September 12th
Today I learned that $\FF_q$ is also a perfect field for prime-powers $q,$ continuing from the Conrad expository paper. Namely, we want to show that no irreducible polynomial $\pi(x)\in\FF_q[x]$ has repeated roots in the algebraic closure $\overline{\FF_q}.$ The outline is to show that $\pi(x)$ divides into a polynomial with no repeated roots, from which the result will follow. Note that\[\frac{\FF_q[x]}{(\pi(x))}\]is a finite field, of order $Q:=q^{\deg\pi}.$ In fact, this is the splitting field of $\pi(x),$ but we don't that strong of information to complete the proof completely. Note if $\pi(x)=x,$ then we're already done; else look at $x$ in the above finite field, which is a nonzero element with multiplicative order dividing $Q-1.$ It follows\[x^Q\equiv x\pmod{\pi(x)}.\]Thus, $\pi(x)\mid x^Q-x.$ However, $x^Q-x$ has no repeated roots in the algebraic closure: its derivative is $Qx^{Q-1}-1=-1$ has $\textit{no roots at all!}$ So it follows that $\pi(x)$ cannot have repeated roots for $x^Q-x$ to also have no repeated roots, which completes the proof.
As an aside, the paper has a pretty clean finish which avoids the derivative. Namely, suppose $\alpha$ is a root of $x^Q-x$ so that we want to know that $(x-\alpha)^2$ is not a root. Well, just write\[x^Q-x=x^Q-x-\left(\alpha^Q-\alpha\right)=(x-\alpha)^Q-(x-\alpha)=(x-\alpha)\left((x-\alpha)^{Q-1}-1\right).\]However, $\alpha$ is not a root of the other factor because it gives $\alpha-\alpha=0,$ so this evaluates to $-1.$ This is roughly isomorphic to taking the derivative (note that the end contradiction $-1\ne0$ is the same), but it's a bit more clever with the $\alpha^Q-\alpha$ trick.