Let $R$ be a PID. Show that every finitely generated $R$-module is a direct sum of a torsion module and of a free module.


There's a theorem that claims that if $M$ is a finitely generated $R$-module (where $R$ is a PID) then $\exists R^\times\ni a_1|\dots|a_k\ne0, r>0$ s.t. $$M\cong R ^r\oplus\bigoplus_{i=1}^k R/\langle a_i\rangle$$

$R^r$ is obviously a free module. We want to show that $N:=\bigoplus_{i=1}^k R/\langle a_i\rangle$ is a torsion moodule: Let $A_i:=\langle a_i\rangle$. Let $N\ni n=(r_1+A_1,\dots,r_k+A_k)$.

IF we find $\forall 1\leq i\leq k$ some $l_i\in R$ s.t. $r_il_i\in A_i$ then we finish the proof (because $n\cdot l_1\cdots l_k=0_N\Rightarrow n\in\operatorname{Tor}(N))$.

Let $1\leq i\leq k$. Let $r:=r_i, A:=A_i$ and $a:=a_i$. We want to find $l,t\in R$ s.t $lr=at$. I'm not sure how.


1 Answer 1


You may take a look at Theorem 4.10.


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