I believe this question is a elementary one, and it may have a very simple answer, of which I'm not aware yet.
Given two non-trivial modules over the same non-trivial ring (or two groups, or two rings, whatever..) is it always possible to find a non-trivial homomorphism between them? Not any special type of homomorphism, just a non-trivial one. If not, could you give me a counter-example?
I am thinking if I can make a commutative diagram like this: be $X$, $Y$ and $Z$ non-trivial modules over the same ring, and be $\lambda: Z \rightarrow X$ a module-homomorphism, whose image $\text{Img}(\lambda)$ is a proper submodule of $X$. So the quotient module $X/\text{Img}(\lambda)$ is a non-trivial module. Can I guarantee existence of a non-trivial module-homomorphism between $X/\text{Img}(\lambda)$ and $Y$, my arbitrary module? And by that, I have a induced non-trivial homomorphism between $X$ and $Y$.
Best Answer
A simple counterexample is the ring $\mathbb{Z}$, and the $\mathbb{Z}$-modules $M=\mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}/3\mathbb{Z}$. There is no non-trivial $\mathbb{Z}$-module homomorphism from $M$ to $N$, because there is no element of $N$ that has order $2$. (One could replace $2$ and $3$ with any relatively prime integers.)
Another counterexample with the ring $\mathbb{Z}$ is given by the modules $M=\mathbb{Q}$ and $N=\mathbb{Z}$. There is no subgroup of $N$ that is divisible (in this sense).