Question

Let $\mathbf{p}\left(x\right)$be irreducible in $\mathbf{F}\left[x\right]$. Prove that $\mathbf{p}\left(x\right)$is separable if and only if ${\mathbf{p}}^{\mathbf{\text{'}}}\left(x\right)\mathbf{\ne }{\mathbf{0}}_{\mathbf{F}}$.

Short answer

p(x) is separable if and only if ${\mathrm{p}}^{\text{'}}\left(\mathrm{x}\right)\ne 0_{\mathrm{F}}$.

Step-by-step solution

Definition of irreducible polynomial

Irreducible polynomial is a non constant polynomial that cannot be factored into the product of two non constant polynomial.

Showing that p(x)  is separable if and only if  p'(x)≠0

Let $f^{\text{'}}\left(x\right)\ne 0$and d(x) be any non trivial common factor of p(x) and $p^{\text{'}}\left(x\right)$. Then $degd\left(x\right)\le deg{p}^{\text{'}}\left(x\right)<degp\left(x\right)$ .

Since f(x) is irreducible and d(x) be proper factor of f(x). So, deg d(x)=0 .

This mean d(x)=k here k is any constant so without loss of generality let k=1 . Thus d(k)=1 .

Hence p(x) and $p^{\text{'}}\left(x\right)$ are relatively prime. Hence all zeroes of p(x) are simple. Therefore p(x) is separable.

Conversely assume that p(x) is a separable polynomial. So, it’s all zeroes are simple.

Let $\alpha$be any zero of p(x) in some splititing field of p(x). Then over this splitting field:

$\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{\alpha }\right)\mathrm{g}\left(\mathrm{x}\right),\mathrm{g}\left(\mathrm{\alpha }\right)\ne 0$

Differentiate:

$p^{\text{'}}\left(x\right)=g\left(x\right)+\left(x-\alpha \right)g^{\text{'}}\left(\alpha \right)$

Therefore$p^{\text{'}}\left(x\right)\ne 0$.

So p(x) is separable if and only if $p^{\text{'}}\left(x\right)\ne 0$.