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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!)

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  • $\begingroup$ Which hypotheses do you think are superfluous? Is there one you didn’t use? $\endgroup$ – Randall Aug 7 '18 at 17:26
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    $\begingroup$ Specifically the $rm \neq 0$ for some $r \in R$ and $m \in M$. I'll edit my post to include that. (Though it's possible I used it somewhere without even realizing it.) $\endgroup$ – rwbogl Aug 7 '18 at 17:28
  • $\begingroup$ Just checking, an "$R$-homomorphism" is defined by $f(x+y)=f(x)+f(y)$ and $f(r\cdot x)=r\cdot f(x)$, right? No fancy attempts to combine them into a single axiom? $\endgroup$ – Henning Makholm Aug 7 '18 at 17:37
  • $\begingroup$ @rwbogl what if the module wasn’t trivial but the ring action was? $\endgroup$ – Randall Aug 7 '18 at 17:38
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    $\begingroup$ @Randall Then the homomorphism $T(x) = rx$ is the zero map, correct? But I don't see how that disagrees with any steps in the proof, or its conclusion. $\endgroup$ – rwbogl Aug 7 '18 at 17:43
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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.

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  • $\begingroup$ This is likely correct. $\endgroup$ – Randall Aug 7 '18 at 18:12
  • $\begingroup$ Ah, I think I see. So why do we want to exclude modules with trivial ring action from being simple? $\endgroup$ – rwbogl Aug 7 '18 at 18:23
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    $\begingroup$ @rwbogl Some authors just do, likely to make some bigger theorems slightly easier to state. I've also seen $\mathbb{Z}_p$ excluded as a simple group, which drives me crazy. $\endgroup$ – Randall Aug 7 '18 at 18:32
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    $\begingroup$ @rwbogl I would advise you to check in Jacobson's Structure of rings. I don't remember clearly why. The one thing I do remember is that $2\mathbb Z/4\mathbb Z$ is a simple module that fails this, and that $4\mathbb Z$ winds up being a maximal ideal which isn't prime, which seems bad. I think that Jacobson eliminates some maximal ideals for a degeneracy condition like this. $\endgroup$ – rschwieb Aug 7 '18 at 20:09

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