Galois Correspondence for Subfields of K:Q

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Homework Statement



i) Find the order and structure of the Galois Group [itex]K:Q[/itex] where
[itex]K = Q(\alpha)[/itex] and

[itex]\alpha = \sqrt{2 + \sqrt{2}}[/itex].

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

Homework Equations



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

I'm told [itex]K:Q[/itex] is normal

The Attempt at a Solution



I get the order of [itex]Gal (K:Q)[/itex] to be 4 with

[itex]Gal (K:Q) = \{ Id, \sigma, \tau, \sigma \tau \}[/itex] where

[itex]\sigma(\sqrt{2 + \sqrt{2}} ) \rightarrow -\sqrt{2 + \sqrt{2}}[/itex]

[itex]\tau(\sqrt{2 + \sqrt{2}} ) \rightarrow \sqrt{2 - \sqrt{2}}[/itex]

and

[itex]\sigma \tau (\sqrt{2 + \sqrt{2}} ) \rightarrow -\sqrt{2 - \sqrt{2}}[/itex]

I get stuck with the 2nd part though. I get the proper subgroups [itex]\{Id, \sigma \} \{Id, \tau\} \{ Id, \sigma \tau \}[/itex] 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 [itex]Gal(K:Q)[/itex]. 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.
 
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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 [itex]\alpha[/itex] isn't fixed by, say, [itex]\sigma[/itex]. But can you arrange for

[itex]\alpha[/itex] + stuff = [itex]\sigma(\alpha)[/itex] + 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.
 
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Hurkyl said:
You know [itex]\alpha[/itex] isn't fixed by, say, [itex]\sigma[/itex]. But can you arrange for

[itex]\alpha[/itex] + stuff = [itex]\sigma(\alpha)[/itex] + other stuff?

Sorry, I don't understand this.

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