Idempotents of Z(n)Date: 10/10/2000 at 00:43:32 From: heather mullin Subject: Idempotents in Z(n) What are the idempotents of Z(n) when n is twice a prime? I know through trial and error the answers are 0,1, n/2, and n/2 + 1, but how do I prove it? Date: 10/10/2000 at 12:28:55 From: Doctor Rob Subject: Re: Idempotents in Z(n) Thanks for writing to Ask Dr. Math, Heather. Let n = 2*p, p a prime number. Then an idempotent is an element x of Z(n) such that x^2 = x. That means that x^2 = x (mod 2*p) x^2 - x = 0 (mod 2*p) First let's assume that p is odd. Then this implies that x^2 - x = 0 (mod p) and x^2 - x = 0 (mod 2) x*(x-1) = 0 (mod p) and x*(x-1) = 0 (mod 2) The solutions to these congruences are, respectively, x = 0 or 1 (mod p) and x = 0 or 1 (mod 2) This you get by finding two roots by inspection, and using the fact that a polynomial of degree d has exactly d roots (counting multiplicity) modulo any prime modulus. So you have them all. These answers have to be put together using the Chinese Remainder theorem to get the solutions modulo n = 2*p. This can be done in four ways, since there are two choices mod p and two choices mod 2. The answers are those you obtained: 0, 1, p, and p+1. Don't forget that we have another case. That is the case when p is even. Now what happens when p = 2, so n = 2^2 = 4? I leave that to you. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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