Re: Galois Field


Jyrki Lahtonen wrote:
> Arturo Magidin wrote:
> > In article <yU71f.327$2i4.115@xxxxxxxxxxxxxxx>,
> > Ricky R. <noemail@xxxxxxxxxxx> wrote:
> >
> >>I have to show that GF(125) is a splitting field for p(x) = x^3+x+1 over
> >>GF(5). I need a hint.
> >
> >
> > You know that it has a root over GF(125), since it is irreducible over
> > GF(5). All you need to do is show that the resulting quadratic also
> > has its roots there; and for that, you just need to show that a
> > certain discriminant is a square in GF(125).
> >
>
> Actually I don't think all of that is necessary, if you know
> a little bit of Galois theory. The Galois group of the
> splitting field (over the prime field) is known to be a
> transitive permutation group of the three roots of the
> irreducible cubic. All the extensions of finite fields have
> cyclic Galois groups, so...
>
> Cheers,
>
> Jyrki Lahtonen, Turku, Finland

To continue this post, let me point out that it is a standard result
that every finite extension of a finite field is a Galois extension
(normal and separable). That means that when you adjoin one root, you
get them all. Now if you take an irreducible polynomial of degree k
over a field of q elements, when you adjoin one root you get a field of
order q^k in which the polynomial splits. In this case, your
polynomial either has three roots, one root or no roots in GF(5). In
the first case, the splitting field is GF(5); in the second GF(25) and
in the third GF(125).

.

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