Re: Algebra with finite field..
- From: "Mariano Suárez-Alvarez" <mariano.suarezalvarez@xxxxxxxxx>
- Date: Tue, 5 Feb 2008 09:19:34 -0800 (PST)
On Feb 5, 1:53 pm, José Carlos Santos <jcsan...@xxxxxxxx> wrote:
On 05-02-2008 15:23, Fatal wrote:
I can't understand (***)part.
Namely, a^p = a, b^p = b.
Why ?
This is false in general.
Indeed, if it where true, every element of F would be a root of the
polynomial X^p-X, thus the cardinality of F should be p: this is not the
case if n>1.
On the other hand, it is true if the hypothesis "F is a field of prime
characteristic _p_" is replaced by the stronger hypothesis "F is a field
with _p_ elements".
Best regards,
Jose Carlos Santos
In any case, the original statement, that
if F is a field of characteristic p, n is a positive
integer and a and b are in F,then (a+b)^(p^n) = a + b
is false, since, by taking b = 0 would imply, this would
imply that for all a in F we have a^p = a, which is absurd
unless F has exactly p elements.
What is true is
(*) if F is a field of characteristic p, n is a positive
integer and a and b are in F,then (a+b)^(p^n) = a^(p^n) + b^(p^n)
which follows trivially from the simpler
if F is a field of characteristic p and a and b are in F,
then (a+b)^p = a^p + b^p
which in turn follows easily from Newton's formula and
the fact that the binomial number binom(p,k) is
divisible by p as soon as 0 < k < p.
(*) might be what mina_world was after...
-- m
.
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