[proofplan] We prove both directions by translating between finite generating sets and maps out of finite free modules. Given generators $m_1,\ldots,m_k$, define the unique $R$-[linear map](/page/Linear%20Map) from $R^k$ sending the standard basis vector $e_i$ to $m_i$, and use the generating property to prove surjectivity. Conversely, a surjection $\pi: R^k \to M$ sends the standard basis of $R^k$ to a finite generating set of $M$. The case $k=0$ is handled by the convention that $R^0$ is the zero module and that the zero module is generated by the empty set. [/proofplan] [step:Construct a surjection from a finite generating set] Assume that $M$ is finitely generated. Choose a finite generating set $m_1,\ldots,m_k \in M$, where $k \geq 0$. If $k=0$, then the empty set generates $M$, so $M=0$. Let \begin{align*} \pi: R^0 \to M \end{align*} be the unique $R$-[module homomorphism](/page/Module%20Homomorphism) from the zero module to $M$. Since $M=0$, this map is surjective. Assume now that $k \geq 1$. For each $i \in \{1,\ldots,k\}$, let $e_i \in R^k$ denote the standard basis vector whose $i$-th component is $1_R$ and whose other components are $0_R$. Define \begin{align*} \pi: R^k \to M,\qquad \pi((r_1,\ldots,r_k))=\sum_{i=1}^{k} r_i m_i. \end{align*} This is an $R$-module homomorphism: for $(r_1,\ldots,r_k),(s_1,\ldots,s_k)\in R^k$ and $a\in R$, \begin{align*} \pi((r_1,\ldots,r_k)+(s_1,\ldots,s_k))=\sum_{i=1}^{k}(r_i+s_i)m_i=\sum_{i=1}^{k}r_i m_i+\sum_{i=1}^{k}s_i m_i=\pi((r_1,\ldots,r_k))+\pi((s_1,\ldots,s_k)). \end{align*} Also, \begin{align*} \pi(a(r_1,\ldots,r_k))=\sum_{i=1}^{k}(ar_i)m_i=a\sum_{i=1}^{k}r_i m_i=a\,\pi((r_1,\ldots,r_k)). \end{align*} Since $m_1,\ldots,m_k$ generate $M$, every $m\in M$ has the form \begin{align*} m=\sum_{i=1}^{k} r_i m_i \end{align*} for some $r_1,\ldots,r_k\in R$. Thus $m=\pi((r_1,\ldots,r_k))$, so $\pi$ is surjective. [/step] [step:Recover generators from a finite free surjection] Conversely, suppose there exist an integer $k\geq 0$ and a surjective $R$-module homomorphism \begin{align*} \pi: R^k \to M. \end{align*} If $k=0$, then $R^0=0$, and surjectivity of $\pi$ implies $M=0$. Hence $M$ is generated by the empty set, so $M$ is finitely generated. Assume now that $k\geq 1$. For each $i\in\{1,\ldots,k\}$, let $e_i\in R^k$ denote the standard basis vector whose $i$-th component is $1_R$ and whose other components are $0_R$. We claim that the finite set \begin{align*} \pi(e_1),\ldots,\pi(e_k) \end{align*} generates $M$ as a left $R$-module. Let $m\in M$. Since $\pi$ is surjective, there exists $(r_1,\ldots,r_k)\in R^k$ such that \begin{align*} \pi((r_1,\ldots,r_k))=m. \end{align*} In the free module $R^k$, we have \begin{align*} (r_1,\ldots,r_k)=\sum_{i=1}^{k} r_i e_i. \end{align*} Using $R$-linearity of $\pi$, we obtain \begin{align*} m=\pi\left(\sum_{i=1}^{k} r_i e_i\right)=\sum_{i=1}^{k} r_i\pi(e_i). \end{align*} Thus every element of $M$ is an $R$-linear combination of $\pi(e_1),\ldots,\pi(e_k)$, so $M$ is finitely generated. [guided] We start with a surjective $R$-module homomorphism \begin{align*} \pi: R^k \to M. \end{align*} The goal is to extract a finite list of elements of $M$ that generates all of $M$. The natural finite list is obtained by applying $\pi$ to the standard basis vectors of the free module $R^k$. First handle the degenerate case. If $k=0$, then $R^0$ is the zero module. Since $\pi:0\to M$ is surjective, every element of $M$ lies in the image of the zero module. The image contains only $0_M$, so $M=0$. The zero module is generated by the empty set, hence $M$ is finitely generated. Now assume $k\geq 1$. For each $i\in\{1,\ldots,k\}$, let $e_i\in R^k$ be the standard basis vector whose $i$-th component is $1_R$ and whose other components are $0_R$. We will show that \begin{align*} \pi(e_1),\ldots,\pi(e_k) \end{align*} generate $M$. Take an arbitrary element $m\in M$. Because $\pi$ is surjective, there exists a vector $(r_1,\ldots,r_k)\in R^k$ such that \begin{align*} \pi((r_1,\ldots,r_k))=m. \end{align*} The defining property of the standard basis of the free left $R$-module $R^k$ gives the expansion \begin{align*} (r_1,\ldots,r_k)=\sum_{i=1}^{k} r_i e_i. \end{align*} Now apply the $R$-linearity of $\pi$ to this finite sum: \begin{align*} m=\pi((r_1,\ldots,r_k))=\pi\left(\sum_{i=1}^{k} r_i e_i\right)=\sum_{i=1}^{k} r_i\pi(e_i). \end{align*} This expresses the arbitrary element $m$ as an $R$-linear combination of the finite list $\pi(e_1),\ldots,\pi(e_k)$. Therefore that list generates $M$, and $M$ is finitely generated. [/guided] [/step] [step:Combine the two implications] The first step shows that every finitely generated left $R$-module admits a surjective homomorphism from some finite free module $R^k$. The second step shows that the existence of such a surjective homomorphism forces $M$ to be generated by the finite set given by the images of the standard basis vectors, with the case $k=0$ giving the zero module. Hence $M$ is finitely generated if and only if there exist an integer $k\geq 0$ and a surjective $R$-module homomorphism $\pi:R^k\to M$. [/step]