[proofplan] We use the canonical quotient map from $M$ to $M/N$ to compare submodules of the quotient with submodules of $M$ that contain $N$. The key point is that a submodule $K \le M/N$ pulls back to a submodule $\pi^{-1}(K)$ of $M$ containing $N$, and a submodule $L$ of $M$ containing $N$ pushes forward to $L/N \le M/N$. Under this correspondence, the zero submodule of $M/N$ corresponds to $N$, and the whole module $M/N$ corresponds to $M$. Thus maximality of $N$ is exactly the assertion that the quotient has no nonzero proper submodules. [/proofplan]

[step:Construct the correspondence between submodules above $N$ and submodules of $M/N$]By [citetheorem:8502], $M/N$ is a left $R$-module. Let $\pi: M \to M/N$ denote the canonical quotient homomorphism, defined by $\pi(m)=m+N$ for every $m \in M$. Let $\mathcal{S}$ denote the set of submodules $L \le M$ satisfying $N \le L$, and let $\mathcal{T}$ denote the set of submodules $K \le M/N$. Define a map \begin{align*} \Phi: \mathcal{S} \to \mathcal{T} \end{align*} by $\Phi(L)=L/N=\{l+N:l \in L\}$ for every $L \in \mathcal{S}$. Define a map \begin{align*} \Psi: \mathcal{T} \to \mathcal{S} \end{align*} by $\Psi(K)=\pi^{-1}(K)=\{m \in M:\pi(m)\in K\}$ for every $K \in \mathcal{T}$. For $L \in \mathcal{S}$, the set $L/N$ is a submodule of $M/N$ because it is closed under subtraction and scalar multiplication inherited from $M/N$: if $x+N,y+N \in L/N$ and $r \in R$, with $x,y \in L$, then \begin{align*} (x+N)-(y+N)=(x-y)+N \in L/N \end{align*} and \begin{align*} r(x+N)=rx+N \in L/N, \end{align*} since $L \le M$. Thus $\Phi$ is well-defined. For $K \in \mathcal{T}$, the preimage $\pi^{-1}(K)$ is a submodule of $M$ because $\pi$ is $R$-linear and $K$ is a submodule of $M/N$. Also $N \subset \pi^{-1}(K)$, since $\pi(n)=0+N$ for every $n \in N$ and $0+N \in K$. Hence $\Psi$ is well-defined. Now let $L \in \mathcal{S}$. For $m \in M$, \begin{align*} m \in \Psi(\Phi(L)) \iff m+N \in L/N. \end{align*} The condition $m+N \in L/N$ means that there exists $l \in L$ such that $m+N=l+N$, equivalently $m-l \in N$. Since $N \le L$, this implies $m=(m-l)+l \in L$. The reverse implication follows by taking $l=m$. Hence $\Psi(\Phi(L))=L$. Let $K \in \mathcal{T}$. For $m+N \in M/N$, \begin{align*} m+N \in \Phi(\Psi(K)) \iff m \in \pi^{-1}(K) \iff m+N \in K. \end{align*} Therefore $\Phi(\Psi(K))=K$. Thus $\Phi$ and $\Psi$ are inverse bijections. Under this bijection, $N \in \mathcal{S}$ corresponds to the zero submodule $\{0+N\}$ of $M/N$, and $M \in \mathcal{S}$ corresponds to $M/N$.[/step]

[guided]The quotient map is the mechanism that records exactly how much information remains after identifying all elements of $N$ with zero. By [citetheorem:8502], $M/N$ is a left $R$-module, so it makes sense to discuss its submodules. Let $\pi: M \to M/N$ be the canonical quotient homomorphism, defined by $\pi(m)=m+N$ for every $m \in M$. We compare two collections. Let $\mathcal{S}$ be the set of all submodules $L \le M$ with $N \le L$, and let $\mathcal{T}$ be the set of all submodules $K \le M/N$. If $L \in \mathcal{S}$, define \begin{align*} \Phi(L)=L/N=\{l+N:l \in L\}. \end{align*} This is a submodule of $M/N$: if $x+N,y+N \in L/N$ and $r \in R$, then $x,y \in L$, and since $L$ is a submodule, \begin{align*} (x+N)-(y+N)=(x-y)+N \in L/N \end{align*} and \begin{align*} r(x+N)=rx+N \in L/N. \end{align*} Thus $\Phi$ is a well-defined map from $\mathcal{S}$ to $\mathcal{T}$. Conversely, if $K \in \mathcal{T}$, define \begin{align*} \Psi(K)=\pi^{-1}(K)=\{m \in M:\pi(m)\in K\}. \end{align*} Because $\pi$ is $R$-linear and $K$ is a submodule, the preimage $\pi^{-1}(K)$ is closed under subtraction and scalar multiplication, so it is a submodule of $M$. It contains $N$ because for every $n \in N$ we have $\pi(n)=0+N$, and the zero element $0+N$ belongs to every submodule $K \le M/N$. Hence $\Psi$ is a well-defined map from $\mathcal{T}$ to $\mathcal{S}$. Now we check that these two constructions undo each other. Let $L \in \mathcal{S}$ and let $m \in M$. Then \begin{align*} m \in \Psi(\Phi(L)) \iff m+N \in L/N. \end{align*} The condition $m+N \in L/N$ says that $m+N=l+N$ for some $l \in L$. Equality of cosets means $m-l \in N$. Since $N \le L$, both $m-l$ and $l$ lie in $L$, hence $m=(m-l)+l \in L$. Conversely, if $m \in L$, then $m+N \in L/N$, so $m \in \Psi(\Phi(L))$. Therefore $\Psi(\Phi(L))=L$. Similarly, let $K \in \mathcal{T}$. For a coset $m+N \in M/N$, \begin{align*} m+N \in \Phi(\Psi(K)) \iff m \in \pi^{-1}(K) \iff m+N \in K. \end{align*} Thus $\Phi(\Psi(K))=K$. The correspondence is therefore a bijection. It sends $N$ to the zero submodule $\{0+N\}$, because $N/N=\{0+N\}$, and it sends $M$ to the whole quotient $M/N$.[/guided]

[step:Use maximality of $N$ to prove simplicity of $M/N$] Assume that $N$ is a maximal submodule of $M$. Since $N$ is proper, choose $m_0 \in M \setminus N$. Then $m_0+N \ne 0+N$, so $M/N$ is nonzero. Let $K \le M/N$ be a submodule. By the correspondence constructed above, $\Psi(K)=\pi^{-1}(K)$ is a submodule of $M$ satisfying $N \le \Psi(K) \le M$. Since $N$ is maximal among proper submodules of $M$, either $\Psi(K)=N$ or $\Psi(K)=M$. Applying $\Phi$ gives either \begin{align*} K=\Phi(N)=\{0+N\} \end{align*} or \begin{align*} K=\Phi(M)=M/N. \end{align*} Thus the only submodules of $M/N$ are $\{0+N\}$ and $M/N$, and $M/N$ is nonzero. Therefore $M/N$ is simple. [/step]

[step:Use simplicity of $M/N$ to prove maximality of $N$] Assume that $M/N$ is simple. Let $L \le M$ be a submodule satisfying $N \le L \le M$. By the correspondence, $\Phi(L)=L/N$ is a submodule of $M/N$. Since $M/N$ is simple, either \begin{align*} L/N=\{0+N\} \end{align*} or \begin{align*} L/N=M/N. \end{align*} If $L/N=\{0+N\}$, then $\Psi(L/N)=\Psi(\{0+N\})=N$, and the inverse correspondence gives $L=N$. If $L/N=M/N$, then $\Psi(L/N)=\Psi(M/N)=M$, and the inverse correspondence gives $L=M$. Hence there is no submodule strictly between $N$ and $M$. Since $N$ is proper by hypothesis, $N$ is a maximal submodule of $M$. Combining the two implications proves that $N$ is maximal if and only if $M/N$ is simple. [/step]