divides it.
Thanks.
Consider the element $[t]\in\mathbb{F}_p[t]/<f>.$ Since the multiplicative group of non-zero elements in $\mathbb{F}_p[t]/<f>$ has $p^{d}-1$ elements and contains $[t]$, we know that $[t]^{p^d-1}=1$. This means precisely that $[t^{p^d-1}-1]=0$ in $\mathbb{F}_p[t]/<f>,$ meaning that $f$ divides $t^{p^d-1}-1$.