Question

Exhibit the Galois correspondence for the extension $\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)}$ of Q.

Short answer

There are seven subgroup of order two and seven subgroup of ordered four.

Step-by-step solution

Definition of the Galois extension.

IfK is a finite dimension normal and separable extension field of the fieldF so Kis a Galois extension ofF .

Step-2: Showing that there are seven subgroup of order two and seven subgroup of order four.

Consider the extension field which contain the number of subgroup field define the Galois group of the extension $\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)}$

There are three extension fields and minimal polynomial of three extension is $\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{,}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{5}\mathbf{)}$ respectively.

Now the Galois group are $\mathbf{(}{\mathbf{I}}_{\mathbf{d}}\mathbf{,}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{,}\mathbf{c}\mathbf{,}\mathbf{d}\mathbf{,}\mathbf{e}\mathbf{,}\mathbf{f}\mathbf{,}\mathbf{g}\mathbf{)}$ since the polynomial are consist 8 subgroup. The number of possible subgroup of order 2 is

$$\begin{gathered} \mathbf{⟨}\mathbf{a}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{)} \\ \mathbf{}\mathbf{⟨}\mathbf{b}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)} \\ \mathbf{}\mathbf{⟨}\mathbf{c}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)} \\ \mathbf{}\mathbf{⟨}\mathbf{d}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)} \end{gathered}$$

And

$$\begin{gathered} \mathbf{⟨}\mathbf{e}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{10}}\mathbf{)} \\ \mathbf{}\mathbf{⟨}\mathbf{f}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{5}}\mathbf{,}\sqrt{\mathbf{6}}\mathbf{)} \\ \mathbf{}\mathbf{⟨}\mathbf{g}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{6}}\mathbf{,}\sqrt{\mathbf{10}}\mathbf{)} \end{gathered}$$

And one subgroup of order one is

$\mathbf{⟨}{\mathbf{I}}_{\mathbf{d}}\mathbf{⟩}\mathbf{=}\mathbf{Q}\mathbf{(}\sqrt{\mathbf{2}}\mathbf{,}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{5}}\mathbf{)}$

Further the subgroup of order four is

$$\begin{gathered} \left\{I_d,a,b,c\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{2}}\mathbf{\right)} \\ \left\{I_d,a,d,e\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{3}}\mathbf{\right)} \\ \left\{I_d,a,f,g\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{3}}\mathbf{\right)} \\ \mathbf{=}\left\{I_d,b,d,f\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)} \end{gathered}$$

And

$$\begin{gathered} \left\{I_d,b,e,g\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)} \\ \left\{I_d,c,d,g\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)} \\ \left\{I_d,c,e,f\right\}\mathbf{\to }\mathbf{Q}\mathbf{\left(}\sqrt{\mathbf{5}}\mathbf{\right)} \end{gathered}$$

Hence there are seven subgroup of order two and seven subgroup of order four.