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/