Question

Find a splitting field of $x^6+x^3+1$over $\mathbb{Q}$.

Short answer

The splitting field of polynomial $f\left(x\right)=x^6+x^3+1$ over $\mathbb{Q}$ field is $\mathbb{Q}\left(\xi \right)$.

Step-by-step solution

Definition of splitting field

There is a existence of a split field K of$f\left(x\right)$overF such that$[K:F]\le nI$ if Fbe a field and$f\left(x\right)$a non-constant polynomial of degree nin$f\left(x\right)$.

Find a splitting field of  x6+x3+1over ℚ.

Factorize the polynomial$f\left(x\right)$by taking $f\left(x\right)=0$.

$\begin{array}{c}{x}^6+x^3+1=0\\ {\left(x^3\right)}^2+x^3+1=0\\ x^3=\frac{-1\pm \sqrt{1-4}}{2}\\ x^3=\frac{-1\pm i\sqrt{3}}{2}\end{array}$

The factors are $x^3=\frac{-1+i\sqrt{3}}{2}$and $x^3=\frac{-1-i\sqrt{3}}{2}$.

Consider .$x^3=\frac{-1+i\sqrt{3}}{2}$

Use the De Moivre’s formula to find $x$as follows.

$\begin{array}{c}{x}^3=\frac{-1+i\sqrt{3}}{2}\\ =-\frac{1}{2}+i\frac{\sqrt{3}}{2}\\ =\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\\ =\cos (\frac{2\pi }{3}+2n\pi )+i\sin (\frac{2\pi }{3}+2n\pi )\end{array}$

Simplify further.

$\begin{array}{c}x={[\cos (\frac{2\pi }{3}+2n\pi )+i\sin (\frac{2\pi }{3}+2n\pi )]}^{\frac{1}{3}}\\ ={[\cos \left(\frac{2\pi +6n\pi }{3}\right)+i\sin \left(\frac{2\pi +6n\pi }{3}\right)]}^{\frac{1}{3}}\\ =\cos \left(\frac{2\pi +6n\pi }{9}\right)+i\sin \left(\frac{2\pi +6n\pi }{9}\right)\mathrm{......}\left(1\right)\end{array}$

Here,$n=0,1,2,\mathrm{...}$.

Calculate values of$x$corresponding to$n=0,1,2,\mathrm{...}$as follows.

For ,$n=0$

$\begin{array}{c}x=\cos \left(\frac{2\pi }{9}\right)+i\sin \left(\frac{2\pi }{9}\right)\\ x=e^{\frac{2\pi }{9}i}\end{array}$

For $n=1$,

$\begin{array}{c}x=\cos \left(\frac{8\pi }{9}\right)+i\sin \left(\frac{8\pi }{9}\right)\\ x=e^{\frac{8\pi }{9}i}\end{array}$

For $n=2$,

$\begin{array}{c}x=\cos \left(\frac{14\pi }{9}\right)+i\sin \left(\frac{14\pi }{9}\right)\\ x=e^{\frac{14\pi }{9}i}\end{array}$

Therefore, the values of $x$are $x=e^{\frac{2\pi }{9}i}$, $x=e^{\frac{8\pi }{9}i}$, and $x=e^{\frac{14\pi }{9}i}$.

Consider .

$x^3=\frac{-1-i\sqrt{3}}{2}$

Use the De Moivre’s formula to find $x$as follows.

$\begin{array}{c}{x}^3=\frac{-1-i\sqrt{3}}{2}\\ =-\frac{1}{2}-i\frac{\sqrt{3}}{2}\\ =\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)\\ =\cos (\frac{4\pi }{3}+2n\pi )+i\sin (\frac{4\pi }{3}+2n\pi )\end{array}$

Simplify further.

$\begin{array}{c}x={[\cos (\frac{4\pi }{3}+2n\pi )+i\sin (\frac{4\pi }{3}+2n\pi )]}^{\frac{1}{3}}\\ ={[\cos \left(\frac{4\pi +6n\pi }{3}\right)+i\sin \left(\frac{4\pi +6n\pi }{3}\right)]}^{\frac{1}{3}}\\ =\cos \left(\frac{4\pi +6n\pi }{9}\right)+i\sin \left(\frac{4\pi +6n\pi }{9}\right)\mathrm{......}\left(2\right)\end{array}$

Here, $n=0,1,2,\mathrm{...}$.

Calculate values of$x$corresponding to $n=0,1,2,\mathrm{...}$as follows.

For $n=0$,

$\begin{array}{c}x=\cos \left(\frac{4\pi }{9}\right)+i\sin \left(\frac{4\pi }{9}\right)\\ x=e^{\frac{4\pi }{9}i}\end{array}$

For $n=1$,

$\begin{array}{c}x=\cos \left(\frac{10\pi }{9}\right)+i\sin \left(\frac{10\pi }{9}\right)\\ x=e^{\frac{10\pi }{9}i}\end{array}$

For $n=2$,

$\begin{array}{c}x=\cos \left(\frac{16\pi }{9}\right)+i\sin \left(\frac{16\pi }{9}\right)\\ x=e^{\frac{16\pi }{9}i}\end{array}$

All the six roots of the equation $x^6+x^3+1=0$are as follows.

$\begin{array}{c}x=e^{\frac{2\pi }{9}i},e^{\frac{8\pi }{9}i},e^{\frac{14\pi }{9}i},e^{\frac{4\pi }{9}i},e^{\frac{10\pi }{9}i},e^{\frac{16\pi }{9}i}\\ =e^{\frac{2\pi }{9}i},e^{\frac{4\pi }{9}i},e^{\frac{8\pi }{9}i},e^{\frac{10\pi }{9}i},e^{\frac{14\pi }{9}i},e^{\frac{16\pi }{9}i}\\ =\xi ,{\xi }^2,{\xi }^4,{\xi }^5,{\xi }^7,{\xi }^8\end{array}$

Here, $\xi =e^{\frac{2\pi }{9}i}$

It implies that following:

$x^6+x^3+1=(x-\xi )(x-{\xi }^2)(x-{\xi }^4)(x-{\xi }^5)(x-{\xi }^7)(x-{\xi }^8)$

Hence, the splitting field of polynomial $f\left(x\right)=x^6+x^3+1$over field$\mathbb{Q}$ is $\mathbb{Q}\left(\xi \right)$.