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 $1_R$