Why can't this happen in fields $K$ with $\operatorname{char}(K)=0$?

Question

Let $f(x) \in K[x]$ irreducible. Then: $f(x)$ separable $\Leftrightarrow $ $(f, f')=1$

$f$ is irreducible so $$ (f,f')=\left\{\begin{matrix} f\\ 1 \end{matrix}\right. $$

$f(x)=a_0+a_1 x + \dots + a_n x^n \\ f'(x)=a_1+2a_2 x+ \dots+ n a_n x^{n-1}$

So, the only case that we have $(f, f')=f$, that means that $f \mid f'$ is when $f' \equiv 0$.

Therefore, in fields such as $\mathbb{Q}, \mathbb{R}, \mathbb{C}$ (that means in fields $K$ with $\operatorname{char}(K)=0$) it cannot happen, so all the irreducible polynomials are separable.

Could you explain to me why this cannot happen in fields $K$ with $\operatorname{char}(K)=0$?

Answer 1

Suppose $f\ne0$ and $ K $ has characteristic zero. Then $f (x)= a_mx^m+... $ where $ a_m \ne0$. So the leading coeff in $ f'(x) $ is $ ma_m $ which is not zero (as $ K $ has characteristic zero) So $ f'$ can't be zero.

Answer 2

$deg(f')<deg(f)$

So $f\nmid f'$ when char F=0

Note that $f'$ is not $0$, as when we talk about an irreducible polynomial, it is non-constant by definition.

$f$ may divide $f'$ in a field of characteristic $p$. (when $f'\equiv 0\pmod{p}$)