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Let $R$ be a ring with identity element. Let $M$ be a finitely generated $R$-module. Show that there is a free $R$-module $F$ and a submodule $K\subseteq F$ such that $M\cong F/K$ as $R$-modules.

My first idea is that I know if $F$ is a $R$-module and $K$ a submodule of $F$ then $F/K$ is an $R$-module.

I also feel like I should be playing around with $M/M_{tor}$ and I know for whatever free module $F$ I need a surjective mapping $\phi$ from $F$ to $M$ and then look at $ker(\phi)$. But beyond that I haven't an idea where to begin.

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3 Answers 3

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Let $I = \{[m]\,|\, m\in M\}$ be an index set indexed by the elements of $M$. Consider $F := \bigoplus_{[m]\in I}R.[m]$, the free $R$-module spanned by the symbols $[m]$ for $m\in M$. Then we have a natural $R$-linear map $$ f\colon F\longrightarrow M,\quad \sum_{[m]\in I} \lambda_{[m]}.[m] \longmapsto \sum_{[m]\in I}\lambda_{[m]}\cdot m, $$ where the sums are finite. Obviously, $f$ is surjective. Setting $K:= \ker(f)$, we obtain by the homomorphism theorem for $R$-modules an isomorphism $F/K\cong \operatorname{im}(f) = M$.

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  • $\begingroup$ Why is F free? What am I missing? (oh i get it now! This is a neat proof btw. Thanks!) $\endgroup$
    – user162089
    Commented Feb 28, 2016 at 9:37
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I think the following may help.

Say $\;M=\langle m_1,...,m_n\rangle_R\;$ , with $\;\mathcal X:=\{m_1,...,m_n\}\;$ a minimal generator set for $\;M\;$ as

$\;R $ - module, and let $\;F:=F(m_1,...,m_n)\;$ be the free $\;R $ - module generated by $\;\mathcal X\;$ .

Define the function $\;f:\mathcal X\to M\;,\;\;f(m_i):=m_i\;$ , so by the universal property of free

$\;R $ - modules, there exists a unique $\;R $ - homomorphism $\;\phi: F\to M\;$ extending $\;f\;$.

Clearly $\;\phi\;$ is surjective and by the isomorphisms theorems we get $\;F/\ker\phi\cong M\;$

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  • $\begingroup$ What does $F(m_{1},...,m_{n})$ mean precisely? $\endgroup$
    – user162089
    Commented Feb 28, 2016 at 9:49
  • $\begingroup$ @user162089 What is written there: the free $\;R $ - module generated by $\;m_1,...,m_n\;$ . If you understood the other answer then this should, I think, be clear. $\endgroup$
    – DonAntonio
    Commented Feb 28, 2016 at 9:51
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Consider the free module $R^n$ and denote by $e_1, e_2,\dots,e_n$ the elements in the standard basis, that is, $$ e_k=(0,\dots,0,\underset{\substack{\uparrow\\k}}{1},0,\dots,0) $$ If $M$ is a module and $x_1,x_2,\dots,x_n$ are elements of $M$, there is a unique $R$-module homomorphism $f\colon R^n\to M$ such that $$ f(e_k)=m_k,\quad k=1,2,\dots,n $$ namely $$ f(r_1,r_2,\dots,r_n)=\sum_{k=1}^n r_km_k $$

Can you think to a case where $f$ is surjective, given that $M$ is finitely generated?

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