# Computing generators for a finitely generated module

I came across this problem yesterday: Let $R$ be a ring and $M$ an $R-$module. $\varphi:R^n\to M$ is a surjective $R-$module homomorphism if and only if $M$ is finitely generated.

Given the set of generators of $M$ as $\{a_i\}_{i=1}^n$, I was able to show that the map $\varphi(r_1,\ldots,r_n)=\sum_{i=1}^nr_ia_i$ defines a surjective $R-$module homomorphism from $R^n$ into $M$.

I am stuck on the converse. Since $\varphi$ is a surjective homomorphism then $\varphi(R^n)=M$. By the first isomorphism theorem we also have $M\cong R^n/\operatorname{ker}\varphi$. Thus $$\varphi(R^n)=M\cong R^n/\operatorname{ker}\varphi.$$

This seems to indicate that there is some way to construct the $n$ generators of $M$ as some sort of preimage in $R^n/\operatorname{ker}\varphi$, but I am not sure how to go about it.

• Maybe I'm completely wrong. But $R^{n}$ is generated as a $R$ Module by the injections of $1 \in R$ into the direct sum $R^n$. Why don't you use the image of that generating set? – Abellan Feb 6 '16 at 14:49
• That was the trick. I don't know what I was thinking before. Thanks for the nudge! – Laars Helenius Feb 6 '16 at 14:59
• Could you post your comment as an answer? That way I can accept it and close out this question. – Laars Helenius Feb 6 '16 at 14:59
• There you have it. – Abellan Feb 6 '16 at 15:00

If $R$ is a ring then clearly is generated as a $R$ Module by $1\in R$. Considering $R^n$ the direct sum of n copies of $R$ we know that there are canonical injections of $i_i: R \hookrightarrow R^n$ sending $r \mapsto (0,0,..r,..)$. We denote $e_i=i_i(1)$ with this is clear that the set $\{e_i\}_{i=1}^n$ generates $R^{n}$
For your question, since you have an epimorphism given $m \in M$ exists some $a \in R^n$ such that $\varphi(a)=m$ but $a=\sum r_i e_i$ so $M$ is generated by $\{ \varphi(e_i) \}$