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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 $Y$.

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up vote 9 down vote accepted

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).

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While I still can't mark it as solved, I would like to add that my line of though was considering only the generators for the module's additive groups. I forgot that you could send only generators to generators and still not end up with a homomorphism, like your exampled showed me. – Henrique Augusto Souza Feb 27 at 4:06
To add to the first example, if $M$ is a simple module, every homomorphism from $M$ is either trivial or an injection. So there is no non-trivial homomorphism $\mathbb{Z}/p \mathbb{Z} \to \mathbb{Z}/n \mathbb{Z}$ for $p$ prime and $p$ does not divide $n$. – Eric Thoma Feb 27 at 4:10

For commutative rings there is a geometric way to think about these things. Every module $M$ over a commutative ring $R$ has a support, which is the set of prime ideals $P$ such that the localization $M_P$ at $P$ is nonzero. This use of "support" is analogous to the notion of support of a function: it's "where the module is nonzero."

For example, when $R = \mathbb{Z}$ the set of prime ideals consists of the zero ideal $(0)$ and the ideals $(p)$ for $p$ a prime. The support of $\mathbb{Z}/p\mathbb{Z}$ consists only of $(p)$; loosely speaking, this module behaves like a "delta function" which is nonzero only at $p$.

Proposition: if $M$ and $N$ are $R$-modules with disjoint support such that $M$ is finitely presented, then the only homomorphism $M \to N$ is the zero homomorphism.

Proof. We want to show that the hom module $\text{Hom}_R(M, N)$ is zero. This condition is local in the sense that a module is zero iff its localizations are, so it suffices to show that the localizations $\text{Hom}_R(M, N)_P$ are zero. Since $M$ is finitely presented, localization commutes with hom in the sense that

$$\text{Hom}_R(M, N)_P \cong \text{Hom}_{R_P}(M_P, N_P)$$

for all prime ideals $P$. But by hypothesis, $M$ and $N$ have disjoint supports, so for any $P$ either $M_P$ or $N_P$ is zero, and hence so is the localization of the hom module at $P$. $\Box$

In other words, again loosely speaking, a homomorphism $M \to N$ must be zero because it is "zero at every point."

The analogy between support for modules and support for functions is tighter if we take tensor products instead of homs: we can drop the finitely presented hypothesis, because localization always commutes with tensor products, and we get that if $M$ and $N$ are $R$-modules with disjoint support then $M \otimes_R N = 0$, an exact analogue of the observation that two functions with disjoint support multiply to zero.

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