The sum of all right ideals isomorphic as modules to a simple module is an ideal I could use some help on the following problem.
Let R be a ring.  
(a) If $r \in R$ and $U$ is a minimal right ideal of $R$, show that either $rU=0$, or that $rU$ and $U$ are isomorphic right $R$-modules.
(b) Now let $S$ be a simple right $R$-module, and let $I_S$ be the sum of all right ideals of $R$ that are isomorphic as right $R$-modules to $S$.  Show that $I_S$ is an ideal of $R$.
I was able to do (a).  If $rU=0$ then $r \in \mathrm{ann}_R(U)$.  Supposing $r \not\in \mathrm{ann}_R(U)$, we can show that $\phi: U \to rU$ given by $\phi(u)=ru$ is an isomorphism.  It's not hard to show that it's a $R$-homomorphism of $R$-groups, and the fact that $r \not\in \mathrm{ann}_R(U)$ and $U$ is a minimal right ideal tells us that $U \cap ra(r)=0$, where $ra(r)$ is the right annihilator of $r$ in $R$, so the map is injective, and it easily follows that it is also surjective.
Now, let $I_S=R_1+R_2+ \cdots$.  Since $S$ is simple, each $R_i \cong S$ is minimal.  For this reason, $R_i \cap R_j = 0$ for all $i \ne j$, and $I_S$ is a direct sum, so multiplication by elements of $R$ is component-wise.  Since $R_i$ is a right ideal, $I_S$ is closed under right multiplication.
Now multiply $I_S$ by $r$ on the left.  By (a), for each component $R_i$, $rR_i=0$ or $rR_i \cong R_i$.  But $rR_i \ne 0$, because then $R_i$ contains a subgroup of the right annihilator of $r$, contradicting minimality.
What I'm having trouble seeing (if it's true at all) is how $rR_i \cong R_i$ shows us that $I_S$ is a left ideal.  Since $I_S$ is a direct sum it is the same if we permute the summands, but I'm not sure if the isomorphism only permutes the summands, or if it yields a new module isomorphic to $I_S$.  I'd appreciate some help.  Thanks.
 A: I think you're overthinking it. You're already at the answer :)
All minimal right ideals of the same isotype sum to $I_S$ (and are hence contained in $I_S$.) You've just said that $rR_i\subseteq I_S$ for each $R_i$ in the isoclass of $S$. So, $rI_S=r(\sum R_i)=\sum rR_i\subseteq I_S$.  
I think I see a little confusion in the second paragraph of the answer (beginning with Now...). While its certainly true that $I_S$ is a direct sum of minimal right ideals, it isn't true that the sum of all mutually isomorphic right ideals is a direct sum. Just think of $M_n(\Bbb R)$: it has infinitely many minimal right ideals, but the whole ring itself is isomorphic to a direct sum of $n$ of them.
I hope that clarifies things enough to see that in the last paragraph, there is no need to worry about "permuting summands." It's already been shown that $I_S$ absorbs multiplication on the left, and that's exactly what's needed.

Added: You asked for the error in the reasoning that the sum of simple submodules is always direct. It is this line: 

For this reason, $R_i \cap R_j = 0$ for all $i \ne j$, and $I_S$ is a direct sum...

It is not enough for the pairwise intersection of any two $R_i$ to be zero. You need $R_i\cap(\sum_{j\neq i} R_j)=\{0\}$. So for example in $\Bbb R^2$, $V_1=\langle(1,0)\rangle$,$V_2=\langle(0,1)\rangle$, and $V_3=\langle(1,1)\rangle$ all have pairwise intersection zero, but of course $V_3\subseteq V_1+V_2$, so $\{0\}\neq V_3\cap (V_1+ V_2)$.
Luckily, we don't need to worry about the direct sum here. The regular old sum is simpler and exactly what we need.
