Finitely generated submodule of a module

Question

Let $M$ be an (infinitely generated) free module over a ring $R$, with generating set(not basis) $\{\alpha_i\}_{i \in \Lambda}$ . Let $N$ be a finitely generated $R$-module such that $f: N \rightarrow M$ is an injective module homomorphism. Can we say that there are finitely many $\{\alpha_i\}$ such that image of $N$ is contained in the submodule generated by them?

Answer 1

Sure, if $\{\beta_j\}_{j=1}^n$ is the finite set that generates $f(N)$ (this set there exists because $f$ restrict to image is an isomorphism and $N$ is finitely generated) than $\beta_j=\sum_{k=1}^{n_j}a_{jk}\alpha_{kj}$ for each $j\in \{1,...,n\}$ and so $f(N)$ is contained by the module generated by $\{\alpha_{kj}: k=1,...n_j, j=1,...n\}$ that is a finite set of $\{\alpha_i\}_{i\in \Lambda}$

Answer 2

Take $\{\beta_j\}$ a finite generating set for $N$. Since $N\subseteq M$, each $\beta_j$ can be written in terms of finitely many $\alpha_i$. Take the collection of all those $\alpha_i$ appearing in the expressions of the $\beta_j$.