# If $x^n$ is a zero-divisor, then $x$ is too.

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?

• Do you know that if $xy$ is a zero-divisor, then either $x$ or $y$ is a zero-divisor? – Eric Wofsey Jun 27 '17 at 22:44
• Well, it's simply associativity $$0=x^n \cdot y = x \cdot (x^{n-1}y)$$ – Crostul Jun 27 '17 at 22:44
• @quid He didn't claim so. But ... induction. – OR. Jun 27 '17 at 22:45
• @User1999: We don't, but what happens if they are zero? – Eric Wofsey Jun 27 '17 at 22:59
• @User1999 IF $yz=0$, then $y$ is a zero divisor. If $xz=0$, then $x$ is a zero divisor. – Sahiba Arora Jun 27 '17 at 23:00

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.