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One can easily see that every finitely generated abelian group has finitely many direct summands up to isomorphism.

Now, assume that $R$ is a ring and $M$ is a finitely generated $R$-module.

Does $M$ have finitely many non-trivial direct summands, up to isomorphism?

  • What is a "summand" to you? Any group can be written as a direct sum with one summand: itself. It doesn't have to be finitely generated or abelian for that. Same for modules. – Arthur Mar 08 '18 at 14:31
  • I mean a submodule $N$ of $M$ is called a direct summand of $M$ if there exists a submodule $K$ so that $M=N\oplus K$. –  Mar 08 '18 at 14:43
  • To answer the question I think you really need $M$ to have a unique decomposition into indecomposable summands, up to permutation and isomorphism. That is the Krull-Schmidt theorem for modules, it holds for modules of finite lenght. I suggest you see if the Wikipedia page for the Krull-Schmidt theorem is any help. – Ögmundur Eiriksson Mar 08 '18 at 14:55

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Let $R$ be a Dedekind domain with an infinite classgroup. If $I$ is a nonzero ideal of $R$, then $R^2\cong I\oplus I^{-1}$ as $R$-modules, where $I^{-1}$ is the fractional ideal inverse to $I$. As $R$-modules, $I\cong J$ iff $I$ and $J$ are in the same ideal classes. Therefore there are infinitely many direct summands in $R^2$ which are not isomorphic as modules.

Angina Seng
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