# Galois correspondence

1. Homework Statement

i) Find the order and structure of the Galois Group $K:Q$ where
$K = Q(\alpha)$ and

$\alpha = \sqrt{2 + \sqrt{2}}$.

ii)Then for each subgroup of $Gal (K:Q)$, find the corresponding subfield through the Galois correspondence.

2. Homework Equations

I get the minimal polynomial to be $f(x): x^4 -4x^2 + 2$ and the four roots of $f(x)$ are $\sqrt{2 + \sqrt{2}} , -\sqrt{2 + \sqrt{2}}, \sqrt{2 - \sqrt{2}}, -\sqrt{2 - \sqrt{2}}$

I'm told $K:Q$ is normal

3. The Attempt at a Solution

I get the order of $Gal (K:Q)$ to be 4 with

$Gal (K:Q) = \{ Id, \sigma, \tau, \sigma \tau \}$ where

$\sigma(\sqrt{2 + \sqrt{2}} ) \rightarrow -\sqrt{2 + \sqrt{2}}$

$\tau(\sqrt{2 + \sqrt{2}} ) \rightarrow \sqrt{2 - \sqrt{2}}$

and

$\sigma \tau (\sqrt{2 + \sqrt{2}} ) \rightarrow -\sqrt{2 - \sqrt{2}}$

I get stuck with the 2nd part though. I get the proper subgroups $\{Id, \sigma \} \{Id, \tau\} \{ Id, \sigma \tau \}$ but I don't see how to find the corresponding subfields.

Through the Galois correspondence, there's a bijective map between the subfields of K and the subgroups of $Gal(K:Q)$. So there'd have to be 3 different subfields. I know these fields have to be generated by the fixed points for each map.

I'd appreciate any help on what the subfields are.

Last edited:

Related Calculus and Beyond Homework Help News on Phys.org
Hurkyl
Staff Emeritus
Gold Member
Well, you know how to find fixed points of any linear transformation, don't you? They're eigenvectors with eigenvalue 1.

Alternatively... finding 1 fixed point is easy. You know $\alpha$ isn't fixed by, say, $\sigma$. But can you arrange for

$\alpha$ + stuff = $\sigma(\alpha)$ + other stuff?

(Just to be clear, you should not be solving any equations with this approach)

I don't know if this automatically gives you a generator, though. You'll have to prove that this is a generator through some other means.

Last edited:
You know $\alpha$ isn't fixed by, say, $\sigma$. But can you arrange for

$\alpha$ + stuff = $\sigma(\alpha)$ + other stuff?
Sorry, I don't understand this.

Is finding the fixed point of $\sigma (\alpha)$ something to do with the sign in between the $2$ and $\sqrt{2}$ not changing?

Hurkyl
Staff Emeritus
Gold Member
(I will use s for sigma, and a for alpha) You want to find "something" such that

something = s(something).

Well, what might appear in "something"? a might. So, we try it out:

a + other stuff = s(a + other stuff) = s(a) + s(other stuff).

Doesn't that suggest what might appear in "other stuff"?