On 16 nov, 06:44, Kusanagi <Kusan...@hotmail.com> wrote:
> Is it true that an extension E of degree 2 over a field F is always Galois extension of F if the characteristic of F is not 2?
Yes it is. Use the following fact : any degree two polynomial X^2+aX+b in characteristic <>2 field F which has a multiple root in some extension E/F is reducible over F. Proof : we know that the multiple root is -b/(2a) (characteristic <> 2 :), this root is in F, q.e.d.
> If yes, why the characteristic of F should not be 2? Any counterexample?
<pierre.herve.bern...@gmail.com> wrote: >On 16 nov, 06:44, Kusanagi <Kusan...@hotmail.com> wrote: >> Is it true that an extension E of degree 2 over a field F is always Galois extension of F if the characteristic of F is not 2?
>Yes it is. >Use the following fact : any degree two polynomial X^2+aX+b in >characteristic <>2 field F which has a multiple root in some extension >E/F is reducible over F. >Proof : we know that the multiple root is -b/(2a) (characteristic <> >2 :), this root is in F, q.e.d.
>> If yes, why the characteristic of F should not be 2? Any counterexample?
>Take k=Z/2Z, F=k(t^2) and E=k(t).
>Pierre
yes, but how find t such as [E:F]=2
(if t is element of F_4 (field with 4 elements) , t not 0, 1 then E=F because t^2=t+1 )
> <pierre.herve.bern...@gmail.com> wrote: > >On 16 nov, 06:44, Kusanagi <Kusan...@hotmail.com> wrote: > >> Is it true that an extension E of degree 2 over a field F is always Galois extension of F if the characteristic of F is not 2?
> >Yes it is. > >Use the following fact : any degree two polynomial X^2+aX+b in > >characteristic <>2 field F which has a multiple root in some extension > >E/F is reducible over F. > >Proof : we know that the multiple root is -b/(2a) (characteristic <> > >2 :), this root is in F, q.e.d.
> >> If yes, why the characteristic of F should not be 2? Any counterexample?
> >Take k=Z/2Z, F=k(t^2) and E=k(t).
> >Pierre
> yes, but > how find t such as [E:F]=2
t is an indeterminate. F is the field of rational functions on t^2 with coefficients in k, E is the field of rational functions with coefficients in k.
(If you prefer, think of F as the field of rational functions, and E as the field obtained by adjoning a square root of the indeterminate.)
There is not t "to find"; it's an indeterminate.
> (if t is element of F_4 (field with 4 elements) ,
On Tue, 17 Nov 2009 08:16:01 -0800 (PST), Arturo Magidin
<magi...@member.ams.org> wrote: >t is an indeterminate. F is the field of rational functions on t^2 >with coefficients in k, E is the field of rational functions with >coefficients in k.
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