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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' U 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?

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  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Open Season Oct 7 '13 at 22:06
  • $\begingroup$ How far did you get? Along the lines of the hint, I mean. $\endgroup$ – rschwieb Oct 8 '13 at 12:37
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    $\begingroup$ I posted what I have so far below $\endgroup$ – Open Season Oct 8 '13 at 22:11
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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, $M’ ∩ M’’ = 0$.

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