Calculating Nil(Z_n) faster

Question

As far as I know, in order to calculate Nil(Z_n) I need to calculate each number raised to different values and see if I get 0. That is a very slow and inefficient way, in my opinion.

Is there any other faster way to do this?

For example, U(Z_n) is equal with the numbers with the property (number, n) = 1.

Is there any similar thing for Nil(Z_n)?

Thanks.

$x^k\equiv0\pmod n$ for some $k \iff $ all the prime factors of $n$ divide $x$ — J. W. Tanner, Nov 17 '20 at 17:38

J. W. Tanner · Accepted Answer · 2020-11-17T18:49:18.653

1

For any integer $n$,

the nilpotent elements of the finite ring $ \mathbb {Z} /n\mathbb {Z}$ are all of the multiples of the radical of $n$,

which is the product of the distinct prime numbers dividing $ n$.

For example, rad$(12)=$rad$(2^2\cdot3)=2\cdot3=6$,

so the nilpotent elements of $\mathbb Z/12\mathbb Z$ are multiples of $6$.

edited Nov 17 '20 at 18:49

answered Nov 17 '20 at 17:49

J. W. Tanner

60,406

Can you help me with an example? Let's say, for n = 12. – Octavian Niculescu Nov 17 '20 at 18:38
Yes, I was thinking I should do an example. rad$(12)=$rad$(2^2\cdot3)=2\cdot3=6$, so the nilpotent elements of $\mathbb Z/12\mathbb Z$ are multiples of $6$ – J. W. Tanner Nov 17 '20 at 18:43
Okay. Thanks a lot. Do you mind telling me if there is any rule for Idem(Z_n)? :) – Octavian Niculescu Nov 17 '20 at 18:51
Also, if we can't calculate the radical (for example n=10 so rad(2*5) ) does that mean that Nil(Z_n) contains only 0? – Octavian Niculescu Nov 17 '20 at 18:52
Yes, and for idempotents, see this; basically, idempotents are $0$ and $1$ for odd prime powers, and use the Chinese remainder theorem – J. W. Tanner Nov 17 '20 at 19:05
Thanks a lot :) – Octavian Niculescu Nov 17 '20 at 19:07

Calculating Nil(Z_n) faster

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