An example of an ideal of order $12$

Question

Provide an example of an ideal in $R=\mathbb{Z}_6\times\mathbb{Z}_{10}$ that has order $12$, and indicate whether the ideal is a principal ideal (if it is, then identify the generator for the ideal).

Let $$I=\{(0,5),(1,0),(2,5),(3,0),(4,5),(5,0),(0,0),(1,5),(2,0),(3,5),(4,0),(5,5)\}.$$ Then $I$ is a principal ideal with $(1,5)$ as the generator. Does this work? If not, what else could possibly work or is there a more elegant solution?

Answer 1

Note that by the Chinese Remainder Theorem we have $$\Bbb{Z}_6 \times \Bbb{Z}_{10} \simeq \Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_3 \times \Bbb{Z}_5$$ The only ideals in a product of fields are the "partial products"; so the example you provided, which as Michael Burr noted is $$I = \Bbb{Z}_6 \times \Bbb{Z}_2 \simeq \Bbb{Z}_2 \times \Bbb{Z}_2 \times \Bbb{Z}_3$$ is the only ideal of order $12$ in $R$.

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It isn't hard to see that $I = \langle (1,0), (0,5) \rangle$, because $1$ has order $6$ in $\Bbb{Z}_6$ and $5$ has order $2$ in $\Bbb{Z}_{10}$. As you observed, it then follows that $I$ is principal and generated by $(1,5)$, because $$ (1,0) (1,5) = (1,0) \quad \text{and} \quad (0,1) (1,5) = (0,5) $$

Answer 2

The set you give is an ideal of order $12$, and as explained in another answer it is the only ideal.

It is in fact also principal, generated by $(1,5)$ as you said correctly, even though the additive group $(1,5)$ generates is only of order $6$.