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No. For example, suppose $N=S$ is simple, so that $\operatorname{soc}N=S$, $M'$ is a non-split extension of $S$ by another simple module $T$, so that $\operatorname{soc}M'=T$ and there is an epimorphism $\alpha:M'\to N$ which is zero on $\operatorname{soc}M'$, and that $M=M'\oplus S$. Then the map $\begin{pmatrix}\alpha&\operatorname{id}_S\end{pmatrix}:... 3 It is true if the ring does not have divisor of zero. Suppose that$m<n$, write$n=pm+r$, you have$a^n=a^{pm+r}=b^{pm+r}=b^{pm}b^r=a^{pm}b^r$. You deduce that$a^{pm}a^r=a^{pm}b^r$and$a^{pm}(a^r-b^r)$and$a^r=b^r$. Thus the assertion is true for$(m,r)$you can repeat this process until the rest is 1. 3 In$D(R)$,$\operatorname{Hom}(R,R[t])=0$for$t\neq0$, so you definitely need the condition that$\operatorname{Hom}(E_i,E_j[t])=0$for$t\neq0$. In my paper "Morita theory for derived categories" (J. London Math. Soc. (2) 39 (1989), no. 3, 436-456) I proved a result for derived categories of rings with a proof along the lines of the idea in your final ... 2 First$(3)$implies$(1)$. Suppose$M \cong R/I$, with$I$a maximal ideal in$R$. Then as a ring$R/I$is a field. Hence the only ideals of$R/I$are the ring itself and$\{0\}$. But this means that the only subgroups of$R/I$that are closed under multiplication by elements of$R$are$R/I$and$\{0\}$, and so$M$is a simple module. For the other ... 1 The annihilator of$0\in M$is$R$, which is always going to be essential. The singular submodule always has at least that. 1 It is enough to find a right nonsingular ring that isn't semiprime. Let$R=\left[\begin{smallmatrix}F&F\\0&F\end{smallmatrix}\right]$, the$2\times 2$upper triangular matrices over a field. Then$I=\left[\begin{smallmatrix}0&F\\0&0\end{smallmatrix}\right]$satisfies$I^2=0$, but$r(\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}...