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11 E

Expert-verifiedFound in: Page 365

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 ${\mathit{f}}{\left(x\right)}{\mathbf{\in}}{\mathit{F}}{\left[x\right]}$. If is the greatest common divisor of $\mathit{f}\left(x\right)$ and ${\mathit{f}}{\mathbf{\text{'}}}{\left(x\right)}$ and ${\mathit{h}}{\left(x\right)}{\mathbf{=}}{\mathit{f}}{\left(x\right)}{\mathbf{/}}{\mathit{d}}{\mathit{x}}{\mathbf{\in}}{\mathit{F}}{\left[x\right]}$, prove **

** **

**(a) ${\mathit{f}}{\left(x\right)}$ and ${\mathit{h}}{\left(x\right)}$ have the same roots in K.**

**(b) ${\mathit{h}}{\left(x\right)}$ is separable**

- It is proved that $f\left(x\right)$ and have same roots in .
- It is proved that is separable.

**A polynomial ${\mathit{p}}{\left(x\right)}$ 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.**

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

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 $\mathit{h}\left(c\right)\mathbf{=}\mathbf{0}$. 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 $K$.

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 $\mathrm{gcd}\left(h\left(x\right),\text{\hspace{0.33em}}h\text{'}\left(x\right)\right)=1$.

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.

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