Every left ideal is complemented on the regular representation module In a proof they used the fact that if R is a ring (with unit) and $_RR$ is the R-module with the action given by multiplication in the ring, then I a left ideal of R is complemented by another left ideal J (complemented in $_RR$, $_RR=I\oplus J$).
The proof also had the hypothesis of $_RR$ being semisimple, but I don't know if that was even used for this statement. I can't seem to figure out why it would necessarily be true (perhaps we need another property on the ring?).
My attempt:
Given the fact that I want an ideal, unique writing and for the sum to be R, my guess would be to consider: $1-i=x_i\ \text{for some}\ i\in I$, $x_i$ clearly doesn't belong to I, otherwise $1\in I$ and then I is R, so I would just consider J={$0$} (WLOG we can also assume I is not the zero-ideal).
Then $J=(x_i)_L$ is a left ideal trivially (and the smallest one whose sum makes R, so as to reduce the chance of intersection). Now I need to see $I\cap J=\lbrace 0\rbrace$.
Let $x\in I\cap J\implies x =y=rx_i=r(1-i)$ for $y\in I$ and $r\in R$.
$$x=r(1-i)=r-ri=y\implies r=ri-y\in I$$
In order for x to be $0$, then r should be a left divisor of $1-i$ and be in I. Yet, I don't see how to choose i so that $I\cap J ={0}$ (or even if I can do so).
 A: This is certainly not true without any hypotheses on $R$. For instance, if $R$ is an integral domain and $I,J<R$ are non-zero ideals, then we have $IJ\neq\{0\}$ and so $I\cap J\geqslant IJ\neq\{0\}$. Thus $R$ cannot be decomposed into a direct sum of any two non-zero ideals.
However, if $_RR$ is semisimple, then the result does hold. In fact, we have the following:
Lemma: For any ring $R$ and any (left) $R$-module, if $M$ is semisimple then $M$ is "completely reducible": ie, for any $A\leqslant M$, there is $B\leqslant M$ such that $A\oplus B=M$.
Proof: Let $A\leqslant M$. Since $M$ is semisimple, we can write $M=\bigoplus_{i\in I}M_i$, where each $M_i\leqslant M$ is simple. Now, let $S=\{N\leqslant M:A\cap N=\{0\}\}$, partially ordered by inclusion. $S$ contains the trivial submodule, and thus is non-empty. Also, if $(N_j)_{j\in J}$ is a chain in $S$, then clearly $N:=\bigcup_{j\in J}N_j$ lies in $S$; indeed, if we have $n\in N\cap A$, then $n\in N_j$ for some $j$, whence $n\in N_j\cap A=\{0\}$ and so $n=0$. Thus we may apply Zorn's lemma to find $B\in S$ maximal with respect to inclusion.
We claim that $C:=A\oplus B=M$. Since $M=\bigoplus_{i\in I}M_i$, it suffices to show $M_i\leqslant C$ for every $i$, so let $i\in I$ and suppose for contradiction that $M_i\nleqslant C$. Since $M_i$ is simple, this means that $C\cap M_i=\{0\}$. In particular, $M_i\cap B=\{0\}$, so $B':=B\oplus M_i$ strictly contains $B$. We claim that $B'\cap A=0$, which will contradict maximality of $B$ in $S$. Indeed, suppose we have $x\in B'\cap A$. Then we have $x=b+m$ and $x=a$ for some $b\in B$, $m\in M_i$, and $a\in A$. This means $a-b=m\in C\cap M_i=\{0\}$, whence $a=b$. Since $A\cap B=\{0\}$ by construction, this means $a=b=0$, and so $x=0$, giving the desired contradiction. $\square$
In fact, the converse of this lemma holds as well; as an exercise, try to prove it! (Hint: first show that a module is semisimple if and only if it is a sum (not necessarily direct) of simple submodules, using Zorn's lemma.) This characterization of semisimple modules is so common that it is sometimes taken as definition! In any case, the (left) submodules of $R$ are precisely the left ideals, so your desired result then follows immediately from the lemma.
