[proofplan] Choose a finite generating family for $M$ and apply $f$ to each generator. The key point is that $R$-linearity carries an $R$-linear expression for an element $m \in M$ into the corresponding $R$-linear expression for $f(m)$. Thus every element of $\operatorname{im} f$ is generated by the finite family of images of the chosen generators. The zero-generator case is included by the usual empty-sum convention. [/proofplan] [step:Choose a finite generating family for $M$] Since $M$ is finitely generated, there exist an integer $k \ge 0$ and elements $m_1, \ldots, m_k \in M$ such that every element $m \in M$ can be written in the form \begin{align*} m = \sum_{i=1}^{k} r_i m_i \end{align*} for some coefficients $r_1, \ldots, r_k \in R$. If $k = 0$, this statement means that the only such sum is the empty sum $0_M$, so $M = 0$. Define elements $n_1, \ldots, n_k \in N$ by \begin{align*} n_i := f(m_i) \end{align*} for each $i \in \{1, \ldots, k\}$. The family $n_1, \ldots, n_k$ is finite. [/step] [step:Show that the images of the generators span $\operatorname{im} f$] Let $y \in \operatorname{im} f$. By the definition of image, there exists $m \in M$ such that \begin{align*} y = f(m). \end{align*} Since $m_1, \ldots, m_k$ generate $M$, choose coefficients $r_1, \ldots, r_k \in R$ such that \begin{align*} m = \sum_{i=1}^{k} r_i m_i. \end{align*} Using additivity and compatibility with the left $R$-action for the $R$-[module homomorphism](/page/Module%20Homomorphism) $f$, we obtain \begin{align*} y = f(m) = f\left(\sum_{i=1}^{k} r_i m_i\right) = \sum_{i=1}^{k} r_i f(m_i) = \sum_{i=1}^{k} r_i n_i. \end{align*} Thus every element of $\operatorname{im} f$ lies in the submodule of $N$ generated by $n_1, \ldots, n_k$. [guided] We want to prove that $\operatorname{im} f$ is finitely generated, so we need to exhibit a finite list of elements of $\operatorname{im} f$ and prove that every element of $\operatorname{im} f$ is an $R$-linear combination of that list. The natural finite list is obtained by applying $f$ to the chosen generators of $M$: \begin{align*} n_i := f(m_i) \end{align*} for each $i \in \{1, \ldots, k\}$. Now take an arbitrary element $y \in \operatorname{im} f$. By the definition of the image of a function, there is some $m \in M$ with \begin{align*} y = f(m). \end{align*} Because $m_1, \ldots, m_k$ generate $M$, this particular element $m$ has an expression \begin{align*} m = \sum_{i=1}^{k} r_i m_i \end{align*} for some $r_1, \ldots, r_k \in R$. The reason this expression is useful is that $f$ is an $R$-module homomorphism. That means $f$ is additive and respects scalar multiplication by elements of $R$. Therefore applying $f$ to the displayed expression for $m$ gives \begin{align*} f(m) = f\left(\sum_{i=1}^{k} r_i m_i\right) = \sum_{i=1}^{k} r_i f(m_i). \end{align*} Since $n_i = f(m_i)$ by definition, we get \begin{align*} y = f(m) = \sum_{i=1}^{k} r_i n_i. \end{align*} This proves that the arbitrary element $y \in \operatorname{im} f$ is generated by the finite family $n_1, \ldots, n_k$. [/guided] [/step] [step:Conclude finite generation of the image] Each element $n_i$ lies in $\operatorname{im} f$ because $n_i = f(m_i)$ with $m_i \in M$. The previous step proves that every element of $\operatorname{im} f$ is an $R$-linear combination of $n_1, \ldots, n_k$. Hence \begin{align*} \operatorname{im} f = Rn_1 + \cdots + Rn_k. \end{align*} Therefore $\operatorname{im} f$ is finitely generated as a left $R$-module. If $k = 0$, then $M = 0$, so $\operatorname{im} f = \{0_N\}$, which is generated by the empty family under the empty-sum convention. Thus the conclusion also holds in the zero-generator case. [/step]