[proofplan] Choose a finite generating list for $M$ and pass it through the canonical quotient map. Every element of $M/N$ is a coset $x+N$ for some $x \in M$, and finite generation of $M$ writes $x$ as a finite $R$-linear combination of the chosen generators. Applying the quotient map turns this expression into a finite $R$-linear combination of their cosets, so those cosets generate $M/N$. [/proofplan] [step:Choose a finite generating list for $M$] Since $M$ is finitely generated as a left $R$-module, there exist an integer $k \geq 0$ and elements $m_1, \ldots, m_k \in M$ such that every $x \in M$ can be written in the form \begin{align*} x = \sum_{i=1}^{k} r_i m_i \end{align*} for some $r_1, \ldots, r_k \in R$. When $k=0$, this is the empty sum, so the assertion means $M=\{0\}$. [/step] [step:Pass the generators to the quotient module] Because $N$ is an $R$-submodule of $M$, the quotient $M/N$ is a left $R$-module. Define the quotient map \begin{align*} \pi: M \to M/N,\quad x \mapsto x+N. \end{align*} The map $\pi$ is an $R$-[module homomorphism](/page/Module%20Homomorphism) by the definition of the [quotient module](/page/Quotient%20Module) structure. We claim that the finite list $\pi(m_1), \ldots, \pi(m_k)$ generates $M/N$. [guided] The goal is not to find new generators from scratch, but to use the generators already available in $M$. Since $M/N$ consists of cosets, the natural way to move elements of $M$ into the quotient is the quotient map \begin{align*} \pi: M \to M/N,\quad x \mapsto x+N. \end{align*} This map is well-defined because $N$ is an $R$-submodule, and the quotient module structure is defined precisely so that $\pi$ is an $R$-module homomorphism. We now take the finite generating list $m_1, \ldots, m_k$ of $M$ and consider its image in the quotient: \begin{align*} \pi(m_1), \ldots, \pi(m_k) \in M/N. \end{align*} The reason this should generate $M/N$ is that every element of $M/N$ has the form $\pi(x)$ for some $x \in M$, and every such $x$ is already controlled by the generators of $M$. [/guided] [/step] [step:Show that the quotient is generated by these images] Let $y \in M/N$. By definition of the quotient module, there exists $x \in M$ such that $y=x+N=\pi(x)$. Since $m_1, \ldots, m_k$ generate $M$, choose $r_1, \ldots, r_k \in R$ such that \begin{align*} x = \sum_{i=1}^{k} r_i m_i. \end{align*} Using $R$-linearity of $\pi$, we obtain \begin{align*} y = \pi(x) = \pi\left(\sum_{i=1}^{k} r_i m_i\right) = \sum_{i=1}^{k} r_i \pi(m_i). \end{align*} Thus every element of $M/N$ is an $R$-linear combination of $\pi(m_1), \ldots, \pi(m_k)$. Hence $M/N$ is finitely generated as a left $R$-module. [/step]