Question

Find the Galois group G of the given polynomial in$\mathbf{\mathbb{Q}}\left[x\right]\mathbf{\mathbb{Q}}\left[x\right]$:

$\left(a\right)\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{6}}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{4}$[Hint: factor]

$$\begin{gathered} \left(b\right)\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{5}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6} \\ \left(c\right)\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{5}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{9}\mathbf{x} \\ \left(d\right)\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{3}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{6} \\ \left(e\right)\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{5}}\mathbf{-}\mathbf{10}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{}\mathbf{[}\mathbf{Hint}\mathbf{:}\mathbf{}\mathbf{see}\mathbf{}\mathbf{example}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{]} \end{gathered}$$

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{The}\mathrm{Galois}\mathrm{group}\mathrm{is}{\mathrm{S}}_3. \\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{Galois}\mathrm{group}\mathrm{is}{\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2. \\ \left(\mathrm{c}\right)\mathrm{The}\mathrm{Galois}\mathrm{group}\mathrm{is}{\mathrm{\mathbb{Z}}}_2. \\ \left(\mathrm{d}\right)\mathrm{The}\mathrm{Galois}\mathrm{group}\mathrm{is}{\mathrm{S}}_4. \\ \left(\mathrm{e}\right)\mathrm{The}\mathrm{Galois}\mathrm{group}\mathrm{is}{\mathrm{S}}_5. \end{gathered}$$

Step-by-step solution

 Step 1: Galois Group

Any Galois group of a group K over F is represented by $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$.

(a)Step 2: Find Galois group of polynomials

Given a polynomial $f\left(x\right)=x^6-4x^3+4$. First, we have to find the roots of the given polynomial:

$$\begin{gathered} x^6-4x^3+4=0 \\ {\left(x^3-2\right)}^2=0 \\ {x}^3=2 \\ x=\sqrt[3]{2} \end{gathered}$$

Here, the splitting field of the given polynomial is the same as the one for $x^3-2$. The splitting field of $x^3-2$is represented by $\mathrm{\mathbb{Q}}\left(\sqrt[3]{2},\omega \right)$, where $\omega$is the proper cube root of unity, which is the root of $x^2+x+1$.

So, from the above calculation, we can say that the Galois group of the given polynomial is $S_3$.

Thus, the Galois group of the polynomial is $S_3$.

(b) Step 3: Find Galois group of polynomials

Given polynomial is $f\left(x\right)=x^4-5x^2+6$. First, we have to find the roots of the given polynomial:

$$\begin{gathered} x^4-5x^2+6=0 \\ \left(x^2-2\right)\left(x^2-3\right)=0 \\ {x}^2=2\mathrm{or}{\mathrm{x}}^2-3 \\ \mathrm{x}=\pm \sqrt{2}\mathrm{or}\mathrm{x}=\pm \sqrt{3} \end{gathered}$$

So, the splitting field of the given polynomial is $\mathrm{\mathbb{Q}}\left(\sqrt{2},\sqrt{3}\right)$.

Now, the Galois group of the given splitting field is of the order 4 that has is ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

Thus, The Galois group is ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$.

(c) Step 4: Find Galois group of polynomials

Given polynomial is $f\left(x\right)=x^5+6x^3+9x$. First, we have to find the roots of the given polynomial:

$$\begin{gathered} {\mathrm{x}}^5+6{\mathrm{x}}^3+9\mathrm{x}=0 \\ \mathrm{x}\left({\mathrm{x}}^4+6{\mathrm{x}}^3+9\right)=0 \\ \mathrm{x}{\left({\mathrm{x}}^2+3\right)}^2=0 \\ \mathrm{x}=0\mathrm{or}{\mathrm{x}}^2+3=0 \\ \mathrm{x}=0\mathrm{or}\mathrm{x}=\sqrt{-3} \end{gathered}$$

So, splitting field of the given polynomial is $\mathrm{\mathbb{Q}}\left(\sqrt{-3}\right)$.

The splitting field $\mathrm{\mathbb{Q}}\left(\sqrt{-3}\right)$is an extension of degree equal to 2. So, its Galois group must be $Z_2$.

(d)Step 5: Find Galois group of polynomials

Given polynomial is $f\left(x\right)=x^4+3x^2-2x-6$. First, we have to find the roots of the polynomial:

$$\begin{gathered} x^4+3x^2-2x-6=0 \\ \left(x+3\right)\left(x^3-2\right)=0 \\ x+3=0\mathrm{or}{\mathrm{x}}^3-2=0 \\ \mathrm{x}=-3\mathrm{or}\mathrm{x}=\sqrt[3]{2} \end{gathered}$$

The given polynomial is irreducible by Eisenstein.

The discriminate of the given polynomial can be obtained as:

$$\begin{gathered} -27\times {\left(2\right)}^2=-27\times 4 \\ =-108 \end{gathered}$$

This discriminate is not a square in $\mathrm{\mathbb{Q}}$. So, the Galois group is $S_4$.

Thus, the Galois group of the polynomial is $S_4$.

(e)Step 6: Find Galois group of polynomials

Given the polynomial $f\left(x\right)=x^5=10x-5$. Here, we have to find the derivative of the given polynomial. Differentiate with respect to x:

$f\text{'}\left(x\right)=5x^4-10$

The derivative of the function has two real roots which means the given function has one relative maximum and one relative minimum.

Also,$f\left(x\right)=x^5-10x-5$ has exactly three real roots. Therefore, its Galois group is $S_5$.

Thus, the Galois group of the given polynomial is $S_5$.