Re: Irreducible
- From: quasi <quasi@xxxxxxxx>
- Date: Sat, 15 Apr 2006 17:07:05 -0400
On Sat, 15 Apr 2006 18:09:30 +0200, Marc Bogaerts
<mbg.DELSPAMnimda@xxxxxxxxx> wrote:
quasi wrote:
On Sat, 15 Apr 2006 01:54:41 EDT, lataianu bogdanSorry I jump in here lately in the discussion but this is what I found up to
<lataianu(with-no-brackets)@math.usask.ca> wrote:
Mr.quasi,
I think Mr. Montgomery refers about degree of alpha as
the degree of the minimal polynomial of alpha over F_{p},
i.e.[F_{p}(alpha]:F_{p}],
I was using degree in the same sense. I don't understand the point you
are trying to make here
it works this way !
What works?
It would help if you quoted the relevant portions of the posts you are
referring to.
My main objection to Montgomery's hint was his implication that if you
find another nonzero polynomial g satisfied by alpha, that the degree
of f must divide the degree of g. You do get that f divides g, but
that only gives you the inequality deg(f) <= deg(g). What I did
instead at that point was to focus on the order of alpha in the
multiplicative group.
quasi
now:
let q:=p^n, p a prime and n any natural number and consider the polynomial
equation
(1) x^q - x + 1 = 0 over F_p.
Let a be a root of this polynomial, considered as an element of the field
k:=F_p^q; the order of a is a divisor of p^q-1 (2), like any other element
of k. Also I recall that in k
(u+w)^p = u^p + w^p and also (u+w)^q = u^q + w^q. (3)
The root a now satisfies:
a^q = a - 1 if we apply (3) to both sides we get
a^(q^2) = a^q - 1 = (a - 1) - 1 = a - 2
Iterating this once again gives:
a^(q^3) = a - 3
and reiterating gives in general
a^(q^k) = a - k,
So for k = p we get:
a^(q^p) = a.
So the order of a, o(a), is a divisor of q^p-1, but, by (2) also of p^q-1.
Substitution of q = p^n gives p^(n*p) - 1 and p^(p^n) - 1 for these
values. We see that we have equality iff n=1, or p=n=2. In all other cases
we have o(a) <= q^p-1 < p^q-1, so a is not a generator of k, so the
polynomial in (1) is reducible.
No, your proof is the same as the flawed proof I posted yesterday, and
the error is the same error.
Just because o(a) is less than p^q - 1 doesn't necessarily imply that
a is not a generator of k. All it implies is a is not a cyclic
generator of the multiplicative group of k.
We know that the multiplicative group of a finite field is cyclic, and
hence any generator of the multiplicative group of k also generates
the field k over F_p, but the converse is not true. A field generator
for k over F_p need not be a generator of the multiplicative group of
k.
quasi
.
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