0
$\begingroup$

Let $R$ be a ring. It is clear that if $x$ is a zero-divisor of $R$, then $x^n$ is also a zero-divisor for $n\ge1$. Why does the converse hold? In other words:

if $x^n$ is a zero-divisor for $n\ge1$, why is $x$ also a zero-divisor?

Would you please help me? Thank you in advance.

$\endgroup$
9
  • 3
    $\begingroup$ Do you know that if $xy$ is a zero-divisor, then either $x$ or $y$ is a zero-divisor? $\endgroup$ Jun 27, 2017 at 22:44
  • 3
    $\begingroup$ Well, it's simply associativity $$0=x^n \cdot y = x \cdot (x^{n-1}y)$$ $\endgroup$
    – Crostul
    Jun 27, 2017 at 22:44
  • 1
    $\begingroup$ @quid He didn't claim so. But ... induction. $\endgroup$
    – OR.
    Jun 27, 2017 at 22:45
  • 1
    $\begingroup$ @User1999: We don't, but what happens if they are zero? $\endgroup$ Jun 27, 2017 at 22:59
  • 1
    $\begingroup$ @User1999 IF $yz=0$, then $y$ is a zero divisor. If $xz=0$, then $x$ is a zero divisor. $\endgroup$ Jun 27, 2017 at 23:00

1 Answer 1

8
$\begingroup$

Let $y \neq 0$ such that $x^ny=0$. Then $x(x^{n-1}y)=0$, so either $x$ is a zero divisor or $x^{n-1}y=0$, that is $x^{n-1}$ is a zero divisor. Then descending induction on $n$ completes the proof.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .