Re: Galois-Theory

From: Arturo Magidin (magidin_at_math.berkeley.edu)
Date: 12/02/04 

Date: Thu, 2 Dec 2004 21:07:53 +0000 (UTC)

In article <319cctF38ce2pU1@individual.net>,
Johannes Beyermann <beyermann.johannes@gmx.de> wrote:
>Hello Arturo,
>
>many thanks for your help!
>
>>>I want to solve the following exercise.
>>>Show that E=Q(sqrt(2),sqrt(3),u) where u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>>normal. Determine Gal E/Q.
>>>
>>>I have no idea, what to do. I know two ways to show, that E/Q is normal.
>>>First: show, that E is the splitting field of a polynomial g in Q[x].
>>>Second: Q=Inv G, where G is a finite Group of automorphisms of E.
>>>
>>>I think the first method needs too many calculations.
>>
>> Not really. You know the polynomial that will give you sqrt(2):
>> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we let F =
>> Q(sqrt(2),sqrt(3)), then F is the splitting field of (x^2-2)(x^2-3).
>
>OK, this is clear to me.
>
>> Now, E = F(u), and is the splitting field, over F, of the polynomial
>>
>> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>>
>> So E must be a splitting field over Q as well. To find the polynomial
>
>Is there a theorem like this?: Let E be a splitting field of f over K, and K
>be a splitting field of g over F, then E is the splitting field of a
>polynomial over F.

Not as such, no; I may have been too hasty there. In fact, you notice
the problem below: it is easy enough to construct a polynomial with
coefficients in Q that has u as a root, but there is no guarantee in
general that K(u) will contain all those roots. In general, normal
over normal is not necessarily normal.

>> explicitly, take
>>
>> x^2 - (9-5sqrt(3))(2-sqrt(2))
>> x^2 - (9+5sqrt(3))(2-sqrt(2))
>> x^2 - (9-5sqrt(3))(2+sqrt(2))
>> x^2 - (9+5sqrt(3))(2+sqrt(2))
>
>I dont see, why the roots of these polynomials is in F(u) too?

Yes, you're right. That is not immediate. Sorry about that. Let me get
back to you.

>> and multiply them together to get a polynomial g(x). Then E is the
>> spliting field over Q of g(x)(x^2-2)(x^2-3). You don't have to
>> calculate g(x) explicitly, just note that it will work.
>>
>>>But i dont know how to
>>>realize the second way. If i assume, that u is not an element of
>>>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of order 8. I think
>>>that G=<r,s,t>, where r:sqrt(2)->-sqrt(2), s:sqrt(3)->-sqrt(3), t:u->-u.
>>
>> There may be some relations between r, s, and t; you should figure out
>> what r and s will do to u (see what they do to u^2). There aren't that
>
>I have the problem, that i cannot see, why such automorphisms exists? Well,
>the identity on F can be extended to an automorphism t of E sending u->-u
>since these are roots of an minimum polynomial m of u over F. But why can i
>extend the automorphisms r',s':sqrt(k)->-sqrt(k) k=2,3 on F to
>automorphisms of E? To show this i must know that r'(m) [transformed of
>Minimum Polynomial of u under r'] has also roots in F(u)?

This will follow once you know that E is a splitting field (in
characteristic zero, anyway); it is one of the basic properties of
splitting fields: if E/k is a splitting field, and F is a subfield of
E containing k, then every automorphism of E that fixes k pointwise
can be extended to an automorphism of F. 

-- 
======================================================================
"It's not denial. I'm just very selective about
 what I accept as reality."
    --- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu

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