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.

$\endgroup$
  • 3
    $\begingroup$ Do you know that if $xy$ is a zero-divisor, then either $x$ or $y$ is a zero-divisor? $\endgroup$ – Eric Wofsey Jun 27 '17 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 '17 at 22:44
  • 1
    $\begingroup$ @quid He didn't claim so. But ... induction. $\endgroup$ – OR. Jun 27 '17 at 22:45
  • 1
    $\begingroup$ @User1999: We don't, but what happens if they are zero? $\endgroup$ – Eric Wofsey Jun 27 '17 at 22:59
  • 1
    $\begingroup$ @User1999 IF $yz=0$, then $y$ is a zero divisor. If $xz=0$, then $x$ is a zero divisor. $\endgroup$ – Sahiba Arora Jun 27 '17 at 23:00
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.