# Irreducible implies Separable in a Finite Field

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 $0$?

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 $x^p$.

• I like this correct proof, but can you address the second question? Is it true that (regardless of whether or not $p$ is irreducible) that $p'$ is either identically $0$ or never $0$? – The Substitute Apr 16 '16 at 21:11
• It's certainly not true if $p(x)$ is reducible. Consider $p(x)=x^2$ in $\mathbb{F}_3$. – carmichael561 Apr 16 '16 at 22:03

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 $x^p$.
• When you write "$p'(x)=0$" do you mean that $p'$ has a root or that it is the zero function? – The Substitute Apr 16 '16 at 3:41
• $p'$ is zero function. – GiantTortoise1729 Apr 16 '16 at 5:06
• I understand the implication, but I am not sure why $p'$ is the zero function. – The Substitute Apr 16 '16 at 5:20
• I'm assuming $p'$ is the zero function. But in general, if $p$ is a polynomial in $x^p$ then the coefficient of every term in the derivative is a multiple of $p$, and thus $p'$ will be zero in your field. – GiantTortoise1729 Apr 16 '16 at 14:30