Galois Correspondence for Subfields of K:Q

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Homework Help Overview

The discussion revolves around the Galois correspondence for the field extension K:Q, where K is defined as Q(α) with α = √(2 + √2). Participants are tasked with finding the order and structure of the Galois group Gal(K:Q) and identifying corresponding subfields for each subgroup.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to determine the order of the Galois group and its structure, noting that Gal(K:Q) has order 4. They express uncertainty about finding the corresponding subfields for the identified subgroups. Some participants suggest exploring fixed points and relationships between transformations, while others question the implications of these transformations.

Discussion Status

The discussion is ongoing, with participants exploring various approaches to identify fixed points and the relationships between elements under the Galois group actions. There is a focus on understanding the nature of the transformations and their effects on the field elements, but no consensus has been reached regarding the subfields.

Contextual Notes

Participants are working under the constraints of the Galois correspondence, which establishes a bijective relationship between subfields of K and subgroups of Gal(K:Q). The original poster has identified the minimal polynomial and roots but is seeking further clarification on the subfields corresponding to the subgroups.

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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.

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

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:
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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 \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:
Hurkyl said:
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?
 
(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"?
 

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