Why is the set that looks like a two sided ideal not necessarily closed under addition? Let $R$ be a ring and $A \subseteq R$ be finite, say $A = \{a\}$. The set $$RaR = \{ras\:\: : r,s\in R\}$$ Why is this not necessarily closed under addition?
Take $r_1as_1$ and $r_2as_2$. Is it because the ring must be commutative in order to guarantee that we can add these two elements and still stay within $RaR$?
Even though I see why commutativity will make this possible, I'm still a bit confused because the very definition of $RaR$ includes all possible combinations of elements of the form $ras$.
 A: Take $R = M_n(F)$ to be the ring of $n \times n$ matrices over a field $F$, and let $a \in R$ be a matrix of rank $1 \le k < n$. Then I claim (and this is a nice exercise) that $RaR$ consists of exactly the matrices with rank $\le k$. And this set of matrices is not closed under addition: in fact their span is all of $M_n(R)$, because every matrix can be written as a sum of matrices of rank $1$.
To be really specific and minimal we can take $R = M_2(\mathbb{F}_2)$ and $a = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]$.
A: The difficulty is that if you are to prove it is closed under addition, and you have $r_1as_1$ and $r_2as_2$, you have to find $r_3$ and $s_3$ such that $r_1as_1+r_2as_2=r_3as_3$.
You can see that here you get stuck.
Contrast this to the case for right ideals where you have $aR:=\{ar\mid r\in R\}$ where any two elements can have the $a$ factored out to the left, so that $aR$ is clearly closed under addition.
Defining it to be finite sums of things of the form $ras$, $r,s\in R$ makes it automatically closed under addition.
