Finding units and zero divisors of the ring $\mathbb Z_6 \times \mathbb Z_8$

Question

As stated in the title, I wish to find all the units and zero divisors of the ring $\mathbb Z_6 \times \mathbb Z_8$.

What I have so far:

The units of $\mathbb Z_6$ are $\{1,5\}$, and the units of $\mathbb Z_8$ are $\{1,7\}$, so the units of $\mathbb Z_6 \times \mathbb Z_8$ are $\{(1,1), (1,7), (5,1), (5,7)\}$. Is that correct?

The zero-divisors of $\mathbb Z_6$ are $\{2, 3, 4\}$, and the zero-divisors of $\mathbb Z_8$ are $\{0, 2, 4, 6\}$. I am not sure how to proceed, however. Thoughts?

I am new to math.stackexchange so please let me know if any part of my question is unclear. Thanks!

Answer 1

For example $\;(2,x)\;$ is a zero divisor for any $\;x\in\Bbb Z_8\;$ , since

$$(2,x)\cdot (3,0)=(0,0)\;,\;\;\text{and none of the elements is the zero element}$$

Answer 2

Actually in a finite commutative ring, a nonzero element is either a unit or a zero divisor. Indeed, let $x$ be an element and consider the successive powers of $x$. Since the ring is finite, two of these powers are equal, say $x^n = x^{n+p}$ for some $n \geqslant 0$ and $p > 0$. It follows that $x^n(1-x^p) = 0$. Thus if $x \not= 0$ and $x$ does not divides $0$, then $1 =x^p$ and thus $x$ is a unit.

Coming back to your question, since you already found the units of your ring, you can deduce immediately the zero divisors.