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

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.

• You can find the zero divisors and units like here. Commented Oct 30, 2016 at 16:52
• – lhf
Commented Jan 15, 2020 at 11:08

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.

• So, the units of Z6 x Z2 are just the different combinations of units of Z6 and units of Z2?
– Max
Commented Oct 30, 2016 at 19:22

There are a few things to look for.

For any two rings $R,S$, and any two elements $r\in R, s\in S$, we have $(r,0)(0,s)=(0,0)$ in $R\times S$. Also, if $r_1,r_2\in R$ are non-zero elements such that $r_1r_2=0$, we have $(r_1,0)(r_2,0)=(0,0)$. Similarly for $s_1,s_2\in S$. You can even combine these three: $(r_1,0)(r_2,s)=(0,0)$, for instance.

As for units, if both $r\in R$ and $s\in S$ are invertible, then clearly $(r,s)$ is. If one of them, say $r$ is not a unit, then there is no $(r',s')\in R\times S$ such that $(r,s)(r',s')=(1,1)$, because $rr'$ cannot be $1$.

In a finite commutative ring, a nonzero element is either a unit or a zero divisor. So it is enough to look for, say, the units. Here we just have $$U(\mathbb{Z}/n\times \mathbb{Z}/m)\cong U(\mathbb{Z}/n)\times U(\mathbb{Z}/m).$$

• Oh, so if I enumerate all the elements of Z6 x Z2, and find all the units, then the zero divisors are just what's left?
– Max
Commented Oct 30, 2016 at 19:24
• Right. Try it first for $\mathbb{Z}/6$ alone, to see what happens. Commented Oct 30, 2016 at 19:50
• @Max The answer is $Z6 X Z2 - {(0,0)} - U(6) X U(2)$
– KON3
Commented Oct 7, 2022 at 3:50