How do I find the zero divisors for $\Bbb Z_{5} \times \Bbb Z_{8}$

Question

I'm trying to find the zero divisors for $\Bbb Z_{5} \times \Bbb Z_{8}$. I know that $\Bbb Z_{5}$ doesn't have any zero divisors but $\Bbb Z_{8}$ has $\{ 2,4,6 \}$ for zero divisors. Does this mean that $\{ 2,4,6 \}$ are the zero divisors for $\Bbb Z_{5} \times \Bbb Z_{8}$? Or since zero is the only element that you can multiply by itself to get zero in $\Bbb Z_{5}$, would the answer be $\{ (0,0),(0,2),(0,4),(0,6) \}$?

Answer 1

The relevant cases are: \begin{array}{ll} \text{Element} & \text{Zero divisor?} \\ \hline (0,0) & \text{zero divisor depending on definition}\\ (0,odd) & \text{zero when multiplied with (1,0) } \\ (0,even \ne 0) & \text{zero when multiplied with (1,0) } \\ (nonzero,0) & \text{zero when multiplied with (0,1) }\\ (nonzero,odd) & \text{only zero when multiplied with (0,0) so not a zero divisor} \\ (nonzero,even \ne 0) & \text{$\exists d\ne 0$ such that zero when multiplied with (0,d)} \end{array} In other words all elements are zero divisors except (a,b) with nonzero a and odd b, and except arguably (0,0). There are $4\cdot 4=16$ such (a,b) on a total of $5\cdot 8=40$ elements. So there are $40-16-1=23$ zero divisors excluding (0,0).

That is, the zero divisors are $\{(0,b)\in \mathbb Z_5\times\mathbb Z_8: b\ne 0\} \cup \{(a,b): a\ne 0\land b\in\{2,4,6\}\}$.