For $\Bbb Z/5\Bbb Z\times \Bbb Z/18\Bbb Z $, I know that $\Bbb Z/5\Bbb Z$ has no zero divisors because $5$ is prime and that the zero divisors of $\Bbb Z/18\Bbb Z=\{[2]_{18},[3]_{18},[4]_{18},[6]_{18},[8]_{18},[9]_{18},[10]_{18},[12]_{18},[14]_{18},[15]_{18},[16]_{18}\} $.

For the zero divisor of $\Bbb Z/5\Bbb Z\times \Bbb Z/18\Bbb Z $, would I say then that it does not exist because there does not exist an zero divisor of the form $([a]_{5},[b]_{18})$ or $([a]_{5},[0]_{18})$? Or, can I say that it does exist because a zero divisor of the form $([0]_{5},[b]_{18})$ exists so the zero divisor of $Z/5\Bbb Z\times \Bbb Z/18\Bbb Z=\{([0]_{5},[b]_{18}):b=2,3,4,6,8,9,10,12,14,15,16\}$? Am I allowed to include $b=0$ into this set?

I am kind of confused about if and why we are allowed to include $[0]_{5}$ even though it is not a zero divisor so if anyone could shed some light onto this, that would be most appreciated. Thanks a lot. :)


It is the latter.

To show that $([0]_5,[2]_{18})$ is a legitimate zero divisor, for example, you must show that you can multiply $([0]_5,[2]_{18})$ by something in $\mathbb Z_5 \times \mathbb Z_{18}$ to give zero. Indeed, $$ ([0]_5,[2]_{18}). ([0]_5,[9]_{18}) = ([0]_5,[0]_{18}).$$ That is how you would write it formally.

And yes, $([0]_5,[0]_{18})$ itself is a zero divisor, because $([0]_5,[0]_{18})$ times anything is zero.

  • $\begingroup$ Okay cool thanks! I thought so, but I felt kinda awkward to do, so I figured I'd ask about it. Thanks :) $\endgroup$ – eulersnumber Mar 7 '17 at 0:27

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