Is there a counterexample for the claim: if $A \oplus B\cong A\oplus C$ then $B\cong C$?

Question

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?

Answer 1

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.

Answer 2

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 $R\oplus R$.

Answer 3

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.

Answer 4

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.