# If $R$ is a commutative ring with unity, then how do I prove: $a \neq 0, ~ b \neq 0 \Longrightarrow a \cdot b \neq 0$?

If $R$ is a commutative ring with unity and $a,b \in R$, then how do I prove that $$a \neq 0, ~ b \neq 0 ~~ \Longrightarrow ~~ a \cdot b \neq 0?$$

Does this also hold for any ring?

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You don’t: it doesn’t. Look at the ring $\Bbb Z/6\Bbb Z$. – Brian M. Scott Feb 7 '13 at 7:33
•means multiplication here. I dunno how to write small circle in Latex.. – Katlus Feb 7 '13 at 7:34
Another example is $\mathbb{R}[X]/(X^2)$. – Julian Kuelshammer Feb 7 '13 at 7:35
I assumed that you meant multiplication. For the record, you can get the composition symbol with \circ. – Brian M. Scott Feb 7 '13 at 7:36
If $R=\Bbb Z/6\Bbb Z$, then $(2x)(3x)=0$ in $R[X]$. You need $R$ to be an integral domain. – Brian M. Scott Feb 7 '13 at 7:39

It’s not true in general: consider $a=2,b=3$ in the commutative ring $\Bbb Z/6\Bbb Z$, which is unital. Rings of the kind that you want are integral domains.