Question

Question: (a) show that $\mathit{\omega }\mathbf{=}\left(-1+\sqrt{3}i\right)\mathbf{/}\mathbf{2}$is a root of${\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{1}$.

(b) Show that$\mathbf{\omega }$and${\mathbf{\omega }}^{\mathbf{2}}$are the root of${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$. Hence$\mathit{Q}\left(\omega \right)$is the splitting field of${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}$.

Short answer

Answer

1. yes,$\omega =\frac{-1\pm \sqrt{3}i}{2}$is a root of$x^3-1$.
2. yes, $Q\left(\omega \right)$is the splitting field.$x^2+x+1$

Step-by-step solution

Explanation

Let the root of$x^3-1$.

i.e$\iff x^3-1=0$

here we factorize it, then we get

$$\begin{gathered} \Rightarrow \left(x-1\right)\left(x^2+x+1\right)=0 \\ \Rightarrow x=1,x^2+x+1=0 \end{gathered}$$

Calculation

$a=1,b=1,c=1$here we factorize $x^2+x+1$.

Here

We know that the formula of D,

$D=b^2-4ac$

Here we find the root,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Put the value a ,b and c here

$$\begin{gathered} =\frac{-1\pm \sqrt{1-4.1.1}}{2.1} \\ =\frac{-1\pm \sqrt{3}i}{2} \end{gathered}$$

The root of is 1,$\frac{-1-\sqrt{3}i}{2}$and $\frac{-1+\sqrt{3}i}{2}$.

Clearly of $\omega =\frac{-1\pm \sqrt{3}i}{2}$is a root of $x^3-1$.

Explanation

Given$g\left(x\right)=x^2+x+1$

We have to show that,$\omega ,{\omega }^2$is a root of.

$\therefore g\left(\omega \right)={\omega }^2+\omega +1$

=0 ( $\omega$ is a cube root of unity )

$\omega$is a root of $g\left(x\right)$.

Solution

Here we know that,

$g\left(\omega \right)={\omega }^2+\omega +1$

Squaring on both sides,

$$\begin{gathered} g\left({\omega }^2\right)={\left({\omega }^2\right)}^2+{\omega }^2+1 \\ ={\omega }^4+{\omega }^2+1 \\ ={\omega }^3.\omega +{\omega }^2+1 \\ ={\omega }^2+\omega +1,\left(\because {\omega }^3=1\right) \end{gathered}$$

Which is show that ${\omega }^2$is also a root of.

splitting field of $g\left(x\right)=x^2+x+1$ over Q.

$Q\left(\omega ,{\omega }^2\right)=Q\left(\omega \right)$

$$\begin{gathered} \left[\because Q\left(\omega \right)=\text{smallest field conatinig}\omega \\ \because \omega \in Q\left(\omega \right)\to \omega .\omega ={\omega }^2\in Q\left(\omega \right) \\ \therefore Q\left(\omega ,{\omega }^2\right)=Q\left(\omega \right) \\ \right] \end{gathered}$$

Hence,$Q\left(\omega \right)$is the splitting field$x^2+x+1$.