# If there exists $n>1$ such that $x^n=x$ for all $x$ in a ring, then there are no nonzero nilpotent elements.

Suppose that there is an integer $$n> 1$$ such that $$x^n=x$$ for all elements $$x$$ of some ring. If $$m$$ is a positive integer and $$a^m= 0$$ for some $$a$$, show that $$a = 0$$.

I have an answer but don't know if it is correct. I made use of the division algorithm to conclude that there exists $$q$$ and $$r$$ such that $$a^n = a^{mq}a^{r}$$ which will yield $$a=(a^m)^q a^r$$ then $$a=0a^r$$ finally, $$a=0$$.

Is this right?

• Your argument is right. Commented Dec 10, 2014 at 13:26
• Are you sure that $n < 1$? Maybe you show that $m < n$ must be true. Then your argument seems OK to me. Commented Dec 10, 2014 at 13:27
• @Ronald Ooops you are right it's $n>1$ I will edit this Commented Dec 10, 2014 at 13:48
• Please try harder to make a more useful title than "ring theory problem" next time. You'd like your question to help other people, right? They will never find the question if it's titled "ring theory problem." Commented Dec 10, 2014 at 13:49
• @rschwieb Thank you for the advice. I am yet a beginner here :) Commented Dec 10, 2014 at 13:51

Your solution works when $n \geq m$ but if $m > n$ then we get $q = 0$ and so just get $a = a^r$.

However since $a^n = a$, we know $a^{n^2} = (a^n)^n = a$, and similiarly $a^{n^s} = a$ for all $s > 0$.

So pick $s$ such that $n^s \geq m$, then we can multiply $a^m = 0$ by $a^{n^s-m}$ to get $a^{n^s} = 0 a^{n^s-m}$ and so $a = 0$. (Or use your solution, but this is easier).

p.s. I'm assuming there's a typo in the question and it should be $n > 1$.

We can make it a little easier with a simplification.

Suppose $a\neq 0$ and let $m$ be minimal with respect to satisfying $a^m=0$. This implies that $m\geq 2$.

If there is a nonzero nilpotent element, then there's a nonzero element whose square is zero.

If $a^2=0$ then there is nothing to do, and if $m>2$, we can always choose a power $a^i=b$ such that $b^2=0$ with $b\neq 0$.

If $b\neq0$ and $b^2=0$, we have a contradiction.

Of course, $n\geq 2$ by hypothesis. Also by hypothesis, $x=x^n=x^2x^{n-2}$ for all $x$. But this implies $b=b^n=0$, a contradiction to the assertion that $b\neq 0$.

Thus there cannot exist such a $b$ with square zero, and moreover there cannot exist a nonzero nilpotent element $a$.

"Suppose that there is an integer $$n > 1,$$ such that $$x^n = x$$ for all elements $$x$$ of some ring",
it is actually means the ring has trival multiplicative/additive subgroup {0},
and all other elements of ring generate multiplicative cyclic group (not really subgroup, cause they do NOT have same identity),
so any nilpotent element must in {0},
that is:

    ∀x ∈ R, x^n = x ⇒
cond1. x = 0 ⇒
0^n = 0 ⇒ <0> = {0}, |<0>| = |{0}| = 1
cond2. x ≠ 0 ⇒
∀x ∈ R, x ≠ 0, <x> = {x,x^2,...,x^(n-1) = ε}, |<x>| | (n - 1)
⇒
∃m ∈ Z+, a^m = 0 ⇒ a^m ∈ <0> ⇒ a = 0,

for example, in ring Z6 = {0,1,2,3,4,5},
when n = 3, x^3 = x ⇒
|<x>| = 1,2 | (3-1),
0^3 mod 6 = 0 ⇒ <0,*> = {0} < Z6,
1^3 mod 6 = 1 ⇒ <1,*> = {1} < Z6,
2^3 mod 6 = 2 ⇒ <2,*> = {2,4} < Z6, the identity is 4
3^3 mod 6 = 3 ⇒ <3,*> = {3} < Z6,
4^3 mod 6 = 4 ⇒ <4,*> = {4} < Z6,
5^3 mod 6 = 5 ⇒ <5,*> = {5,1} < Z6, then identity is 1