# Homework Help: Galois correspondence

1. Jan 10, 2007

### ElDavidas

1. The problem statement, all variables and given/known data

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. Relevant 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: Jan 10, 2007
2. Jan 10, 2007

### Hurkyl

Staff Emeritus
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: Jan 10, 2007
3. Jan 10, 2007

### ElDavidas

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?

4. Jan 10, 2007

### Hurkyl

Staff Emeritus
(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"?