# Irreducible inseparable polynomial of prime degree has derivative zero

Let $F$ be a field and let $f(x)$ be irreducible polynomial of degree $p$ "prime" in $F[x]$ having splitting field $K$. Let $[K,F]=pt$, for some integer $t$, then:

1) $f(x)$ is irreducible over $F$

2) If $t$ > $1$, then $f(x)$ is separable over $F$

My question: In part (2), we prove it by contradiction, so we suppose $f(x)$ is not separable, and since it is irreducible, its derivative is zero. and so $f(x)= a_px^{p}+a_0$. What I don't understand is the derivative of $f(x)$ being zero and the coefficients between $a_n$ and $a_0$ being zeros. Could someone explain?

• Write out a general polynomial $\sum_{i=0}^n a_ix^i$ and its derivative. – Hoot Aug 21 '15 at 3:14
• I don't think I understand (1). Isn't $f(x)$ an element of $F(x)$? And $f$ is by assumption irreducible over $F$. So what exactly is to be proved in part (1)? – whacka Aug 21 '15 at 3:28
• @Hoot $f(x)=a_px^{p}+...+a_0$ and $f'(x)= pa_px^{p-1} +(p-1)a_{p-1}x^{p-2}+....+a_1$. Why is $f'(x)=0$ ? – MATH Aug 21 '15 at 3:29
• @whacka you are right, thank you. – MATH Aug 21 '15 at 3:31
• This is an important field theory fact: a polynomial $f$ has repeated roots $\iff \gcd(f,'f)\ne1$. If $f$ is irreducible, then $f$ and $f'$ can only share a nontrivial factor if $f'=0$. – whacka Aug 21 '15 at 3:46

If $K/F$ is not seperable, then $f(x)$ is not. Now if $f(x)$ is irreducible and inseperable then $F$ is an infinite field of prime characteristic, say $q$, avoiding confusion with $p$. And so, you can find a polynomial $g(x)$ in $\mathbb{F}[x]$ such that $f(x)=g(x^{q})$. Setting the degrees equal, $q$ divides $p$ so $q=p$. Now ( I will be assuming $f(x)$ is monic), we can write $f(x)=x^{p}-a$ for some $a \in \mathbb{F}$. Now let $\alpha \in K$ such that $\alpha^{p}=a$. So, $f(x)=(x-\alpha)^{p}$ (char $p$), so $K=\mathbb{F}(\alpha)$ which has degree $p$ over $F$ so $t=1$.