Irreducible implies Separable in a Finite Field

Question

Proposition 37 on page 549 of Abstract Algebra, 3rd Ed. by Dummit and Foote claims that irreducible implies separable over a finite field.

Suppose $p(x)$ is irreducible over a finite field of characteristic $p$ and (for the purpose of arriving at a contradiction) suppose $p(x)$ is not separable.

From this I can only conclude that $p'(x)$ is zero for at least one value. However, the authors make the claim that the derivative is identically zero (equivalently, each element in the domain is a multiple root) by claiming that $p(x)$ is a polynomial in $x^p$.

Why is $p(x)$ necessarily a polynomial in $x^p$? And why is it the case that over a finite field we have that either $p'(x)$ is never zero or is identically $0$?

Answer 1

If $p(x)$ has a multiple root, then $p(x)$ and $p^{\prime}(x)$ have a common factor. But since $p(x)$ is irreducible and $p^{\prime}(x)$ has degree strictly smaller than $p(x)$, this implies that $p^{\prime}(x)$ is the zero polynomial. From this it then follows that $p(x)$ is a polynomial in $x^p$.

Answer 2

For your first question:

If $p(x) = \sum_{j=1}^n a_jx^j$ then $p'(x) = \sum_{j=1}^nja_jx^{j-1}$. If $p'(x) = 0$, then each of its coefficients must lie in the ideal $(p)$, that is, $p|ja_j$ for each $j$. Thus, $a_j$ is either $0$ mod $p$ or $p|j$, so for each $a_jx^j$ with $a_j$ non-zero we must have $p|j$. So $p$ is a polynomial in $x^p$.

Maybe you can use this to work out the rest.

Answer 3

If $p(x)$ is non-separable (i.e., has a root of multiplicity $\geq 2$), then $p(x)$ and $p'(x)$ are not relatively prime -- that is, they must share a root $u$ in some extension $\mathbb{K}$ of the finite field of characteristic $p$ (call it $\mathbb{F}_p$). We can make $p(x)$ monic to satisfy the definition of the minimal polynomial of $u$ (let monic $p(x)$ be notated $p_1(x)$). We then have that any other polynomial $q(x)\in\mathbb{F}_p[x]$ with the same root $u$ must be divisible by $p_1(x)$; that is, $$p_1(x) \mid q(x).$$ Since monicness does not change divisibility, the above is interchangeable with $$p(x)\mid q(x).$$ In the case $q(x)=p'(x)$, we have necessarily that $p'(x)=0$ as the divisibility equation is not true otherwise. From this, it follows that $p(x)\in\mathbb{F}[x^p]$ (as described by @GiantTortoise1729), and then by the Freshman's Dream lemma for commutative rings of characteristic $p$ and Fermat's Little Theorem (which is a consequence of the Frobenius automorphism), $p(x)$ is reducible, a contradiction.

Edit: The Freshman's Dream lemma has a proof that can be altered to be applied to polynomials with coefficients in a commutative ring of characteristic $p$ (as everything simplifies the same way with the binomial coefficient). That is why it is applicable in this setting.