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


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

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

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


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

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.

  • 4
    $\begingroup$ What about $\Bbb Q(i\sqrt2)$? $\endgroup$ – Lord Shark the Unknown Nov 16 '18 at 21:02
  • $\begingroup$ 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. $\endgroup$ – 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 $H_3$?

  • 1
    $\begingroup$ So, its the one mentioned by @Lord Shark the Unknown in the coments.above. Well, now i get some idea of how to look into this subfields just by looking at the roots that are fixed on the automorphism, thank you! $\endgroup$ – Eduardo Silva Nov 16 '18 at 22:10

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