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4

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}...


1

Not to be gauche, but I ended up finding a good answer to my own question: Varadarajan's Supersymmetry for Mathematicians: An Introduction



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