[Math] Finding Zero Divisors and Units of Rings $\mathbb{Z}_n \times \mathbb{Z}_m$

Question

How do I exactly find the zero divisors and units of a ring in the form of $\mathbb{Z}_n \times \mathbb{Z}_m$? For instance, how does one calculate zero divisors and units of $\mathbb{Z}_6 \times \mathbb{Z}_2$? I understand that a zero divisor is a non-zero element of $R$ such that $rs = 0$ for some other non-zero element $s$ in $R$. But I don't really quite understand how to apply that to a ring of the form $\mathbb{Z}_n \times \mathbb{Z}_m$. Also, the units of $\mathbb{Z}_6 \times \mathbb{Z}_2$ would be elements that send some element $s$ in $\mathbb{Z}_n \times \mathbb{Z}_m$ to the multiplicative identity, right? (which would be 1)? How do I apply these concepts to "product" rings? Thank you.

Answer 1

Every element of the form $(x,y)$ with either $x$ or $y$ a zero divisor is a zero divisor.

Indeed, suppose $x$ is a zero divisor. Then $ax=0$ for some $a\ne0$, so $(a,0)(x,y)=(0,0)$. Similarly if $y$ is a zero divisor.

Conversely, if $(x,y)$ is a zero divisor, then either $x$ or $y$ is.

Indeed, if $(a,b)(x,y)=(0,0)$, with $(a,b)\ne(0,0)$, then either $a\ne0$ or $b\ne0$. If $a\ne0$, then $(a,0)(x,y)=(0,0)$ and similarly if $b\ne0$.

An element $(x,y)$ is a unit if and only if both $x$ and $y$ are units.

Since every element of $\mathbb{Z}_k$ is either a zero divisor or a unit, you should be able to finish.