necessary and sufficient condition

[tamser0630]

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:

[matt grime]

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?

[tamser0630]

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!

[matt grime]

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.

[tamser0630]

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!