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Here's a nice little problem distantly related to some work I did way back when for my honours thesis.

In ring theory, an idempotent element is one which is its own square. In a ring with unity, 0 and 1 are always idempotents, but there may or may not be other idempotent elements.

Now the product of idempotents is not necessarily an idempotent. In my honours thesis I was concerned with characterising the elements that could be expressed as the product of idempotents. Here's a little problem relating this to the simple ring Z_n (n any integer ≥ 2).

a) Prove that in Z_n, the product of two idempotents is also an idempotent.

b) If n = p^r where p is prime, show that the only idempotents of Z_n are 0 and 1.

c) If n = pq where p and q are distinct primes, show that Z_n has exactly four idempotents and characterise them.

d) For arbitrary n ≥ 2, characterise the number of idempotents in Z_n in terms of the prime power factorisation of n.

2007-09-26 15:53:24 · 2 answers · asked by Scarlet Manuka 7 in Science & Mathematics Mathematics

2 answers

a) multiply the congruences x^2=xmodn, y^2=ymodn to get
(xy)^2=xymodn

b) x^2=xmod(p^r) prime p implies p^r | x or p^r |x-1
since x,x-1 are coprime, giving x=0,1 mod p^r

c) x^2=xmod pq solves via x=0modpq,x=1modpq,
x=0modp and x=1modq giving x=p(p') modpq where
p' is the inverse of p, modulo q and lastly x=q(q')modpq
where q' is the inverse of q,modulo p.

d)We may distribute the prime powers in n over the two
factors x and x-1 in 2^k ways where k is the number of powers
each allocation giving a distinct solution to x^2=xmod n.

2007-09-26 17:09:05 · answer #1 · answered by knashha 5 · 0 0

This one brings back so much memory.... but it is way too long for me to spend time formulating the answers again (except for d) after 3 years and type them all out :P yes i m lazy. (a) is trivial unless you are reeeaaally anal about the showing one step per definition/axiom. (b) and (c) are simple applications of the properties of primes in Z mod n. As for (d)... yea, i'd skip that one for now. I might come back later if no one answers it.

2007-09-26 16:14:50 · answer #2 · answered by W 3 · 0 0

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