Cardinal number of $U(\Bbb Z / 4 \Bbb Z \times \Bbb Z/5 \Bbb Z)$ and zero divisors of $\Bbb Z / 45 \Bbb Z \times \Bbb Z / 27 \Bbb Z$

Question

I'd like to know how many units does $ \Bbb Z / 4 \Bbb Z \times \Bbb Z / 5 \Bbb Z $ has. And how many zero divisor has $ \Bbb Z / 45 \Bbb Z \times \Bbb Z / 27 \Bbb Z $. I think I should work with $\Bbb Z /‭1215 \Bbb Z‬$ and $\Bbb Z /‭20 \Bbb Z‬$ using the chinese remainder theorem and that the function $$\begin{array}{rccl} f \colon & \Bbb Z / nm \Bbb Z & \longrightarrow & \Bbb Z / n \Bbb Z \times \Bbb Z / m \Bbb Z \\ &[x]_{nm} & \longmapsto & ([x]_n, [x]_m) \end{array}\\ $$ where $n,m\in \Bbb Z^+$ are coprimes is bijective, but I have no idea about how to do it. Any help?

Answer 1

Let $m$ and $n$ be any two positive integers.

Units in the ring $R=\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$:

$([x]_m,[y]_n)$ is a unit in $R$ if and only if ($[x]_m$ is a unit in $\mathbb{Z}/m\mathbb{Z}$ and $[y]_n$ is a unit in $\mathbb{Z}/n\mathbb{Z}$).

Zero divisors in the ring $R=\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$:

Let $([x]_m,[y]_n)$ be a zero divisor in $R$. Then there exists $([a]_m,[b]_n)\neq 0$ in $R$ such that $([xa]_m,[yb]_n)=([0]_m,[0]_n)$. Thus, all the zero divisors are given by $([x]_m,[y]_n)$ where either $[x]_m$ is zero divisor of $\mathbb{Z}/m\mathbb{Z}$ or $[y]_n$ is a zero divisor of $\mathbb{Z}/n\mathbb{Z}$.

Answer 2

Hints: Let $R,S$ be unital rings. Then prove the identity

$$U(R \times S) = U(R) \times U(S)$$

What are the units of $\mathbb{Z}/4\mathbb{Z}$, what are the units of $\mathbb{Z}$?

For the second question,

$$(x_1,y_1)(x_2, y_2) = (0,0)\iff x_1x_2 = 0 \quad \land \quad y_2y_2 = 0 $$

so finding the zero divisors of $\mathbb{Z}/45\mathbb{Z}\times \mathbb{Z}/27\mathbb{Z}$ also reduces to finding the zero divisors of both rings separately.