Question

Let $\mathbf{K}$ be an Exercise Prove that ${\mathbf{Gal}}_{\mathbf{Q}\mathbf{\left(}\mathbf{i}\mathbf{\right)}}\mathbf{K}\mathbf{\cong }{\mathbf{Z}}_{\mathbf{4}}$.

Short answer

Thus, the${\mathrm{Gal}}_{\mathrm{Q}\left(\mathrm{i}\right)}\mathrm{K}\cong {\mathrm{Z}}_4$ is proved .

Step-by-step solution

Define the abelian group .

The abelian group is commutative. This is satisfied the law for all $\mathbf{a}\mathbf{,}\mathbf{b}$in field extension is $\mathbf{a}\mathbf{*}\mathbf{b}\mathbf{=}\mathbf{b}\mathbf{*}\mathbf{a}$. If the group is not satisfied with the condition of the commutative is called a non-abelian group.

Consider the subgroup of the 4 degree polynomial.

The subgroups of the polynomial have 4 degrees, and minimal polynomial is ${\mathrm{x}}^4-2$.

All group has expected 4 degrees and it also consists of the abelian group.

So here possible groups are $⟨⟨1234⟩⟩,⟨⟨13⟩⟩,⟨⟨24⟩⟩,⟨\left(13\right)\left(24\right)⟩,⟨\left(12\right)\left(34\right)⟩\mathrm{and}⟨\left(14\right)\left(23\right)⟩$.

Find here isomorphism.

We know that the non-cyclic subgroup of order 4 must have 3 elements of order 2 these are isomorphism.

Let's consider the two subgroups here $\{{\mathrm{H}}_1,{\mathrm{H}}_2\}$.

Here the element of the subgroup are;

${\mathrm{H}}_1=\{\mathrm{e},\left(13\right),\left(24\right),\left(13\right)\left(24\right)\}$.

${\mathrm{H}}_2=\{\mathrm{e},\left(12\right)\left(34\right),\left(14\right)\left(23\right),\left(13\right)\left(24\right)\}$.

Since any subgroup contains two non-identity elements from ${\mathrm{H}}_1$this will contain all of ${\mathrm{H}}_1$the same subgroup ${\mathrm{H}}_2$.

Thus,

localid="1659608424617" $\begin{array}{l}{\mathrm{H}}_1\le \mathrm{H}\text{or}{\mathrm{H}}_2\le \mathrm{H}\\ {\mathrm{H}}_1=\mathrm{H}\\ {\mathrm{H}}_2=\mathrm{H}\end{array}$

So, both subgroups have only two subgroups' isomorphism of ordered $2$.

Now here written subgroup of degree 4 is :

The Galois correspondence of all the degrees are localid="1659608430438" $\sqrt[4]{2},\sqrt[4]{2}\mathrm{i},-\sqrt[4]{2},\mathrm{and}-\sqrt[4]{2}\mathrm{i}$for the degree $1,2,3,and\text{4}$.

Here the field of the degree $4$is $⟨e⟩$from the splitting field $\mathrm{Q}(\sqrt[4]{2},\mathrm{i})$.

Find automorphism of the four-degree fixed filed.

Here $\mathrm{\sigma }=\left(1234\right)$and $\mathrm{r}=\left(13\right)$.

So,

$\begin{array}{c}\mathrm{\sigma }\left(\sqrt[4]{2}\right)=\sqrt[4]{2}\mathrm{i}\\ \mathrm{\sigma }\left(\mathrm{i}\right)=\frac{\mathrm{\sigma }\left(\sqrt[4]{2}\right)}{\sqrt[4]{2}}\\ =-\frac{\sqrt[4]{2}}{\sqrt[4]{2}\mathrm{i}}\\ =\mathrm{i}\end{array}$

Now for all elements localid="1660231175722" $\mathrm{\sigma }(\mathrm{a}+\mathrm{b}\sqrt[4]{2}+\mathrm{c}\sqrt[4]{4}+\mathrm{d}\sqrt[4]{8}+\mathrm{ei}+\mathrm{g}\sqrt[4]{4}\mathrm{i}+\mathrm{h}\sqrt[4]{8\mathrm{i}})=(\mathrm{a}+\mathrm{b}\sqrt[4]{2}-\mathrm{c}\sqrt[4]{4}-\mathrm{d}\sqrt[4]{8}+\mathrm{ei}-\mathrm{g}\sqrt[4]{4}\mathrm{i}-\mathrm{h}\sqrt[4]{8\mathrm{i}})$

Thus, the fixed elements must be in the form $\mathrm{a}+\mathrm{ei}$.therefore the fixed field is $Q\left(i\right)$.

Hence, the Galois group of the fixed field is ordere of $4$, so it is proved${\mathrm{Gal}}_{\mathrm{Q}\left(\mathrm{i}\right)}\mathrm{K}\cong {\mathrm{Z}}_4$.