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In Robert Ash's Abstract Algebra book, chapter 4 section 4.6.3 (Fundamental Decomposition Theorem), it states that for any finitely generated module $M$ over PID $R$, there exist ideals such that $M \cong \bigoplus_{i \in N} R/I_i$, where $N=\lbrace1,2,3,...,n\rbrace$. Thus $M$ is a direct sum of cyclic modules.

I don't understand why $R/I_i$ is cyclic module. The definition of cyclic module given in this book is $M=Rx=\lbrace rx:r \in R \rbrace$ where $x$ is the generator of $M$. I don't see the relation between $R/I_i$ and $Rx$. Can anyone guide me?

Remark: Here is the link to the softcopy of the book.

EDIT: In the book, it states that each $R/<a_i>$ is a torsion submodule. Is it possible that direct sum of torsion module is still torsion?

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1 Answer 1

An $R$-module is cyclic if it is spanned by a single (not necessarily unique) element $x\in M$.

Any quotient $R/I$ of $R$ as a module is spanned by $1+I\in R/I$, so these are cyclic.

Moreover if $M=Rx$ ($x\in M$) is cyclic then there is a surjective $R$-module morphism $f:R\to M$ given by $r\mapsto rx$ (not unique if $x$ isn't unique though), so $M\cong R/I$ with $I=\ker f$.


For the edit: suppose $x,y\in M$ are torsion. Then $rx=0=sx$ for some nonzero $r,s\in R$ by definition of torsion. Can you think of something in $R$ that annihilates $x+y$? (Hint: find something that annihilates both $x$ and $y$.) What does this mean for sums of torsion modules?

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