Question

Let K be as in exercise 11 exhibit the Galois correspondence for this extension among the intermediate field $\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}\mathbf{}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{\left)}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{\right)}$.

Exercise: $\mathbf{K}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{,}\mathbf{i}\mathbf{)}$ is a splititing field of ${\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{2}$ over Q .

Short answer

$\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}\mathbf{}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{)}\mathrm{and}\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{)}\mathbf{}\sqrt[\mathbf{4}]{\mathbf{2}}\mathbf{)}\mathbf{}$ splititing field.

Step-by-step solution

Definition of the Galois extension.

If K is a finite dimension normal and separable extension field of the fieldK soFis a Galois extension ofF .

Step-2: Showing that Q((1+i) 24) and  Q((1-i) 24)  are splitting field.

consider the extension fieldF and normal intermediate fieldKnow the Galois group of the corresponding of this extension among the intermediate field is $\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{)}\mathbf{}\sqrt{\mathbf{2}}\mathbf{)}\mathbf{}$ and second is $\mathbf{Q}\mathbf{\left(}\mathbf{\right(}\mathbf{1}\mathbf{-}\mathbf{i}\mathbf{)}\mathbf{}\sqrt{\mathbf{2}}\mathbf{)}\mathbf{}$.

Here both subgroup intermediate field having the same minimal polynomial that is ${\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{8}$ so the Galois correspondence is depends on the root of the polynomial now the root are $\mathbf{\pm }\sqrt[\mathbf{4}]{\mathbf{8}}\mathbf{.}$.

Now the possible subgroup are

$$\begin{gathered} {\mathbf{I}}_{\mathbf{d}}\mathbf{:}\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}}\mathbf{\to }\sqrt[\mathbf{4}]{\mathbf{8}} \\ \mathbf{a}\mathbf{:}\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}}\mathbf{\to }\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}} \\ \mathbf{b}\mathbf{:}\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}}\mathbf{\to }\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}} \\ \mathbf{c}\mathbf{:}\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}}\mathbf{\to }\sqrt[\mathbf{4}]{\mathbf{-}\mathbf{8}} \end{gathered}$$

Since these subgroup are the ordered of one and two. Here order one is $\left\{I_d\right\}$ subgroup and left subgroup are the ordered two.