A finitely related module is the direct sum of a free module and a finitely presented module

Question

Prove that any finitely related module may be expressed as the direct sum of a finitely presented module and a free module.

Hint: If $M$ is generated by $X = X' \cup X''$, where $X'$ is the finite subset of $X$ involved in the defining relations, then $M$ is the direct sum of :

• the free module over $X''$
• the module generated by $X'$ with the defining relations for $M$

How do I prove that the module generated by $X'$ is finitely presented?

How do I prove that the direct sum of these modules is M?

Answer 1

What I have:

Let $M=\langle X | K \rangle$ where $K = \{r_1,r_2,...,r_t\}$ is a finitely related module. Let $F$ be the free module on $X$. Since $K≤F$, each $r_j∈K$ can be expressed as a linear combination of finitely many elements of $X$. Let $X’ ⊂ X$ denote the finite collection of elements involved in expressing our $r_j$’s as linear combinations of elements of $X$. Let $M'=\langle X' | K \rangle$. Let $G$ be the free module on $X’$. Then $G$ is finitely generated because $X’$ is finite. Since $K$ is finite, $M’$ is finitely presented. Let $X’’ = X \ X’$. Let $M’’$ be the free module over $X’’$. We have $M=∑_{λ∈X}λR=∑_{λ∈X'}λR+∑_{λ∈X''}λR=M’+M’’$. Since $M’$ and $M’’$ have no generators in common, $M’ ∩ M’’ = 0$.