We have $\operatorname{Hom}(R/I,M)\simeq (0:_MI)$ and this implies $\operatorname{Hom}(R/I,R/J)\simeq (J:I)/J$.

As a consequence we get
$$\operatorname{Hom}(\mathbb Z_m,\mathbb Z_n)\simeq\mathbb Z_d,$$ where $d=\operatorname{gcd}(m,n)$

and
$$\operatorname{Hom}_{K[X]}(K[X]/(X^m),K[X]/(X^n))\simeq (X^n):(X^m)/(X^n).$$ We have two cases: $m\ge n$, and in this case $(X^n):(X^m)/(X^n)=K[X]/(X^n)$ or $m<n$ and now we get $(X^n):(X^m)/(X^n)\simeq K[X]/(X^m)$.

Here is another approach.

If $M$ is an $R$-module, and $N$ is a submodule of $M$, when we want to define a homomorphism with domain $M/N$, is sufficient define a homomorphism with domain $M$ whose kernel contains $N$. Explicitly, if $P$ is an $R$-module, and $\alpha \colon M \to P$ is a homomorphism with $N \subseteq \ker \alpha$, then we can define $\widetilde\alpha \colon M/N \to P$ via $\widetilde\alpha(m+N) := \alpha(m)$ for all $m \in M$. Sometimes, this is stated (in a very little enlightening way) as follows:

If $P$ is an $R$-module, then *for every* homomorphism $\alpha \colon M \to P$ with $N \subseteq \ker \alpha$ *there exists a unique* homomorphism $\widetilde\alpha \colon M/N \to P$ such that $\widetilde\alpha \circ \pi = \alpha$, where $\pi \colon M \to M/N$ is the canonical quotient map.

Note that the words "*for every ... there exists a unique*" is a fancy way to say that the function
$$
\begin{align*}
\operatorname{Hom}_R(M/N,P) & \longrightarrow \{\alpha \in \operatorname{Hom}_R(M,P): N \subseteq \ker \alpha\} \\
\beta & \longmapsto \beta \circ \pi
\end{align*}
$$
is bijective.$^1$ Futhermore, one can check that the set in the right is a subgroup of $\operatorname{Hom}_R(M,P)$, and that above function is a group homomorphism, hence a group isomorphism.

One more thing that we need to remember is the group isomorphism $\operatorname{Hom}_R(R,M) \cong M$ that identifies $\alpha \in \operatorname{Hom}_R(R,M)$ with $\alpha(1) \in M$.

As an exercise, I will left to you to think about the following chain of group isomorphisms:
$$
\begin{align*}
\operatorname{Hom}_R(R/I,M) & \cong \{\alpha \in \operatorname{Hom}_R(R,M) : I \subseteq \ker \alpha\} \\
&= \{\alpha \in \operatorname{Hom}_R(R,M): \forall i \in I,\, \alpha(i)=0\} \\
&= \{\alpha \in \operatorname{Hom}_R(R,M): \forall i \in I,\, i\alpha(1)=0\} \\
&\cong \{m \in M : \forall i \in I,\, im=0\} \\
&= \operatorname{Ann}_I(M).
\end{align*}
$$

$^1$ A function $f \colon X \to Y$ is bijective if for every $y \in Y$ there exists a unique $x \in X$ such that $f(x)=y$, isn't?

## Best Answer

In the case that $R$ is a field, $M$ and $F$ vector spaces, one typically shows injectivity and then uses a dimension counting argument to show that this map is an isomorphism.

One way to prove it in this setting would be to pick a basis $\{e_i\}_{i\in I}$ of $F$ (this can be done, as $F$ is free!). Then $\text{Hom}_R(F,R)$ is also free, with basis denoted by $\{e^i\}_{i\in I}$, where the duality pairing is defined by $e^j(e_i) = \delta^j_i$.

Then given a map $f:M \to F$, the inverse of the map $\phi$ can be written as $$ \phi^{-1}(f)(m) = \sum_{i\in I} e^i(f(m))e_i, $$ for $m\in M$, where the sum only has finitely many nonzero terms.