Prove / Disprove : if $a \neq 0$ and is not invertible so $a$ is a zero divisor

Question

R is a commutative ring with an identity element. Prove or disprove that if

$a \in R ,\; a \neq 0 \;$and is not invertible so $a$ is a zero divisor.

I tried to pick $\;x \in R,\; x\neq 0\;$ s.t $\;ax = 0$

My first try led me to pick $\;x = ax-x$

$ax = a(ax-x) = a^2x - ax\;$ Then I thought that if i could prove that $\;a\;$ is idempotence I could prove that $\;ax = 0$

my bad , will edit now – Eliran Turgeman Jan 03 '20 at 14:18 — Eliran Turgeman, Jan 03 '20 at 14:18
Awesome, thanks for taking the time to fix that. – rschwieb Jan 03 '20 at 14:19 — rschwieb, Jan 03 '20 at 14:19

score 3 · Accepted Answer · answered Jan 03 '20 at 14:02

3

Take $R=\mathbb{Z}$ an element distinct of $-1$ and $1$ and $0$ is not invertible and is not a zero divisor

answered Jan 03 '20 at 14:02

Tsemo Aristide

if I understand correctly you pick $a = 0$ which is against one of the demands – Eliran Turgeman Jan 03 '20 at 14:21
1

@EliranTurgeman Take $a=2$, if you prefer. But the answer states to take any element not among $1$, $-1$ and $0$. – egreg Jan 03 '20 at 14:24

1 Answers1