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Based upon these two older questions: Show a direct product is not finitely generated. and $R^\mathbb{N}$ is not finitely generated as an $R$-module, I would like to know the answer to the following questions.

  • Given an infinite family $\{M_i\}_{i \in I}$ of non-trivial left $R$-modules (where $R$ is a ring), can $\prod M_i$ be a finitely generated $R$-module ?
  • Given an infinite family $\{M_i\}_{i \in I}$ of non-trivial left $R$-modules (where $R$ is a ring), can $\bigoplus M_i$ be a finitely generated $R$-module ?

It is suggested in the second link to prove that $\prod_{i \in P_n} M_i$ ($P_n$ is a subset of $I$ of cardinality $n$) requires at least $n$ generators.

user26857
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Leyv Zahav
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1 Answers1

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For direct sums, this is easy. By definition, any element of $\bigoplus M_i$ has only finitely many nonzero coordinates. So, if you have finitely many elements, there are only finitely many coordinates on which any of them are nonzero, and the same is true of the entire submodule they generate. So, since infinitely many of the $M_i$ are nontrivial, that submodule cannot be all of $\bigoplus M_i$.

Surprisingly, although direct products are "bigger" than direct sums, it actually is possible for an infinite direct product of nontrivial modules to be finitely generated. For instance, let $V$ be an infinite-dimensional vector space and let $R$ be the ring of endomorphisms of $V$. Then $V$ is a left $R$-module in the obvious way. If you pick a basis $B$ for $V$, then the infinite product $V^B$ is actually isomorphic as an $R$-module to the cyclic module $R$, by sending each map $B\to V$ to its unique extension to a linear map $V\to V$.

For another class of examples (which includes some commutative rings), take any infinite family $(R_i)$ of nonzero rings and let $R=\prod R_i$. Then each $R_i$ can be considered as an $R$-module via the projection map, and the product of these $R$-modules is the cyclic module $R$.

user26857
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Eric Wofsey
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  • Thanks. So, when is $\prod M_i$ necessarily non finitely generated ? – Leyv Zahav Apr 29 '20 at 22:30
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    It is always true over a Noetherian ring, since $\bigoplus M_i$ is a submodule of $\prod M_i$. – Eric Wofsey Apr 29 '20 at 23:20
  • I accidentaly stumbled upon this question, and its first part makes absolutely zero sense (in addition to being false). Consider $CFM(Q)$, ring of column finite matrices over arbitrary ring $Q$. Then $CFM(Q) = End(\bigoplus_{\Bbb N}Q)$, and $\bigoplus_{\Bbb N}Q \cong \bigoplus_{\Bbb N}(\bigoplus_{\Bbb N}Q)$ – xsnl Aug 15 '22 at 21:21
  • @xsnl: $\bigoplus_{\Bbb N}Q \cong \bigoplus_{\Bbb N}(\bigoplus_{\Bbb N}Q)$ as $Q$-modules but not as $CFM(Q)$-modules. – Eric Wofsey Aug 15 '22 at 22:24
  • @EricWofsey ...would you argue with the statement that $CFM$ is isomorphic as a module over itself to $CFM \oplus CFM$? – xsnl Aug 16 '22 at 00:08
  • @xsnl: I would not. I'm not sure what your point is. – Eric Wofsey Aug 16 '22 at 00:50