Re: Galois-Theory

From: Johannes Beyermann (beyermann.johannes_at_gmx.de)
Date: 12/02/04 

Date: Thu, 02 Dec 2004 22:37:21 +0100

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.

Did you exchange E an F? Then I know this theorem, but the thing is, that we
dont know, whether E is a splitting field over Q, otherwise it follows
directly, that E is normal over Q, since Q is perfect.
Or do you mean another way to apply the above property?

Many thanks in advance
Johannes

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