Existence of a sub-module such that a certain isomorphism holds

Question

Let $R$ be a noetherian ring and $M$ be a finitely generated $R$ - Module.

Prove that there exists $n\in \mathbb N$ such that there is a finitely generated $R$ sub-module $N\subset R^n$ such that

$$R^n / N \cong M$$

Any ideas ? Maybe use the isomorphism theorem ?

Answer 1

Let $a_1, \ldots, a_n$ be generators of $M$ and define the (surjective) homomorphism $$\varphi : R^n \to M, \quad x = (x_1, \ldots, x_n) \mapsto \sum_{i = 1}^n x_i a_i.$$ Now $M \cong R^n / N$ with $N = \ker(\varphi)$ by the isomorphism theorems. Since $R$ is noetherian, $R^n$ is noetherian and thus $N$ is finitely generated.

Answer 2

Let $u_1,...,u_n$ generators of $M$, defined $f:R^n\rightarrow M$ by $f(e_i)=u_i$ where $e_1,...,e_n$ is the canonical basis of $R^n$; ($e_1=(1,0,...,0), e_2=(0,1,..),...$) and denote by $N$ the kernel of $f$.