# Irreducible polynomial does not imply separable polynomial

Let $F$ be a field. I want to find an irreducible polynomial $f(X) \in F[X]$ such that is not separable. If $Char F = 0$, it is impossible to find (all the irreducible polynomial in a field of null characteristic are separable polynomials) and I have the next statement too:

Let $F$ be a field and $f(X) \in F[X]$ an irreducible polynomial over $F$. Then $f(X)$ is a separable polynomial if and only if $f'(X) \neq 0$.

So, for instance, I have thought about a polynomial $f(X)$ in ${\mathbb{Z}}_2[X]$ such that $f'(X) = 0$ and it should work as a counterexample. For example $f(X) = X^2 + 1$, as $f'(X) = 2 X = 0$. It looks irreducible over ${\mathbb{Z}}_2[X]$, but how can I state that this polynomial is not separable? Which is the field of descomposition of $f(X)$ over $F$?

• $\mathbb{F}_2$ won't work. All finite fields (and, as you have seen yourself, all characteristic 0 fields) are perfect, meaning that all irreducible polynomials over them have distinct roots. You need an imperfect field like in mathphys' hint, i.e. you need to consider polynomials in $\mathbb{F}_p(t)[X]$. – M.G. Jun 13 '18 at 16:45
• How is ${\mathbb{F}}_p(t)[X]$ defined? – joseabp91 Jun 13 '18 at 19:35
• The coefficients of polynomials from $\mathbb{F}_p(t)[X]$ are rational functions (=fractions of polynomials) in the variable $t$ with coefficients in $\mathbb{F}_p$. – M.G. Jun 13 '18 at 20:09
• Can you define it as a set? I don't understand what you mean with ''in the variable $t$''. And I don't know what is ${\mathbb{F}}_p$. – joseabp91 Jun 13 '18 at 20:13
• $\mathbb{F}_p:=\mathbb{Z}/(p)$, you probably denote it as $\mathbb{Z}_p$, which is kinda bad notation in algebra since $\mathbb{Z}_p$ also has another meaning. Then $\mathbb{F}_p(t) = \{P/Q: P,Q\in\mathbb{F}_p[t], Q\neq 0\}$. It is a little bit odd that you don't know the notation $\mathbb{F}_p$, but are asked about irreducibility and separability of polynomials. – M.G. Jun 13 '18 at 20:28

## 1 Answer

Hint: consider the field $\Bbb{F}_p(t)$ where $p$ is prime ($\Bbb{F}_p$ is a finite field) and $t$ is a variable.

Expanding on my hint earlier, $\Bbb{F}_p(t)$ is just the finite field of $p$ elements with $t$ a variable adjoined. You're probably familiar with something like $\Bbb{R}[x]$, the real numbers with a variable $x$ adjoined. This is the ring of all polynomials with coefficients in the real numbers. $\Bbb{F}_p(t)$ is similar, but here we are also allowed to "divide" by the variable $t$. So this is not only a ring, but a field too. For the formal definition, see July's comment.

Anyhow, consider the polynomial $f(x) = x^p - t \in \Bbb{F}_p(t)[x]$. Let $\alpha$ be a root of this. Then consider $f(x) \in \Bbb{F}_p(t)(\alpha)[x]$. Since $\text{char}(\Bbb{F}_p(t)(\alpha)) = p$ prime, we have that $$f(x) = x^p - t = x^p - \alpha^p = (x-\alpha)^p.$$ It can be shown that $f$ is irreducible in $\Bbb{F}_p(t)[x]$, and so since $\alpha$ is a multiple root, we do not have separability.

• How is $\mathbb{F}_p(t)$ defined? – joseabp91 Jun 13 '18 at 18:27
• Please see my recent edit. – mathphys Jun 13 '18 at 20:39
• This is incorrect. $x^p - t^p$ is reducible in $\mathbb F_p(t)[x]$ -- it factors as $(x-t)^p$. I guess the polynomial you were looking for was $x^p-t$. – Mathmo123 Jun 13 '18 at 20:54
• Indeed, thanks. – mathphys Jun 13 '18 at 20:59
• Why do you deduce that char$({\mathbb{F}}_p(t)(\alpha)) = p$? – joseabp91 Jun 13 '18 at 21:11