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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?

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closed as off-topic by Strants, Mostafa Ayaz, José Carlos Santos, Xander Henderson, Taroccoesbrocco Jun 27 '18 at 5:59

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    $\begingroup$ write down the image of generators of $N$, see what happens $\endgroup$ – chí trung châu Jun 26 '18 at 15:45
  • $\begingroup$ @AlexFrancisco Thanks, your comment forced me to write the idea in mind rigorously. I basically wanted to make sure there isn't some sort of flaw in that. $\endgroup$ – m_ath Jun 26 '18 at 16:02
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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 $\{\alpha_i\}_{i\in \Lambda}$

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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 $\beta_j$.

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