Question

(a) Let $\mathit{\omega }\mathbf{=}\left(-1+\sqrt{3}i\right)\mathbf{/}\mathbf{2}$be a complex cube root of 1. Find the minimal polynomial p(x) of $\mathit{\omega }$over$\mathbf{\mathbb{Q}}\mathbf{}\mathbf{\mathbb{Q}}$and show that ${\mathit{\omega }}^{\mathbf{2}}$ is also a root of p(x).

(b) What islocalid="1657970981929" $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}\left(\omega \right)$?

Short answer

(a)The minimal polynomialp(x)of $\mathit{\omega }$over$\mathrm{\mathbb{Q}}$is ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$.

(b) localid="1657972365701" role="math" $Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}\left(\omega \right)$is a cyclic field of a power unit and root is $\mathit{\omega }$.

Step-by-step solution

Minimal polynomial of a Galois group.

The minimal polynomial of the Galois group is defined by simplified the root of the polynomial is equal to zero.

Finding the minimal polynomial.

(a)

Consider the root of the polynomial is $\mathit{\omega }\mathbf{=}\mathbf{\left(}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}\mathbf{i}}}{\mathbf{2}}\mathbf{\right)}$, it is a complex root of 1.

Now, the minimal polynomial$\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is a function of variablex, it is the element of the Galois group.

So,

$\mathit{x}\mathbf{=}\mathbf{\left(}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}\mathbf{i}}}{\mathbf{2}}\mathbf{\right)}$

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}\mathbf{i}} \\ \mathbf{2}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{=}\sqrt{\mathbf{3}\mathbf{i}} \end{gathered}$$

Take square both sides,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{=}\sqrt{\mathbf{3}\mathbf{i}} \\ \mathbf{}\mathbf{}\mathbf{}{\left(2x+1\right)}^{\mathbf{2}}\mathbf{}\mathbf{=}{\left(\sqrt{3i}\right)}^{\mathbf{2}} \\ \mathbf{4}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{=}\mathbf{-}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{0} \end{gathered}$$

Hence, the minimal polynomial of the given complex root is $\mathbf{}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{x}\mathbf{+}\mathbf{1}\mathbf{-}\mathbf{0}$.

Further such that ${\mathit{\omega }}^{\mathbf{2}}$is also the root of the minimal polynomial.

Put $\mathit{\omega }\mathbf{=}{\mathit{\omega }}^{\mathbf{2}}$.

So, equation is:

$$\begin{gathered} {\mathit{x}}^{\mathbf{2}}\mathbf{=}\left(\frac{-1+\sqrt{3i}}{2}\right) \\ \mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\sqrt{\mathbf{3}\mathbf{i}} \\ \mathrm{Take}\mathrm{square}\mathrm{both}\mathrm{sides}, \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\left(2x^2+1\right)}^{\mathbf{2}}\mathbf{=}{\left(\sqrt{3i}\right)}^{\mathbf{2}} \\ \mathbf{4}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{3} \\ \mathbf{4}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}\mathbf{} \end{gathered}$$

Hence, the${\mathit{\omega }}^{\mathbf{2}}$is also the root of polynomial which is satisfied this equation .

${\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$

This equation is similar as the previous equation here $\mathit{x}\mathbf{=}{\mathit{x}}^{\mathbf{2}}$.

The meaning of Galℚℚ(ω).

(b)

The Galois group is consisted the element of field extension, so the minimal polynomial p(x) has a root $\mathit{\omega }$.

Now, the Galois group$Gal{}_{\mathrm{\mathbb{Q}}}\mathrm{\mathbb{Q}}\left(\omega \right)$ is defined it is a cyclic field of power unity and root is $\mathit{\omega }$

Since, the minimal polynomial of the given Galois group is:

$$\begin{gathered} {\mathit{x}}^{\mathbf{n}}\mathbf{=}\mathit{\omega } \\ \mathbf{}\mathbf{}\mathit{n}\mathbf{=}\mathbf{1} \\ \mathbf{}\mathbf{}\mathit{x}\mathbf{=}\mathit{\omega } \end{gathered}$$

Hence, the given Galois group is the element of the cyclic group which contain the complex element under the one multiplication.