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Here $A$, $B$ and $C$ are $R$-modules. Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$? And what if $B$ and $C$ are finitely generated?

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  • $\begingroup$ This question gets asked from time to time, but it looks like the duplicates are hard to track. This one also adds the question about finite generation, so maybe others will become duplicates of this. $\endgroup$
    – rschwieb
    Apr 23 '15 at 18:49
  • $\begingroup$ Related: math.stackexchange.com/questions/1075709/… $\endgroup$
    – egreg
    Apr 26 '15 at 0:08
  • $\begingroup$ Suppose $R, A, B, C$ are all finite? What then? $\endgroup$
    – GEdgar
    Apr 26 '15 at 0:33
  • $\begingroup$ I assume $R$ is a ring. You are correct that the OP didn't say. All the answers so far are infinite. $\endgroup$
    – GEdgar
    Apr 26 '15 at 0:38
  • $\begingroup$ Why on earth would anyone vote this down? Do you really want everyone to 'show effort'? Isn't this kind of questions valuable in itself? $\endgroup$ Jun 9 '15 at 12:50
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Here's another one. Let $R$ be a ring which has any non-free finite projective module $A$ (where here $A$ finite projective means there exists $B$ with $A \oplus B\cong R^k$ for a finite $k$). Then if we denote $R^\infty$ the countable direct sum of copies of $R$, we have $A\oplus R^{\infty}\cong R \oplus R^{\infty}\cong R^\infty$ by the Eilenberg-Mazur swindle. In particular, $A \oplus R^\infty\cong A \oplus (B\oplus A) \oplus (B \oplus A) \ldots \cong (A\oplus B) \oplus (A \oplus B) \ldots \cong R^{\infty}$

For a simple example of such an $R$, let $k$ be a field and let $R=k\times k$ and $A=k$, where the module action is just multiplication from the first coordinate.

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Let $R$ be the ring of smooth functions on $S^2$. Then $R^3 \cong \operatorname{Vect}(S^2)\oplus R$ where $\operatorname{Vect}(S^2)$ is the $R$-module of smooth vector fields on $S^2$. $\operatorname{Vect}(S^2)$ is not isomorphic to $R^2$ by the hairy-ball theorem, and the standard smooth embedding $S^2\rightarrow \Bbb R^3$ gives the first isomorphism.

Also note the splitting $R^3 \cong \operatorname{Vect}(S^2)\oplus R$ gives a surjective map from $R^3$ to $\operatorname{Vect}(S^2)$, so $\operatorname{Vect}(S^2)$ is finitely generated.

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Without using any topology, you can take $R$ the ring of endomorphisms of an infinite-dimensional vector space $V$, say over $\mathbb{R}$. Then as a (left) module over itself, $R=R\oplus 0$ is isomorphic to $R\oplus R$.

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Let $R=\mathbb Z$, and $A=\mathbb Z\oplus\mathbb Z\oplus\cdots$. Now set $B=\mathbb Z$ and $C=0$.

Moreover, if you also want $A$ finitely generated, then there still are counterexamples, but maybe not that easily found; see e.g. here.

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