Why $(xyR)$ elements are of the form $\sum (u_i x)(v_i y)$? This should be simple but I can't some up with a more heuristic answer to it.
$xR$ is the set of $ux$ elements, and $yR$ the set of $vy$ elements, in both cases with $u,v\in R$. It seems natural to me that elements of $xyR$ will have the form $uv(xy)$ but they are in fact given the more general form $\sum (u_i x)(v_i y)$.
 A: Er, it looks like you got a few expressions backwards in the post, so I'll try to straighten it out. If I have misunderstood your intention I'll do my best to correct myself.
If you're working with noncommutative rings, the set $xR=\{xv\mid v\in R\}$  and the set $yR=\{yv\mid v\in R\}$ are right ideals.
The product of two right ideals $(xR)(yR)$ is described just like the product of two two-sided ideals: the set of elements of the form $\sum_{i=1}^n a_ib_i$ where $n$ can be any natural number, $a_i\in xR$ and $b_i\in yR$.  That is how you get elements of the form $\sum xr_iys_i$.
Without commutativity, there is no reason to expect you can rewrite these as $\sum xyt_i $ for some $t_i\in R$. But with commutativity, you can of course do that.
In general, $xyR\subseteq (xR)(yR)$, sometimes without equality. For example, in $R=M_2(\mathbb R)$, with $x=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $y=\begin{bmatrix}0&0\\0&1\end{bmatrix}$, you have $xyR=\{0\}$ but $(xR)(yR)=xR$
If you have any doubts about the ideal product, you could take a look at this post.
