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 $F$?
Best 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.