[Math] Ideal generated by a set in a commutative ring without unity

Question

In a commutative ring with unity $1$, call it $R$, the the ideal generated by the set $S=\{a_1,…,a_n\}$ is the smallest ideal of $R$ containing $S$. It can be proven that this ideal is

$$
(a_1,…,a_n)=\left\{\sum_{k=1}^{n} r_ka_k \, : \, r_k \in R \right\}
$$

I think to have proved this fact by the standard way: clearly any such ideal containing $S$ must contain this set, and this set is itself an ideal. But if the commutative ring does not have a unit, i cannot see why the same proof does not apply. That is, where we use the fact of a unity's existence to show this?

Thanks.

Answer 1

If $R$ has no multiplicative identity, $(a)$ need not contain $a$ if you take it to be $aR$. You want the ideal generated by a set $S$ to contain $S$.