Re: Congruence question.

rob@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <y8zhcbyz2pj.fsf@xxxxxxxxxxxxxxxxxxxx>,
Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx> wrote:
Saysero <saysero@xxxxxxxxx> writes:

Prove that for any prime p, if a^p - b^p is divisible by p
then a^p - b^p is also divisible by p^2.

n n
x - y n-1 n
HINT In any ring ------- = n x = D (x ) if x = y
x - y

See also my prior posts [1] on the algebraic derivative for polynomials

[1] http://google.com/groups?selm=y8z64ry3745.fsf%40nestle.csail.mit.edu
http://google.com/groups?selm=y8zr6b6fd92.fsf%40nestle.csail.mit.edu

If we allow division by x - y when x = y, then we allow the following
common fallacy:

Assume x = y, then

2
xy = y

2 2 2
x - xy = x - y

2 2
x(x-y) x - y
------ = -----
x-y x - y

x = 2x

Then either all x = 0, or 1 = 2. Both of these restrict the ring
to one element.

The algebraic formalism you cite above works if we carry out the
division first, without assuming x = y, then assume x = y in the
resulting quotient. However, if we let x = y before the division,
problems can arise. In most rings, 0/0 is reasonably undefined.

You've completely misunderstood - please read the cited posts.

--Bill Dubuque

.

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