An exercise paraphrased from Herstein's Topics in Algebra (2nd edition, Chapter 4, §4.5, problem 12):
Let $M$ be an irreducible left $R$-module, where $R$ is an arbitrary ring and $rm \neq 0$ for some $r \in R$ and $m \in M$. Prove that if $T \colon M \to M$ is an $R$-homomorphism, then it is either the zero map or an isomorphism.
This exercise is marked as more difficult than others, but I don't understand why, or why all of the hypotheses are necessary. In particular, I don't understand why the $rm \neq 0$ for some $r \in R$ and $m \in M$ matters. I know that assuming that implies that $M$ is cyclic, but not why that would be helpful here. Disagreeing with the text makes me think I've completely missed some subtlety of modules.
Here's my attempted proof:
Proof: The kernel of $T$ is a submodule of $M$. Since $M$ is irreducible, it can be either $\{0\}$ or $M$. If the kernel is $M$, then $T$ is the zero map, and we're done; if it's $\{0\}$, then $T$ is injective. Further, if the kernel is $\{0\}$ and $M$ has at least one nonzero element, then $T(M) \neq \{0\}$. Applying the irreducibility of $M$ to the submodule $T(M)$ yields $T(M) = M$, so $T$ is also surjective. $\blacksquare$
Have I gone wrong somewhere? If not, why would the extra hypotheses be added? (They appear in nearly every exercise following this one as well!)
Not having the book leaves me at a disadvantage, but I think I have a pertinent comment about the extra conditions. At any rate, I think your work in this exercise does not require the bit about $rm\neq 0$.
I believe I recognize the condition from other texts I've seen on module theory for rings without identity. The idea is to exclude modules with trivial action from being considered as simple modules. In rings with identity, of course, this happens automatically.
You will also see definitions of "simple ring (without identity)" as being "a nonzero ring $R$ having only trivial ideals, and also $R^2\neq 0$" or sometimes just "$R^2=R$."
So my feeling is that it is just a nondegeneracy condition that was included by habit.