Describing elements in a principal ideal in a ring that does not have a $1$.

Question

Let $R$ be any ring that does not contain $1$. For $a\in R$, describe the elements of $(a)$.

Breaking this down piece by piece. Since the ring $R$ does not contain $1$, then the ring does not contain an identity element. Therefore, to describe the principal ideal $(a)$, how should I approach the problem or what should I think about? Any help would be great. Thanks

Answer 1

The smallest ideal that contains $a$ must contain:

1. $a$, $-a$, $a+a$, $-(a+a)$, etc. That is, the additive cyclic subgroup of $R$ generated by $a$. This is the set $\{na\mid n\in\mathbb{Z}\}$.

2. All elements of the form $ra$ with $r\in R$. Note that this is already closed under sums and differences, since $ra-sa = (r-s)a$.

3. All elements of the form $as$ with $s\in R$. This is also closed under sums and differences.

4. All elements of the form $ras$ with $r,s\in R$. This is not closed under sums and differences, so you also need sums and differences of such elements.

5. Sums of elements of each of the previous forms.

That means that, at the very least, $(a)$ must contain all elements of the form $$ na + ra + as + \sum_{i=1}^m r_ias_i$$ where $n\in\mathbb{Z}$, $r,s,r_i,s_i\in R$, $m\geq 0$.

Now the question is: does it contain any more elements, or do these elements already suffice to give you an ideal?