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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:

  • $a=(1,1,0)$

  • $b=(1,0,1)$

  • $c=(0,1,1)$

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.

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  • $\begingroup$ This question is an exact duplicate of math.stackexchange.com/q/601835 which should have been found during a search. Since there is nothing to distinguish the questions, I compared the contents of the solutions and concluded it would be a little more beneficial for that one to be the duplicate of this one. Still, this question should never have been asked in the first place. Please use the search before asking next time. $\endgroup$ – rschwieb Jul 11 '18 at 13:05
  • $\begingroup$ Ok. Sure....... $\endgroup$ – blue boy Jul 11 '18 at 14:42
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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 $c=(0,1,1)$.

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    $\begingroup$ Oh.........i got it now. Just started ring theory couple of days back. So struggling. $\endgroup$ – blue boy Jul 11 '18 at 5:09
  • $\begingroup$ @blueboy Looks good for just a couple of days in! $\endgroup$ – Benjamin Dickman Jul 11 '18 at 5:14
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It is worth pointing out that in a commutative ring, if $x$ is a zero divisor, then so is $xy$, unless $xy=0$. The reason is that, if there is some $z$ such that $xz=0$, then $(xy)z=(xz)y=0$. This means that the condition "$abc$ is not a zero divisor" means that $abc=0$. Further, since $ab, bc, ac$ are all nonzero, this means that $a,b,c$ are all zero divisors.

Since the way an element of $\mathbb Z^3$ is a zero divisor is if one of its coordinates is zero, what matters in an example is which coordinates are zero. One might ask "are there any examples that are not of the form "$(x,y,0),(z,0,w),(0,s,t)$" (as such examples are fundamentally relying off of the same key idea as the given example). In fact, there are not!

Between $a,b,c$, we need to have a zero in each of the three coordinates, and each term must have at least one zero. The question is, can we have an example where, say, $a=(1,0,0)$? The answer is no! Because $ab\neq 0$, we can't have $0$ in the first slot of $b$. Similarly, we couldn't have a zero in the first slot of $c$. But then the first slot of $abc$ has to be nonzero too (because $\mathbb Z$ has no zero-divisors).

So not only does the given example work, but it is essentially the only example, conceptually speaking.

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