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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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9  
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
1  
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

1 Answer 1

up vote 8 down vote accepted

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.

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