Question

Prove that $\mathbf{f}\left(x\right)\mathbf{\in }\mathbf{F}\left[x\right]$is separable if and only if $\mathbf{f}\left(x\right)$and $\mathbf{F}\left[x\right]$are relatively prime.

Short answer

It is proved that $f\left(x\right)=F\left[x\right]$ is separable if and only if$f\left(x\right)$ and$\mathrm{F}\left[\mathrm{x}\right]$ are relatively prime.

Step-by-step solution

Definition of polynomial

A polynomial $p\left(x\right)$ over a given field K is separable if its root are distinct in an algebraic closure of K. That is the number of root is equal to the degree of the polynomial.

Showing that f(x)∈F[x] is separable

Let $\mathrm{f}\left(\mathrm{x}\right)\in \mathrm{F}\left[\mathrm{x}\right]$be a nonzero polynomial and $f\left(x\right)$is separable and $\alpha$be any root of $f\left(x\right)$. Then:

$\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{x}-\mathrm{\alpha }\right)\mathrm{h}\left(\mathrm{x}\right)$withrole="math" localid="1659078573097" $h\left(\alpha \right)\ne 0$

This implies that $\alpha$is not a root of $f^{\text{'}}\left(x\right)$. Therefore $f\left(x\right)$and $f^{\text{'}}\left(x\right)$have no common roots.

So, they have no common factors in $F\left[x\right]$. Hence they are relatively prime. Now let $f\left(x\right)$is not separable. So, it has repeated roots in a splititing field over F.

That root is also a root of $f^{\text{'}}\left(x\right)$when $f\left(x\right)={\left(x-\alpha \right)}^{2}g\left(x\right)$.

Differentiate$f\left(x\right)$,

$$\begin{gathered} {\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)={\left(\mathrm{x}-\mathrm{\alpha }\right)}^2{\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)+2\left(\mathrm{x}-\mathrm{\alpha }\right)\mathrm{g}\left(\mathrm{x}\right) \\ {\mathrm{f}}^{\text{'}}\left(\mathrm{\alpha }\right)=0 \end{gathered}$$

Since $f\left(x\right)$and $f^{\text{'}}\left(x\right)$ are not relatively prime in $F\left[x\right]$. Now take contrapositive to the above statement. If $f\left(x\right)$and $f^{\text{'}}\left(x\right)$ are relatively prime $F\left[x\right]$ then $f\left(x\right)$has no repeated root and hence $f\left(x\right)\in F\left[x\right]$ is separable.