[proofplan] Unpack the finite presentation of $M$ as an exact sequence from finite free left $R$-modules ending in $M$. Exactness gives a surjective $R$-[module homomorphism](/page/Module%20Homomorphism) from one of those finite free modules onto $M$. The standard finite generating set of that finite free module maps to a finite generating set of $M$, so $M$ is finitely generated. [/proofplan] [step:Unpack the finite presentation as a finite free presentation] Since $M$ is finitely presented as a left $R$-module, there exist integers $m,n \geq 0$ and $R$-module homomorphisms \begin{align*} \alpha: R^m \to R^n \end{align*} and \begin{align*} \pi: R^n \to M \end{align*} such that the sequence of left $R$-modules \begin{align*} R^m \xrightarrow{\alpha} R^n \xrightarrow{\pi} M \to 0 \end{align*} is exact. Exactness at $M$ says that $\pi$ is surjective. [/step] [step:Push the standard generators of $R^n$ onto $M$] For each $i \in \{1,\ldots,n\}$, let $e_i \in R^n$ denote the element whose $i$-th coordinate is $1_R$ and whose other coordinates are $0_R$. Define the finite subset \begin{align*} G := \{\pi(e_i) : i \in \{1,\ldots,n\}\} \subset M. \end{align*} If $n = 0$, then $R^n = 0$, so the surjectivity of $\pi: 0 \to M$ implies $M = 0$, and the empty subset generates $M$. Assume now that $n \geq 1$. Let $x \in M$. Since $\pi$ is surjective, there exists $v \in R^n$ such that $\pi(v) = x$. Write \begin{align*} v = \sum_{i=1}^{n} r_i e_i \end{align*} for uniquely determined elements $r_1,\ldots,r_n \in R$. Since $\pi$ is $R$-linear and all modules are left $R$-modules, we have \begin{align*} x = \pi(v) = \pi\left(\sum_{i=1}^{n} r_i e_i\right) = \sum_{i=1}^{n} r_i \pi(e_i). \end{align*} Thus every element of $M$ lies in the left $R$-submodule generated by $G$. [guided] The finite presentation gives more structure than we need: it gives both finitely many generators and finitely many relations. For finite generation, only the finite generating side matters. We have an exact sequence \begin{align*} R^m \xrightarrow{\alpha} R^n \xrightarrow{\pi} M \to 0, \end{align*} where $\pi: R^n \to M$ is an $R$-module homomorphism. Exactness at the final term means precisely that $\pi$ is surjective, so every element of $M$ is the image of some element of $R^n$. For each $i \in \{1,\ldots,n\}$, define $e_i \in R^n$ to be the vector with $1_R$ in the $i$-th coordinate and $0_R$ in every other coordinate. These elements generate the finite free left $R$-module $R^n$: if $v \in R^n$, then there are elements $r_1,\ldots,r_n \in R$ such that \begin{align*} v = \sum_{i=1}^{n} r_i e_i. \end{align*} The coefficients appear on the left because $R^n$ is being used as a left $R$-module. Now define \begin{align*} G := \{\pi(e_i) : i \in \{1,\ldots,n\}\} \subset M. \end{align*} This is a finite subset of $M$. To show that it generates $M$, take an arbitrary element $x \in M$. Since $\pi$ is surjective, choose $v \in R^n$ with $\pi(v)=x$. Express $v$ in the standard generators: \begin{align*} v = \sum_{i=1}^{n} r_i e_i. \end{align*} Applying $R$-linearity of $\pi$ gives \begin{align*} x = \pi(v) = \pi\left(\sum_{i=1}^{n} r_i e_i\right) = \sum_{i=1}^{n} r_i \pi(e_i). \end{align*} Thus $x$ is an $R$-linear combination of elements of $G$. Since $x \in M$ was arbitrary, $G$ generates $M$. If $n=0$, the same conclusion is the degenerate case: $R^0=0$, surjectivity of $\pi:0\to M$ forces $M=0$, and the empty set generates the zero module. [/guided] [/step] [step:Conclude finite generation of $M$] The set $G$ is finite and generates $M$ as a left $R$-module. Therefore $M$ is finitely generated. [/step]