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

Question

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?

Answer 1

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 $n > 1$.

Answer 2

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 $a$.

Answer 3

"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