Irreducible polynomials have distinct roots? 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.
 A: 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 $\mathbb{F}_p(t^{1/p})$.
A: Consider a field $F$ of characteristic $p$.
A polynomial has multiple roots only if it has a root in common with its (formal) derivative; that is, the multiple roots of $f$ are the roots of $\gcd(f,f')$. Since $f$ is irreducible, multiple roots can occur only if the $\gcd$ is $f$ itself, that is $f'$ is a multiple of $f$. And that is only possible if $f'=0$, that is, all monomials in $f$ have degree a multiple of $p$, so $f(x)=g(x^p)$ for some polynomial $g$. 
If $F$ is finite, then $\phi\colon a\mapsto a^p$ is an automorphism of $F$ (and also of the splitting field $E$ of our polynomial), and there exists $h(x)$ such that $\phi(h)=g$. Then for $\alpha\in E$ with $f(\alpha)=0$ also $h(\alpha)=0$ (because $\phi(h(\alpha))=\phi(h)(\phi(\alpha))=g(\alpha^p)=f(\alpha)=0$). Since $h$ is of smaller degree than $f$, $f$ is not irreducible.
As this proof shows, one has to look for cases where $\phi$ is not an automorphism to find a counterexample (such as in Andreas Carantis comment).
A: The result holds over any finite field. One way of seeing this is that if $h(x) \in \Bbb{F}_{q}[x]$ (where $q$ is a power of the prime $p$) is irreducible over $\Bbb{F}_{q}$, and $\alpha$ is one of its roots, then $\alpha$ is also algebraic over $\Bbb{F}_{p}$. If $f(x) \in \Bbb{F}_{p}[x]$ is the minimal polynomial of $\alpha$ over $\Bbb{F}_{p}$, then $h(x)$ divides $f(x)$, and you know that the latter has distinct roots.
There are examples, though, of irreducible polynomials of degree $> 1$ over an infinite field of positive characteristic which have only one root.
