Problem.) Let $R$ be a commutative ring with unity, and $0\rightarrow L\rightarrow R^n\rightarrow M\rightarrow0$ an exact sequence. Prove if $M$ is finitely presented, $L$ is finitely generated.
Solution from a book.) Assume $M$ is finitely presented; say $R^l\rightarrow R^m\rightarrow M\rightarrow0$ is a finite presentation. Let $L'$ be the image of $R^l$. Then $L'\bigoplus R^n\cong L\bigoplus R^m$ by Schauel's Lemma. Hence
$L$ is a quotient of $R^l\bigoplus R^n$.
Thus $L$ is finitely generated.
Question.) Shouldn't the line, '$L$ is a quotient of $R^l\bigoplus R^n$' be '$L$ is a quotient of $L'\bigoplus R^n$'? Then since $L'$ is
the image of a finitely generated module($R^l$), $L'$ is finitely
generated and $L$ is finitely generated as $L$ is a quotient of
finitely generated module($L'\bigoplus R^n$). I think $R^l$ is not necessarily isomorphic to $L'(\subset R^m)$.
Best Answer
Since $L'$ is the image of $R^\ell$, it's a quotient of $R^\ell$ by the first isomorphism theorem. It follows that $L$ is a quotient of $R^\ell \oplus R^n$ by the third isomorphism theorem.