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Let $R$ be a nonzero ring and assume $a\in R$. Here we do not assume $R$ has a unity or is commutative. According to most of the references (e.g. Dummit and Foote), the principal ideal $(a)$ has a closed form: $$(a)=RaR=\left\{r_1ar_1'+\cdots+r_nar_n'\mid n\geq1, r_i,r_i'\in R\right\}.$$ It can also be shown easily that $RaR$ is a two-sided ideal:

Proof. Set $r=r'=0$, we have $rar'=0a0=0\in RaR$, so $RaR$ is nonempty. Secondly, $$(r_1ar_1'+\cdots+r_nar_n')-(s_1as_1'+\cdots+s_mas_m')=r_1ar_1'+\cdots+r_nar_n'+(-s_1)as_1'+\cdots+(-s_m)as_m'\in RaR.$$ Finally, for every $x\in R$, $$x(r_1ar_1'+\cdots+r_nar_n')=(xr_1)ar_1'+\cdots+(xr_n)ar_n'\in RaR.$$ $$(r_1ar_1'+\cdots+r_nar_n')x=r_1a(r_1'x)+\cdots+r_na(r_n'x)\in RaR.$$ Therefore, $RaR$ is a two-sided ideal of $R$.

Furthermore, we can also show that $RaR$ is contained in every two-sided ideal of $R$ containing $a$.

Proof. Let $I$ be a two-sided ideal of $R$ containing $a$. Then for every $r,s\in R$, we always have $ra\in I$ and hence $ras\in I$. Therefore, $r_1ar_1'+\cdots+r_nar_n'\in I$ as well, implying that $RaR\subseteq I$.

It seems reasonable to conclude that $(a)=RaR$, but I found a problem. If $R$ is not assumed to be commutative or contain a unity $1$, what about the elements of the form $ra$ and $ar$ for some $r\in R$ and $n\cdot a$ for some $n\geq 1$? It seems that they cannot be easily represented easily in the form of elements in $RaR$.

I also referred to this question: In general rings, what do principal ideals look like? and it made me more confused. So I hope if anyone have good ideas or suggestions on this. Any help will be appreciated.

  • I believe my previous proofs are fine. But there is one implicit question: Is $a\in RaR$? As mentioned by @rschwieb below, if the identity is not assumed, we will not be able to conclude $a\in RaR$, although $RaR$ is indeed a two-sided ideal. –  Mar 21 '20 at 00:08

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If $R$ is not assumed to be commutative or contain a unity $1$, what about the elements of the form $ra$ and $ar$ for some $r\in R$ and $n\cdot a$ for some $n\geq 1$?

Missing commutativity does not cause any problem at all.

$ra=ra1$ and $ar=1ar$, obviously.

You are right that removing identity changes things.

Then you need to modify to sums of the form $na+\sum r_ias_i$ where $n$ is any integer, so that $a\in (a)$.

rschwieb
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  • Thanks a lot for your response! The identity is really essential in this manipulation. I am reading Dummit&Foote. The authors defines rings without identity, but in most of the following discussions, they implicitly assumes the ring contains an identity, which makes it very confusing :) –  Mar 20 '20 at 17:20
  • When learning the basics of ring theory, it is not really that bad an idea to assume identity until you are pretty competent with the theory. Exploring rings without identity is really better done once you appreciate what all breaks down (since you'd be familiar with what rings with identity give you.) – rschwieb Mar 20 '20 at 17:24
  • You are right. Most of the examples I worked out contain an identity. They are really helpful. But personally I still like general results ^_^ –  Mar 20 '20 at 17:27