Intermediary Extensions of $\mathbb{K}=\mathbb{Q}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})$.

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

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

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.

• What about $\Bbb Q(i\sqrt2)$? – Lord Shark the Unknown Nov 16 '18 at 21:02
• Surely this is the splitting field of $x^4+1$ over $\Bbb{Q}$ as opposed to that of $x^2+1$. Also $$\frac{\sqrt2}2+i\frac{\sqrt2}2=\cos\frac\pi4+i\sin\frac\pi4=e^{i\pi/4}$$ is an eighth primitive root of unity. – Jyrki Lahtonen Nov 17 '18 at 4:33

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 $$G\cong(\Bbb{Z}/2\Bbb{Z})^2$$. The intermediary extensions of $$\Bbb{Q}\subset K$$ then correspond bijectively to the normal subgroups of $$G$$. More precisely, each subgroup $$H\subset G$$ 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 $$L:=\{x\in K:\ (\forall\sigma\in H)(\sigma(x)=x)\}.$$
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 $$H_3$$?