Re: Algebra with finite field..

In article <fo9udg$ekn$1@xxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> wrote:
Hello sir~

If F is a field of prime characteristic p,
then (a + b)^(p^n) = a + b for all a, b in F
and all positive integers n.

As has been pointed out, this is false as stated. To get a specific
counterexample, recall that any finite subgroup of the multiplicative
group of nonzero elements of a field must be cyclic. Pick any field of
order p^m greater than p^n, and let a be a generator of the multiplicative
group F^* and let b=0. Then (a+b)^{p^n} = a^{p^n}, which is different
from a, since a is not of order p^n-1, but rather p^m - 1.

On the other hand, the result is true if F is a field with p^k
elements, where k|n.

pf)
Let a, b in F.
Applying the binomial theorem to (a + b)^p,
we have
(a + b)^p = a^p + (p).a^(p-1).b + {p(p-1)/2}.a^(p-2).b^2
+ ... + p.a.b^(p-1) + b^p
= a^p + 0.a^(p-1).b + 0.a^(p-2).b^2 + ... + 0.a.b^(p-1) + b^p
= a^p + b^p
= a + b (***)

And this is false unless F has order p.

Proceeding by induction on n,
suppose that we have (a + b)^(p^(n-1)) = a + b.
Then (a + b)^(p^n) = [(a + b)^(p^(n-1))]^p = (a + b)^p = a + b.

-----------------------------------------
I can't understand (***)part.

Good, because the "proof" is wrong.

Namely, a^p = a, b^p = b.
Why ?

If F has characteristic p, then by induction we have prove that
(a+b)^{p^m} = a^{p^m} + b^{p^m}, by doing the case m=1 as above and
stopping at (***), then proceeding by induction.

Then, if F has order p^k, then it is straightforward to check that
r^{p^k}=r for all r in F.

Then r^{p^{2k}} = (r^{p^k})^{p^k} = r^{p^k} = r; proceeding
inductively,

r^{p^{k(s+1)}} = (r^{p^{ks}})^{p^k} = r

so the result holds if F has p^k elements, and k|n.

It does not hold if you just have k<=n. For example, say F has p^2
elements. Pick a of multiplicative order p^2-1, and compute
(a+0)^{p^3}. Then we have a^{p^3}; this is equal to a if and only if
p^3 - 1 = 0 (mod p^2 - 1). This is not true in general (e.g., p=2 you
have 7 = 0 (mod 3); p=3 you have 26 = 0 (mod 8); etc).


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.

Relevant Pages

  • Re: Galileos Paradox and the Project of the Reals
    ... since Peano's axiom of induction is ... the positive integers but on the side remarking about how each ... of successions, but not for any iterations beyond those corresponding ... to the finite naturals. ...
    (sci.math)
  • question about the induction axiom and principle of mathematical induction
    ... induction and in my research of it i've come across both an induction ... axiom and a principle of mathematical induction. ... let S be a subset of positive integers satisfying: ...
    (sci.math)
  • Re: Algebra with finite field..
    ... and all positive integers n. ... group of nonzero elements of a field must be cyclic. ... and let a be a generator of the multiplicative ... stopping at, then proceeding by induction. ...
    (sci.math)
  • Re: x+y is a factor of x^(2n) - y^(2n) ?
    ... but where is the induction procedure? ... The question was how to use induction. ... It suffices to prove that ... Therefore f_n is divisible by for all positive integers n. ...
    (sci.math)
  • Re: Mathematical induction
    ... just a quick question regarding mathematical induction. ... It's usually called just "Backwards Induction". ... If Pis true for infinitely many positive integers n, ...
    (sci.math)