In a commutative ring without identity, is $(a)(b)\subset (ab)$ or $(ab)\subset (a)(b)$?

Question

Let $R$ be a commutative ring without unity. Consider an ideal $(a)$ generated by $a\in R$. Note that $(a)=\{ra+na : r\in R, n\in \textbf Z\}$ since $R$ has no identity. I wonder if $(a)(b)\subset (ab)$ or $(ab)\subset (a)(b)$. It seems a simple question but I'm confused because I always deal with a ring with unity. Please prove one of these and give a counterexample of the other.

Answer 1

The two ideals are in fact equal.

Let $ra + na \in (a)$ and $sb + mb \in (b)$ where $r, s \in R$ and $n, m \in \mathbb{Z}$, then

\begin{align*} (ra + na)(sb + mb) &= rsab + mrab + nsab + nmab\\ &= (\underbrace{rs + mr + ns}_{\in\ R})ab + (\underbrace{nm}_{\in\ \mathbb{Z}})ab \in (ab). \end{align*}

As $(a)(b)$ consists of finite sums of elements of the form $(ra + na)(sb + mb)$ and $(ab)$ is closed under addition, $(a)(b) \subseteq (ab)$.

For the other inclusion, note that $rab + nab = (ra + na)b \in (a)(b)$ so $(ab) \subseteq (a)(b)$.