Let $R$ be a commutative ring and let $R^{\omega}$ denote the direct sum of $\omega$ copies of $R$, as a $R$-module.
Is $(R^{\omega})^{\omega}\approx R^{\omega}$, as $R$-modules?
On countable direct sums: do we have $(R^{\omega})^{\omega}\approx R^{\omega}$?
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$\begingroup$
commutative-algebra
modules
free-modules
1 Answers
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Yes.
We know that $(R^{\omega})^{\omega} \cong R^{\omega^2}$ as $R$-modules. But $\omega$ and $\omega^2$ both have countable cardinality, and for any commutative ring with unity, we have $$R^{(I)} \cong R^{(J)} \iff |I|=|J|,$$ where $R^{(X)} = \bigoplus_{x \in X} R$, and $|X|$ denotes the cardinality of $X$. [To see why $\implies$ is true, you can tensor by $R/m$ over $R$ for any fixed maximal ideal $m$ of $R$, and then use that the dimension of isomorphic $R/m$-vector spaces is the same].
In particular, $R^{\omega^2} \cong R^{\omega}$ as $R$-modules.
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0The isomorphism $R^{(I)} \cong R^{(J)}$ is an isomorphism as $R$-modules. – 2017-02-08
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0Notice that $\require{cancel}\cancel{R^n \cong R^m \text{ as rings } \implies n=m}$ is wrong. For instance, take $R = \Bbb Z^{\Bbb N} \cong R^2$ as rings (but not as $R$-modules). – 2017-02-08
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0Moreover, for any ring $S$, let $R = \mathrm{End}_S(S^{(\Bbb N)})$. Then $R^n \cong R^m$ as $R$-modules, for any $n,m \geq 1$ (but here $R$ is not a commutative ring). Indeed, let $M=S^{(\Bbb N)}$. Then $M \cong M \oplus M$ as $S$-modules, so that $R = \mathrm{Hom}_S(M,M) \cong \mathrm{Hom}_S(M,M \oplus M) \cong R \oplus R$ as $R$-modules. – 2017-02-08