If the Quotient $M/M'$ of a finitely generated module $M$ is a free module then $M'$ is finitely generated.

Question

I am new to modules and I want to show that "If the Quotient $M/M'$ of a finitely generated module $M$ is a free module then $M'$ is finitely generated". Please help me.

Answer 1

Since $M/M^\prime$ is free, the quotient map $M \to M/M^\prime$ admits a section (by sending the generators of $M/M^\prime$ to one of the elements in their preimages). This implies that $M \cong M^\prime \oplus M/M^\prime$. Since $M$ is finitely generated, $M^\prime$ must also be.

(for a proof of the fact used, see for example here: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/splittingmodules.pdf)

Answer 2

You have an ses $0 \to M'\to M \to M/M' \to 0$ that splits because $M/M'$ is free and hence projective. Thus by the splitting lemma $M \cong M/M' \oplus M'$. Thus upon taking the projection of $M$ onto $M'$ we see that $M' $ is finitely generated.

Answer 3

Find a section of the Short exact sequence $0\rightarrow M'\rightarrow M\rightarrow M/M'$ and deduce that $M$ is isomorphic to the direct sum of $M/M'$ and $M'$. Now use the fact that $M/M'$ is free and $M$ is finitely generated. Hope that helps !