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For some reason this hasn't been asked before and I can't seem to find an example. Here $R$ is a commutative ring.

Find a non-zero $R$-module $M$ such that $M \oplus M \cong M$

soggycornflakes
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  • How about $R^{\infty} \cong R^{\infty} \oplus R^{\infty}$ where $R^{\infty} = \oplus_{\mathbb{Z}} R$? – Alvin Jin Jan 13 '24 at 22:52
  • I thought of that too but was unable to find an isomorphism. But I think I've found one now. Take the map $M \oplus M \to M$ that sends everything in the first copy of $M$ to the 'even' places in $M$ and the second copy to the 'odd' places. – soggycornflakes Jan 13 '24 at 22:59
  • @AnneBauval less so than Alvin's comment. – soggycornflakes Jan 14 '24 at 00:01
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    The construction of an isomorphism: $f:( \oplus_{i=1}^{\infty}R)\oplus (\oplus_{i=1}^{\infty}R)\to \oplus_{i=1}^{\infty}R$ is easy. A construction is: \begin{equation} (r_1,r_2,\cdots)\oplus(s_1,s_2,\cdots)\mapsto (r_1,s_1,r_2,s_2,\cdots). \end{equation} That is just like constructing a bijection from $\mathbb{Z}+\times{0,1}$ to $\mathbb{Z}+$. – Asigan Jan 14 '24 at 06:07
  • @Asigan Yes, that is what I said, isn't it? – soggycornflakes Jan 14 '24 at 14:44
  • @soggycornflakes Ah, yes. Sorry maybe I did not see your comment while replying. – Asigan Jan 14 '24 at 15:37

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