Question

Show that ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$ and ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\in }\mathbf{Q}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ have the same Galois group.

Short answer

${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$ and ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\in }\mathbf{Q}\mathbf{\left[}\mathbf{x}\mathbf{\right]}$ have the same Galois group.

Step-by-step solution

Definition of the Galois extension.

Let F be a field and K be an extension of F . A F-automorphism of K is an isomorphism$\mathit{\sigma }\mathbf{:}\mathit{K}\mathbf{\to }\mathit{K}$ which fixes the F element wise. All the set of F-automorphism ofK is called Galois group of K over F .

Step-2: Showing that x2-3 and x2-2x-2∈Q[x]  have the same Galois group.

If ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$and ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}$have same splititing field then they are same Galois group. Here find the splititing field of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$.

First find the root of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$as follows

$$\begin{gathered} {\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{\pm }\sqrt{\mathbf{3}} \end{gathered}$$

So splititing field of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}$ over $\mathit{Q}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{}$ is $\mathit{Q}\mathbf{\left[}\sqrt{\mathbf{3}}\mathbf{\right]}$.

Noe find the splititing field of${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}$ . So, again find the root of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}$ as follows

$$\begin{gathered} {\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{(}\sqrt{\mathbf{3}}{\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{3}}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{)}\mathbf{=}\mathbf{0} \\ \mathbf{x}\mathbf{=}\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{3}} \end{gathered}$$

So now the root of role="math" localid="1657965986375" ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}$are $\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{3}}$. So, splititing field of ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{2}$over role="math" localid="1657966040092" $\mathit{Q}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{}$is role="math" localid="1657966063934" $\mathit{Q}\mathbf{\left[}\sqrt{\mathbf{3}}\mathbf{\right]}$ .

Thus the splititing field of both are same. So they are same Galois group