Rings isomorphic to $\mathbb{Z}_6\times\mathbb{Z}_{10}$

Question

What are five ring properties that hold for each ring that is isomorphic to $R=\mathbb{Z}_6\times\mathbb{Z}_{10}$, but not for every ring?

Suppose $Q\approx R$. Then $Q$ has unity, $Q$ is not a field, $Q$ is commutative, $Q$ has no zero divisors, and $Q$ is an integral domain. Do these work? Are there others?

Answer 1

Let $Q \cong R$.

• $Q$ has zero divisors. $(2,2)$ is one.
• $Q$ is commutative.
• $Q$ is not an integral domain, because a ring is an integral domain by definition iff it's commutative and has no zero divisors.
• $Q$ is not a field, because every field is an integral domain.
• $Q$ has unity (assuming your definition of ring doesn't already require it).

I apologize if I haven't understand what is meant by a "ring property". $Q$ also has finitely many elements.

Answer 2

$\mathbb Z_6\times\mathbb Z_{10}\cong \mathbb Z_2\times\mathbb Z_{30}$. It might be easier to characterize that ring.

One characterization is in terms of an idempotent[*]:

The ring is commutative and there is an idempotent element $e\neq 0$ such that $e+e=0$, with $1-e$ having additive order $30$, and $e$ and $1$ generate the entire ring additively. In $\mathbb Z_6\times\mathbb Z_{10}$, you get $e=(3,0)$. Then $(1,0)=(3,0)+10(1,1)$ and $(0,1)=(3,0)+21(1,1)$.

Or:

The ring is commutative with identity and $60$ elements, with an idempotent $e\neq 0$ such that $e+e=0$ and $e,1$ generate the entire ring, additively. Same $e,1$ for this case.

Or:

The ring is commutative with identity and there is an idempotent $e$ of additive order $6$ with $1-e$ having additive order $10$ and $e$ and $1$ generate the entire group additively. Here, $e=(1,0)$.

[*] An idempotent element of a ring is an element $e$ such that $e\cdot e=1$.