Find elements $a,b,$ and $c$ in the ring $\mathbb{Z}×\mathbb{Z}×\mathbb{Z}$ such that $ab, ac,$ and $bc$ are zero divisors but $abc$ is not a zero divisor.
Work:
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$a=(1,1,0)$
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$b=(1,0,1)$
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$c=(0,1,1)$
Why this works: because $ab=(1,0,0)\neq(0,0,0)$.
Definition of zero divisor. A zero divisor is a non-zero element $a$ of a commutative ring $R$ such that there is a non-zero $b \in R$ with $ab=0$.
Amy hint or suggestion will be appreciated.
Best Answer
The $a, b,$ and $c$ that you suggest seem to work, although your reasoning is somewhat missing. Depending on where you are in your studies, it appears that you want to prove your choice of these three elements in $R = \mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ yield zero divisors $ab$, $ac$, and $bc$, but for which $abc$ is not a zero divisor. To show all of this in accordance with the definition that you have provided, you need to know that $R$ is a commutative ring, and that its zero element is $(0,0,0)$. Whether that observation requires proof is a function of your course; let us assume it is so for the present purpose.
As to your suggested elements: You have correctly computed $ab$. What about $ac$? What about $bc$? Finally, what is it about $abc$ that makes it not a zero divisor?
For completeness, it may help to prove that the specified elements really are zero divisors when relevant. For example, you calculated that $ab = (1, 0, 0) \neq (0, 0, 0)$; but, this only shows that $ab$ is nonzero. To prove that it is a zero divisor, you will still need to prove that there is a nonzero element of $R$ that, multiplied by $ab$, yields the zero element $(0,0,0)$. In this particular case, you can conveniently find such an example with your nonzero element $c=(0,1,1)$.