0

Let A be a finite commutative ring with unity. Prove that if $a\neq 0$ is not a divisor of zero then some positive power of a is equal to 1.

So A is finite so A=$\{a,a^2,a^3,...a^n\}$ and one of these $a^i=1$. I'm not sure where to take this proof if someone could help me out.

K Math
  • 1,245
  • 1
  • 11
  • 21

3 Answers3

1

Consider the set $X = \{ a, a^2, a^3, \dots \} \subset A$, being $A$ finite there must be two distinct indices $i,j \geq 1$ such that $a^i = a^j$ , which implies $a^i - a^j = 0$. Now suppose $i > j$, then: $a^j ( a^{i-j} - 1) = 0$. We suppose that $a$ is not a divisor of zero so $a^j$ is not a divisor of zero and it is not zero, therefore $a^{i-j} - 1 = 0$, equivalently: $a^{i-j} = 1$.

Federico
  • 1,755
1

You have no reason to believe that $A=\{a^i\mid i=1,2,\ldots\}$, but you do know the right hand side is a subset of $A$, and hence finite.

So the elements of the set of powers cannot be all distinct (it would be absurd if infinitely many distinct elements existed inside a finite set.)

So there exists $a^n=a^m$ for some $m < n$. Now look at $a^m(a^{n-m}-1)=0$ and ask yourself how the property that $a$ isn't a zero divisor will help you.

rschwieb
  • 153,510
0

Hint:

The set of positive powers of $a$ is finite, so there exist positive integers $r, s$, $\;r<s$, such that $a^r=a^s$.

Bernard
  • 175,478