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### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624

# Assume F has characteristic 0 and K is a splitting field of ${\mathbit{f}}\left(x\right){\mathbf{\in }}{\mathbit{F}}\left[x\right]$. If is the greatest common divisor of $\mathbit{f}\left(x\right)$ and ${\mathbit{f}}{\mathbf{\text{'}}}\left(x\right)$ and ${\mathbit{h}}\left(x\right){\mathbf{=}}{\mathbit{f}}\left(x\right){\mathbf{/}}{\mathbit{d}}{\mathbit{x}}{\mathbf{\in }}{\mathbit{F}}\left[x\right]$, prove (a) ${\mathbit{f}}\left(x\right)$ and ${\mathbit{h}}\left(x\right)$ have the same roots in K.(b) ${\mathbit{h}}\left(x\right)$ is separable

1. It is proved that $f\left(x\right)$ and have same roots in .
2. It is proved that is separable.
See the step by step solution

## Describe the condition for separable field

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

## Prove that f(x) and h(x) have the same roots in K.

Claim: $f\left(x\right)$ and $h\left(x\right)$ have same roots in $\mathbit{K}$.

Let $c$ be the root of $h\left(x\right)$. This implies that $h\left(c\right)=0$.

Consider $h\left(x\right)=\frac{f\left(x\right)}{d\left(x\right)}$

Substitute $x=c$ in $h\left(x\right)$.

$h\left(c\right)=\frac{f\left(c\right)}{d\left(c\right)}$

Since $\mathbit{h}\left(c\right)\mathbf{=}\mathbf{0}$. Therefore, $\frac{f\left(c\right)}{d\left(c\right)}=0$.

That means $f\left(c\right)=0$

Thus, $f\left(c\right)=0$ indicates that $c$ is the root of $f\left(x\right)$.

Therefore, $f\left(x\right)$ and $h\left(x\right)$ have same roots in $K$.

## Prove that is h(x) separable

Claim: $h\left(x\right)$ is separable.

So, just show that $h\left(x\right)$ and $h\text{'}\left(x\right)$ both are relatively prime.

That is $\mathrm{gcd}\left(h\left(x\right),\text{ }h\text{'}\left(x\right)\right)=1$.

Consider $h\left(x\right)=\frac{f\left(x\right)}{d\left(x\right)}$.

Differentiate $h\left(x\right)$ with respect to .

$\begin{array}{c}h\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)d\left(x\right)-d\text{'}\left(x\right)f\left(x\right)}{{\left(d\left(x\right)\right)}^{2}}\\ =\frac{f\text{'}\left(x\right)}{d\left(x\right)}-\frac{f\left(x\right)}{d\left(x\right)}\frac{d\text{'}\left(x\right)}{d\left(x\right)}\end{array}$

Suppose on the contrary $h\left(x\right)$ and $h\text{'}\left(x\right)$ both are not relatively prime.

Let is the common root of $h\left(x\right)$ and $h\text{'}\left(x\right)$, this implies that $h\left(c\right)=0$ and $h\text{'}\left(c\right)=0$.

Since $h\left(c\right)=0$ implies that $f\left(c\right)=0$. Now,

$h\text{'}\left(c\right)=\frac{f\text{'}\left(c\right)d\left(c\right)-d\text{'}\left(c\right)f\left(c\right)}{{\left(d\left(c\right)\right)}^{2}}$

So $h\text{'}\left(c\right)=0$ if $f\text{'}\left(c\right)=0$. But $f\text{'}\left(c\right)=0$ this contradicts the fact that $d\left(x\right)$ is the greatest common divisor of $f\left(x\right)$ and $f\text{'}\left(x\right)$.

Therefore, the assumption is wrong.

Thus, $h\left(x\right)$ and $h\text{'}\left(x\right)$ both are relatively prime.. so, claim follows and $h\left(x\right)$$h\left(x\right)$ is separable.

Hence, it is proved that is separable.