Necessary and sufficient condition

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Discussion Overview

The discussion revolves around identifying the necessary and sufficient conditions for an integer \( n \) such that the ring \( \mathbb{Z}/n\mathbb{Z} \) contains a nilpotent element. The scope includes theoretical exploration and mathematical reasoning related to prime factorization and nilpotency.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about the conditions on \( n \) for \( \mathbb{Z}/n\mathbb{Z} \) to have a nilpotent element, noting difficulty in finding this information in texts.
  • Another participant suggests considering \( n \) as the product of distinct primes and questions if the product itself is nilpotent.
  • A participant reflects on whether \( n \) being a product of primes implies nilpotency, stating that \( n^k \equiv 0 \mod n \) for \( k > 0 \) and seeks clarification on the role of prime factorization.
  • One participant proposes that nilpotent elements must be divisible by every prime dividing \( n \), expressing uncertainty about deriving a necessary condition from this.
  • Another participant clarifies that they meant to consider one of each prime factor of \( n \) rather than \( n \) itself, suggesting that this leads to a necessary and sufficient condition.
  • A participant hypothesizes that the prime factorization of \( n \) must contain a repeating factor, using \( n = 18 \) as an example to illustrate their reasoning.

Areas of Agreement / Disagreement

Participants express various viewpoints on the conditions for nilpotent elements in \( \mathbb{Z}/n\mathbb{Z} \), with no consensus reached on a definitive necessary condition. The discussion includes both agreement on certain aspects and ongoing uncertainty regarding the implications of prime factorization.

Contextual Notes

Participants note the importance of prime factorization and divisibility in determining nilpotency, but the discussion remains open-ended regarding the exact necessary conditions and their implications.

tamser0630
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Could someone please tell me what condition on n is necessary and sufficient such that Z/nZ contains a nilpotent element?

I can't seem to find this in any of my texts.

By the way, I accidently posted this same question in the homework help section first. Sorry to anyone who's wondering why I'm cluttering up multiple boards with my quandries. :rolleyes:
 
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suppose that n is the product of distinct primes p_i, with multiplicity n_i (i ranges form 1 to r), then is p_1p_2...p_r niplotent? can you see how to use that idea to find a necessary condition?
 
okay, so let's see..
you're asking me whether n itself is a nilpotent element of Z/nZ.

yes, I think n = (p_1)(p_2)...(p_r) is nilpotent by definition because:

n^k is congruent to 0 (mod n), for k > 0 ...right?

I'm not sure how your hint regarding the prime factorization can be used though. Can you elaborate? I did manage to find something while I was reading that I thought might help me:

if x^k is congruent to 0 (mod n) ---> then every prime dividing n divides x also.

*this works as far as I can tell (just by trying a few examples)
though I'm not sure why. but it seems to be along the same
lines as your hint (because "every prime dividing n" is the prime
factorization of n ..which you told me to consider).*


so I guess all the nilpotent elements of Z/nZ would have to be divisible by every single prime which divides n.


I still don't see how any of this leads me to a necessary condition for n though.

please help!
 
as n is zero that is not what i meant, I said take one of each prime factor of n, not n. eg in 18=2*3*3, consider 2*3. If you think about this you will obtain a neccesary and sufficent condition. Indeed it appears you have, but just don't realize it. You've proven that it is necessary for every prime factor of n to divide a nilpotent element (which is also sufficient), so you need to figure out when that gives you a non-zero possible nilpotent element.
 
Last edited:
hey, how's this - I think I have an answer..

does the prime factorization of n have to contain a repeating factor?

this makes sense to me because, for example if n = 18 then..

18 = (2)(3)(3) <---- there are two 3's here (3 is a repeating factor)

so if I divide by the repeating factor I get 6!.. which is nilpotent because the prime factorization of 6 contains all the prime factors of 18 (minus one of the 3's which I had one of to spare anyway)!

it seems good to me, what do you think?!
and thank you for all the help!
 

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