[proofplan] We first fix a basis $B$ of the free module $F$ and push finite $R$-linear combinations through the quotient map $q: F \to F/K$. This proves that the images of the basis elements generate the quotient. Then we compute the kernel of $q$ directly from the definition of equality of cosets, which identifies the relations with $K$. For the converse, we take a generating set for an arbitrary module $M$, build the free module on that set, map each basis vector to the corresponding generator, and identify $M$ with the quotient by the kernel of this surjection. [/proofplan] [step:Push the basis of $F$ through the quotient map] Let $B$ be the given basis of the free left $R$-module $F$. Define \begin{align*} \overline{B} := q(B) = \{q(b) : b \in B\} \subset F/K. \end{align*} We prove that $\overline{B}$ generates $F/K$. Let $y \in F/K$. By the definition of the [quotient module](/page/Quotient%20Module), there exists $x \in F$ such that $y = q(x)$. Since $B$ is a basis of $F$, there are a finite subset $B_x \subset B$ and coefficients $r_b \in R$ for each $b \in B_x$ such that \begin{align*} x = \sum_{b \in B_x} r_b b. \end{align*} Because $q$ is an $R$-[module homomorphism](/page/Module%20Homomorphism), it preserves finite sums and scalar multiplication. Therefore \begin{align*} y = q(x) = q\left(\sum_{b \in B_x} r_b b\right) = \sum_{b \in B_x} r_b q(b). \end{align*} Thus every element of $F/K$ is a finite $R$-linear combination of elements of $\overline{B}$, so $\overline{B}$ generates $F/K$. [guided] Let us unpack why the basis of $F$ automatically gives generators for the quotient. Define \begin{align*} \overline{B} := q(B) = \{q(b) : b \in B\} \subset F/K. \end{align*} To prove that $\overline{B}$ generates $F/K$, we must start with an arbitrary element of the quotient and express it as a finite $R$-linear combination of elements of $\overline{B}$. So let $y \in F/K$. By definition of a quotient module, every element of $F/K$ is a coset of the form $x + K$ for some $x \in F$; equivalently, there exists $x \in F$ such that $y = q(x)$. Now use freeness. Since $B$ is a basis of $F$, the element $x$ has a unique expression as a finite $R$-linear combination of basis elements. Thus there are a finite subset $B_x \subset B$ and coefficients $r_b \in R$ for each $b \in B_x$ such that \begin{align*} x = \sum_{b \in B_x} r_b b. \end{align*} The quotient map $q: F \to F/K$ is an $R$-module homomorphism, so it is additive and respects left scalar multiplication by elements of $R$. Applying $q$ to the displayed finite sum gives \begin{align*} y = q(x) = q\left(\sum_{b \in B_x} r_b b\right) = \sum_{b \in B_x} r_b q(b). \end{align*} This is exactly a finite $R$-linear combination of elements of $\overline{B}$. Since $y \in F/K$ was arbitrary, $\overline{B}$ generates $F/K$. [/guided] [/step] [step:Compute the kernel of the quotient map] We show that $\ker q = K$. By definition, \begin{align*} \ker q = \{x \in F : q(x) = 0_{F/K}\}. \end{align*} The zero element of $F/K$ is the coset $K = 0_F + K$. Hence, for $x \in F$, \begin{align*} q(x) = 0_{F/K} \iff x + K = K \iff x \in K. \end{align*} Therefore $\ker q = K$. Consequently, if a relation among the generators $q(B)$ means an element $x \in F$ whose image under $q$ is zero, then the set of all such relations is \begin{align*} \{x \in F : q(x) = 0_{F/K}\} = \ker q = K. \end{align*} Thus the relations among the chosen quotient generators are precisely the elements of $K$. [/step] [step:Construct a free module mapping onto an arbitrary module] Let $M$ be a left $R$-module. Let $S$ be the underlying set of $M$, and let $P$ be the free left $R$-module with basis $\{e_s : s \in S\}$. Define the $R$-module homomorphism \begin{align*} \varphi: P \to M, \quad \sum_{s \in S_0} r_s e_s \mapsto \sum_{s \in S_0} r_s s, \end{align*} where $S_0 \subset S$ is finite and $r_s \in R$ for each $s \in S_0$. This formula is well-defined because each element of the free module $P$ has a unique finite expansion in the basis $\{e_s : s \in S\}$. The map $\varphi$ is surjective. Indeed, if $m \in M$, then $m \in S$, and \begin{align*} \varphi(e_m) = m. \end{align*} Define \begin{align*} N := \ker \varphi \subset P. \end{align*} Since $\varphi$ is an $R$-module homomorphism, $N$ is an $R$-submodule of $P$. [/step] [step:Identify the arbitrary module with the quotient by the kernel] Define \begin{align*} \widetilde{\varphi}: P/N \to M, \quad p + N \mapsto \varphi(p). \end{align*} This map is well-defined: if $p + N = p' + N$, then $p - p' \in N = \ker \varphi$, so $\varphi(p - p') = 0_M$, and hence $\varphi(p) = \varphi(p')$. The map $\widetilde{\varphi}$ is an $R$-module homomorphism because $\varphi$ is an $R$-module homomorphism. It is surjective because $\varphi$ is surjective. It is injective because if $\widetilde{\varphi}(p + N) = 0_M$, then $\varphi(p) = 0_M$, so $p \in \ker \varphi = N$, and therefore $p + N = N$, the zero element of $P/N$. Thus $\widetilde{\varphi}: P/N \to M$ is an isomorphism of left $R$-modules. Hence every left $R$-module is isomorphic to a quotient of a free left $R$-module. [/step]