# Why does Herstein add all these extra hypotheses for a “simple” modules proof?

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!)

• Which hypotheses do you think are superfluous? Is there one you didn’t use? – Randall Aug 7 '18 at 17:26
• 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.) – rwbogl Aug 7 '18 at 17:28
• 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? – hmakholm left over Monica Aug 7 '18 at 17:37
• @rwbogl what if the module wasn’t trivial but the ring action was? – Randall Aug 7 '18 at 17:38
• @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. – rwbogl Aug 7 '18 at 17:43

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 $rm\neq 0$.
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 "$R^2=R$."
• @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. – Randall Aug 7 '18 at 18:32
• @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. – rschwieb Aug 7 '18 at 20:09