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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 $1_R$

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    $\begingroup$ 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 $N=R$. $\endgroup$ – Berci Nov 17 at 2:39
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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.

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

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  • $\begingroup$ 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. $\endgroup$ – childishsadbino Nov 17 at 2:47

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