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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 $\sum (u_i x)(v_i y)$.

Cure
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  • WHat do you mean by $(xyR)$? The ideal generated by the right ideal $xyR$? At first I had thought you were talking about the product of the right ideals $xR$ and $yR$, but because so many things were out of order I figured I must be reading something wrong. – rschwieb Sep 20 '17 at 16:26
  • It would stand for $(xy)R$. I had it mind because of a proof I was reading. The proof showed that $(xR)(yR)=(xy)R$, and it simply stated that elements of $(xy)R$ had the form $\sum a_i b_i x_i y_i$ before showing that these elements are in $(xR)(yR)$ (to show the mutual inclusion of sets) – Cure Sep 20 '17 at 16:27

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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.

rschwieb
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