Determine if this is an ideal of $\Bbb Z \times \Bbb Z$:
$$E=\{(2n, 3m): n,m \in \Bbb Z \}$$
So to show this is an ideal, I must show that it is closed under addition, closed with respect to negatives, and that it absorbs products.
So for addition: Take $(2n,3m),(2a,3b) \in E$ Then $(2n,3m)+(2a,3b) = (2(n+a),3(m+b))$.
And for negative: $-(2n,3m)=(-2n,3m)=(2(-n),3(-m))$. So that checks out as well.
Now I'm confused about how to show absorbs products. I'm pretty sure it does... $$\text{Take} \ (a,b) \in \Bbb Z. \ \text{Then} \ (2n,3m)(a,b)=(2(na),3(mb)) \in E$$
That one's easy. But showing that $(a,b)(2n,3m) \in E$ is less clear, since $E$ is not necessarily commutative: $$(a,b)(2n,3m)=(a2n,b3m)=(2(na),3(mb)) \ ?$$
$\mathbb{Z\times Z}$ is commutative - in fact, it's an Abelian group. Hence, $a2n=2na$ and $b3m=3mb$.