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.
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$\begingroup$ I'm confused, aren't they obviously equal? $\endgroup$– Olivier BégassatCommented May 3, 2014 at 0:01
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$\begingroup$ @OlivierBégassat Depending on your definition of obvious. They are, in fact, equal, though the first containment takes a few steps to show. $\endgroup$– Karl KroningfeldCommented May 3, 2014 at 1:51
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1 Answer
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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 $(ab) \subseteq (a)(b)$.
answered Jul 2, 2015 at 16:04