# Maps from Cyclic R Modules to other R Modules

Let $$M$$ be a cyclic $$R$$ module generated by $$x ∈ M, x\not= 0$$. Let $$N$$ be another $$R$$ module and $$v ∈ N, v\not=0$$.

Prove that if $$R$$ is a field, there exists a module homomorphism $$ϕ : M → N$$ with $$ϕ(x) = v$$.

MY ATTEMPT: I know $$M = Rx$$, so any $$m$$ in $$M$$ can be represented by $$r.x$$, for some $$r$$ contained in the ring. So, if I make a map $$ϕ(c.x) = c.v$$ shouldn't this do the trick?

For $$c=1_R$$, the map will take $$x$$ to $$v$$, and it satisfies the conditions to be a module homomorphism. Where do I need that $$R$$ must be a field?

By the way: the assumptions are that $$R$$ is commutative and contains $$1_R$$

• You need $\phi$ to be well defined: $cx=0\implies cv=0$. If $R$ has an ideal $I$, then $R/I$ is a cyclic $R$-module, and take $N=R$. – Berci Nov 17 at 2:39

In general, there is a bijective correspondence between module homomorphisms $$R/I \to N$$ and elements $$n \in N$$ for which $$I \subseteq \operatorname{Ann}(n)$$ for any ring $$R$$, any ideal $$I$$ of $$R$$, and any $$R$$-module $$N$$.

Now, suppose that $$R$$ is a field and $$M$$ is a nonzero cyclic $$R$$-module generated by $$m$$. Then, the annihilator of $$m \in M$$ is a proper ideal of $$R$$, which must therefore be the zero ideal because $$R$$ is a field. Hence, the homomorphism $$\phi:cx \mapsto cv$$ will always be well-defined.

• Ah, well-definedness was the issue I was missing. Thank you! – childishsadbino Nov 17 at 2:39

The place where you need to use the fact that $$R$$ is a field is in proving that the map you've given is well defined.

For a general ring $$R$$ it might be possible for a given element of $$M$$ to be represented by both $$c_1 x$$ and $$c_2 x$$, with $$c_1 \neq c_2$$. Then if $$c_1 v \neq c_2 v$$ in $$N$$, the map you try to define gives two different answers depending on whether you represent the element as $$c_1 x$$ or $$c_2 x$$, even though they are the same in $$M$$. (Now you know what you are looking for, you should be able to find some examples e.g. using $$M = \mathbb{Z}/2$$ as a cyclic $$\mathbb{Z}$$ module.)

This can't happen if $$R$$ is a field. If $$c_1 x = c_2 x$$, then either $$c_1 = c_2$$ so the representative is the same, or $$c_1 - c_2$$ is invertible and therefore $$x=0$$ contradicting the setup. Therefore if $$R$$ is a field each element of $$M$$ has a unique representation as $$c x$$ and the map is well defined.

• Yes, the example I used is $M= Z_n$ and $R=N=Z$. Then, $M$ is generated by $1_n$, but if I have a homomorphism that takes 1 to 7, say, then it won't be well-defined. – childishsadbino Nov 17 at 2:47