Jtdylktuy

First, I must prove this extensions is Galois, by some algebra I proved that:

$mathbb{Q}(frac{sqrt{2}}{2}+ifrac{sqrt{2}}{2})=mathbb{Q}(sqrt{2},isqrt{2})=mathbb{Q}(sqrt{2},i)$.

And since this last extensions is the splitting field of the irreducible polynomial $X^2+1 in mathbb{Q}[x]$ it is Galois with degree $4$.

Then, by the Galois correspondece the only intermediary fields are those with order $2$ with $mathbb{K}$.

explicitly $G=Gal(mathbb{K},mathbb{Q})={Id, sigma_1,sigma_2,sigma_3}$, such that

$sigma_1(i)=i$, $sigma_1(sqrt{2})=-sqrt{2}$

$sigma_2(i)=-i$, $sigma_2(sqrt{2})=sqrt{2}$

and

$sigma_3(i)=-i$, $sigma_3(sqrt{2})=-sqrt{2}$

All the $sigma's$ are of order 2.

By the tower rule we can see that $mathbb{Q(i)}$ and $mathbb{Q(sqrt{2})}$

are extensions with degree $2$ with $mathbb{K}$, since they are degree 2 with $mathbb{Q}$.

I would like to say that those two are the only fields that i'm looking for but i can't prove what are they respective images under the Galois Correspondence Theorem, i.e, what normal groups of $G$ they represent.

You have shown that $G={operatorname{id},sigma_1,sigma_2,sigma_3}$ and that $sigma_i^2=operatorname{id}$ for all $i$. It follows that $Gcong(Bbb{Z}/2Bbb{Z})^2$. The intermediary extensions of $Bbb{Q}subset K$ then correspond bijectively to the normal subgroups of $G$. More precisely, each subgroup $Hsubset G$ corresponds to the subfield of elements of $K$ that are fixed by each element of $H$. That is to say, $H$ corresponds to the subfield
$$L:={xin K: (forallsigmain H)(sigma(x)=x)}.$$

In this way the whole field $K$ corresponds to the trivial subgroup ${operatorname{id}}$, and the subfield $Bbb{Q}$ corresponds to the entire group $G$. This leaves precisely three subgroups of $G$; the subgroups $H_i:={operatorname{id},sigma_i}$ for $i=1,2,3$. These are subgroups of index $2$, so they correspond to extensions of $Bbb{Q}$ of degree $2$.

For $sigma_1$ we have $sigma_1(i)=i$, so the subfield $Bbb{Q}(i)$ is invariant under $sigma_1$. Because $[Bbb{Q}(i):Bbb{Q}]=2$ it follows that $Bbb{Q}(i)$ is the subfield corresponding to $H_1$.

Similarly, for $sigma_2$ we have $sigma_2(sqrt{2})=sqrt{2}$, so the subfield $Bbb{Q}(sqrt{2})$ is invariant under $sigma_2$. Because $[Bbb{Q}(sqrt{2}):Bbb{Q}]=2$ it follows that $Bbb{Q}(sqrt{2})$ is the subfield corresponding to $H_2$.

So what is the subfield corresponding to $H_3$?