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Idempotent help

asked 2010-12-09 13:56:12 -0500

CENTO gravatar image

updated 2011-04-28 08:59:57 -0500

Kelvin Li gravatar image

Hello my fellow Sage users. I'm trying to come up ith a formula in sage that relates the number of idempotents in Zn with the number of distinct prime factors that n has. Talking about idempotent rings in my class has gotten a few students confused and this formula will help me better explain the concept. Thank you. Any pointers in the right direction would help greatly.

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answered 2010-12-12 08:24:09 -0500

Jason Bandlow gravatar image

Here is some code for this:

for n in [2..100]:
    print n,': (',len(n.prime_factors()), sum(a.is_idempotent() for a in Zmod(n)),')'

For a fixed n, len(n.prime_factors()) gives the number of prime factors of n and sum(a.is_idempotent() for a in Zmod(n)) gives the number of idempotents.

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answered 2010-12-10 15:25:43 -0500

niles gravatar image

I guess the number of idempotents should be basically the same as the number of direct summands -- is that right? If so, then maybe you're looking for the direct-sum decomposition of Z/n. Does the fundamental theorem of finitely generated abelian groups give you enough information?

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Asked: 2010-12-09 13:56:12 -0500

Seen: 203 times

Last updated: Dec 12 '10