Number of zero divisors of a ring

Question

I am given the ring $\mathbb{Z}_8 \times \mathbb{Z}_{10}$, and I am asked to find the number of zero divisors. I have counted $23$, not manually, but taking into account several aspects such as what happens if both elements are $0$ and so. In $\mathbb{Z}_8$ there are $3$ zero divisors, and in $\mathbb{Z}_{10}$ there are $5$. Therefore, one has $3\cdot 5$ zero divisors at the moment. However, the element $(0,a)$ is also a zero divisor if $a$ is, and the same for $(b,0)$, so there are $8$ more, and thus there are a total of $23$ zero divisors. Something seems wrong for me, so I would be grateful if someone could help me.

Answer 1

$x$ is zero divisor if $x \neq 0$ and there is $y \neq 0$ s.t. $xy = 0$.

Lets count number of elements $(a, b)$ s.t. $(a, b) \cdot y = 0$ has a non-zero solution in $y$ - these are zero divisors and $0$. For this, we count number of elements s.t. this equation doesn't have a non-zero solution, and subtract it from $80$.

$(a, b) \cdot (c, d) = 0$ has no non-zero solutions (in $c$ and $d$) iff neither $a\cdot c = 0$ nor $b \cdot d$ have no non-zero solutions. There are $4$ such $a$ and $4$ such $b$, for a total of $16$ pairs. So, we have $80 - 16 - 1 = 63$ zero divisors.

You can, of course, count zero divisors directly, but it's harder to consider all the cases.

Element $(a, b)$ is zero divisor, if:

1. $a$ is $0$ and $b$ is arbitrary non-zero: $9$ elements.
2. $b$ is $0$ and $a$ is arbitrary non-zero: $7$ elements.
3. $a$ is zero divisor and $b$ is arbitrary non-zero (we already counted $b = 0$): $3 \cdot 9 = 27$.
4. $b$ is zero divisor and $a$ is arbitrary non-zero and not zero divisor: $5 \cdot 4 = 20$.

For a total of the same $63$. But it's much harder to consider all cases and avoding double-counting.