TITLE: Irreducible polynomials have distinct roots?
QUESTION [16 upvotes]: I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.
Theorem 7.3 page 27
seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me.
Does the result hold for $\Bbb F_q$ where $q$ is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots?
Please help.

REPLY [4 votes]: As pointed out, to get an example of an irreducible polynomial with multiple roots, it cannot be finite.  Here is an example where this happens in an infinite field of characteristic $p$.
Consider the irreducible polynomial $f(x)=x^p-t\in \mathbb{F}_p(t)[x]$ in the extension field $\mathbb{F}_p(t^{1/p})$ of the field $\mathbb{F}_p(t)$  of rational functions over $\mathbb{F}_p$.
Notice that $(x - t^{1/p})^p = x^p + (-1)^p t = x^p - t$, where the first equality follows from the binomial theorem and the second is obviously true for all odd $p$, and when $p=2$ we have $t=-t$ so it holds there as well.
Thus $f$ has one root of multiplicity $p$ in $\mathbb{F}_p(t^{1/p})$.