Question

Find $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\left(\sqrt{2},i\right)$ .

Short answer

The answer is 4 .

Step-by-step solution

Definition of Galois group.

The set of all $\mathit{F}$automorphism of $\mathit{K}$is an extension of $\mathit{F}$ then Galois group is denoted by role="math" localid="1657957616090" $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{F}}\mathit{K}$it is the Galois group of $\mathit{K}$over $\mathit{F}$.

Step-2: Showing that every element in a finite field can be written as the sum of two squares.

Given $\mathit{G}\mathit{a}{\mathit{l}}_{\mathbf{Q}}\mathit{Q}\left(\sqrt{2},i\right)$concluded the extension field it can be obtain by the sequence of simple extension.

$\mathit{Q}\mathbf{⊑}\mathit{Q}\left(\sqrt{2}\right)\mathbf{⊑}\mathit{Q}\left(\sqrt{2}\right)\left(i\right)\mathbf{=}\mathit{Q}\left(\sqrt{2},i\right)$

Now the minimal polynomial $\mathit{p}\left(x\right)\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}$of $\sqrt{\mathbf{2}}$over $\mathit{Q}$so the ordered of this polynomial is 2 $\left[Q\left(2\right):Q\right]\mathbf{=}\mathbf{2}$and further next polynomial $\mathit{p}\left(x\right)$is ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$whose coefficients are in $\mathit{Q}\left(\sqrt{2}\right)$ is the minimal polynomial ofi over $\mathit{Q}\left(\sqrt{2}\right)$because $\mathbf{+}\mathit{i}$are not in the $\mathit{Q}\left(\sqrt{2}\right)$so second polynomial ${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$is irreducible over $\mathit{Q}\left(\sqrt{2}\right)$now order is

$\mathit{Q}\left(\sqrt{2}\right)\left(i\right)\mathbf{:}\mathit{Q}\left(\sqrt{2}\right)\mathbf{=}\mathbf{2}$

Then

$$\begin{gathered} \left[Q\left(\sqrt{2},i\right):Q\right]\mathbf{=}\left(Q\left(\sqrt{2}\right)\left(i\right)\right)\left(Q\left(\sqrt{2},i\right):Q\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{2} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{4} \end{gathered}$$

Thus the finitely generated extension is finite dimensional and hence algebraic over Q and ordered 4 .