Originally posted by SPMarsRings , ideals, abelian groups ? I will have to go back to the text-books. I would better go out of this domain, till I update my abstract algebra skills..
Z is the ring of integers (an integral domain, actually).
pZ is the subset of Z consisting of those integers divisible by p. Actually, pZ is an (principal) ideal in the ring Z. Therefore we can form the quotient ring
R = Z/pZ
(look up any book on abstract algebra). This ring R has p elements, and can be thought of as the set of integers with + and ...[text shortened]... mod p. R is a field (and therefore, by definition (R,+) and (R/ { 0 } , x) are abelian groups.)
Originally posted by ranjan sinhaYeah, sarathian asked a simple question about the difference between 'necessary' and 'sufficient'. Instead, all he got was a bunch of irrelevant stuff about group theory.
Rings , ideals, abelian groups ? I will have to go back to the text-books. I would better go out of this domain, till I update my abstract algebra skills..
Originally posted by ThudanBlunderActually, I thought sarathian asked for a proof of Wilson's Theorem.
Yeah, sarathian asked a simple question about the difference between 'necessary' and 'sufficient'. Instead, all he got was a bunch of irrelevant stuff about group theory.
But I wasn't answering that question. Instead I was explaining what the notation Z/pZ means.
Originally posted by SPMarsOK, I stand corrected.
Actually, I thought sarathian asked for a proof of Wilson's Theorem.
But I wasn't answering that question. Instead I was explaining what the notation Z/pZ means.
I thought he asked why it doesn't work both ways. But in fact it does:
iff p is prime then p divides (p-1)! + 1
If p is a prime number, then
it always divides (p - 1)! +1. But for any other (non-prime) number n,
(n - 1) ! + 1 is not necessarily ( rather never) divisible by the first prime which is greater than n.
And of course it is never divisible by n too.
Why?
But Wilson's Theorem is a theorm about primes, not non-primes. I think he will have to elucidate.