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#1 2010-04-22 02:15:07

laipou
Member
Registered: 2009-09-19
Posts: 24

cyclotomic extension

Let

(primitive 11th root of unity).
Let
(primitive 5th root of unity).
Determine all intermediaye fields between
and
.

I have shown that actually

,(
(primitive 5th root of unity)),
and
.
So after a little calcualtion,we have
,which is isomorphic to
.
Hence,the subgroups of
are
and
.
Now comes the main problem,how to find the fixed fields of these two subgroups?

Thanks for any help.

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#2 2010-04-22 14:04:13

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: cyclotomic extension

For any subgroup H, the corresponding subfield is all the elements which are fixed by any permutation in H.  Hence if you take H = G you get the base field (assuming the extension is Galois), and if you take H = 1, you get the entire field extension.  These are just the trivial cases.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2010-04-22 21:10:55

laipou
Member
Registered: 2009-09-19
Posts: 24

Re: cyclotomic extension

Is it possible to write down the fixed field explictily(something like

)?

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