Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Necessary and sufficient condition

  1. May 9, 2004 #1
    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:
     
  2. jcsd
  3. May 9, 2004 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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?
     
  4. May 9, 2004 #3
    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!
     
  5. May 10, 2004 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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: May 10, 2004
  6. May 10, 2004 #5
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?