[proofplan] The quotient module $M/N$ has zero element equal to the coset $N=0_M+N$. Therefore an element $m \in M$ lies in $\ker \pi$ precisely when its coset $m+N$ equals $N$. The defining [equivalence relation](/page/Equivalence%20Relation) for quotient modules says that $m+N=N$ exactly when $m \in N$, which gives both inclusions at once. [/proofplan] [step:Identify the zero element of the quotient module] By definition of the quotient module $M/N$, its elements are cosets $m+N$ with $m \in M$, and its zero element is the coset of the zero element $0_M \in M$: \begin{align*} 0_{M/N}=0_M+N=N. \end{align*} Thus, for $m \in M$, the condition $m \in \ker \pi$ is equivalent to $\pi(m)=N$. [/step] [step:Compute the kernel by reducing coset equality to membership in $N$] By definition of the kernel of a module homomorphism, \begin{align*} \ker \pi=\{m \in M : \pi(m)=0_{M/N}\}. \end{align*} Using $\pi(m)=m+N$ and $0_{M/N}=N$, this becomes \begin{align*} \ker \pi=\{m \in M : m+N=N\}. \end{align*} We now show that $m+N=N$ if and only if $m \in N$. If $m+N=N$, then $m=m+0_M \in m+N=N$, so $m \in N$. Conversely, if $m \in N$, then for every $n \in N$ we have $m+n \in N$ because $N$ is a submodule, so $m+N \subset N$. Also, for every $n \in N$, since $-m \in N$, we have \begin{align*} n=m+((-m)+n), \end{align*} with $(-m)+n \in N$, so $n \in m+N$. Hence $N \subset m+N$, and therefore $m+N=N$. It follows that \begin{align*} \ker \pi=\{m \in M : m \in N\}=N. \end{align*} [guided] We want to identify exactly which elements of $M$ are sent to zero by the quotient map. The kernel is defined by \begin{align*} \ker \pi=\{m \in M : \pi(m)=0_{M/N}\}. \end{align*} In the quotient module $M/N$, the zero element is not the element $0_M$ itself, but the zero coset \begin{align*} 0_{M/N}=0_M+N=N. \end{align*} Since the quotient homomorphism is defined by $\pi(m)=m+N$, the kernel condition becomes \begin{align*} m \in \ker \pi \iff m+N=N. \end{align*} It remains to translate the equality of cosets into membership in the submodule. Suppose first that $m+N=N$. Since $0_M \in N$, the element $m=m+0_M$ lies in the coset $m+N$. Therefore $m \in N$. Conversely, suppose $m \in N$. We prove equality of sets. If $x \in m+N$, then $x=m+n$ for some $n \in N$. Since $m \in N$, $n \in N$, and $N$ is closed under addition, we get $x \in N$. Thus $m+N \subset N$. For the reverse inclusion, let $x \in N$. Since $m \in N$ and $N$ is closed under additive inverses, $-m \in N$. Hence $(-m)+x \in N$, and \begin{align*} x=m+((-m)+x). \end{align*} So $x \in m+N$, proving $N \subset m+N$. Therefore $m+N=N$. Combining the two implications gives \begin{align*} m \in \ker \pi \iff m \in N. \end{align*} Hence the two subsets of $M$ are equal: \begin{align*} \ker \pi=N. \end{align*} [/guided] [/step]