List the distinct principal ideals in $\mathbb{ℤ}_5 \times \mathbb{ℤ}_{12}$

Question

If I want to accomplish this, do I need to first list all the elements of $\mathbb{ℤ}_5 \times \mathbb{ℤ}_{12}$ and then multiply each individual pair by all other 48 pairs to see in the end the unique (distinct) pairs possible, or am I missing some trick that would allow a more efficient process?

Answer 1

Hint: First note that $\mathbb Z_5$ is a field so if $(a, b)$ generates an ideal and $a \neq 0$ then $(a^{-1}, 1)$ is a unit so multiplying I get that $(1, b)$ generates the ideal as well. So your generators are of the form $(0, b)$ or $(1, b)$.

Also note that the Chinese remainder theorem tells you that $\mathbb Z_{12} \simeq \mathbb Z_3 \times \mathbb Z_4$. Since $\mathbb Z_3$ is also a field you can reduce your options even further.

Answer 2

By CRT you have the cyclic ring $\Bbb Z_{60}$. The principal ideals are all the ideals of the form $n\Bbb Z_{60}$, where $n|60$.