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I am reading a fairly cumbersome proof showing that some homology group $M$ (over $\mathbb{Z}$) is actually zero. In the argument we have a Mayer-Vietoris sequence containing the fragment $$0 \to M \to M \oplus M \to 0,$$ which implies that $M \cong M \oplus M$. Isn't this enough to conclude that $M$ is zero?

More in general: if we have a module (over a commutative ring with $1$) that is isomorphic to the direct sum of two (or more) copies of itself, when can we conclude that it is necessarily zero?

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    Obviously not. For example $R^{\infty} \cong R^{\infty} \oplus R^{\infty}$ (where $R^{\infty} = \bigoplus_{i \in \Bbb{Z}}R$). – 2017-01-19
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    Sorry, the title is not exactly what I wanted. I know of such cases, I would like to know if there are some situations in which it's true. For example, for finitely generated free modules it would be the case. And homology groups are quotients of (in general infinitely generated) free modules. – 2017-01-19
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    I think that under the assumption that $M$ is finitely generated, or Noetherian, then it should be true. – 2017-01-19
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    This is certainly false in the category of modules over any unital ring $R$ that does not have the IBN property (Invariant Basis Number), as in those rings $R^n \cong R^m$ as left $R$-modules for any $m,n \geq 1$. – 2017-01-19

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If $M$ is a finitely generated $R$-module ($R$ commutative with $1$, but I believe not necessarily Noetherian), it should be true.

By localizing at a prime $\mathfrak{p}$, we have $M_\mathfrak{p} \simeq M_\mathfrak{p}\oplus M_\mathfrak{p}$. If one proves that $M_\mathfrak{p}=0$ for all primes $\mathfrak{p} \subseteq R$, then $M=0$. Thus, without loss of generality we may assume that $R$ is local (denote by $\mathfrak{m}$ its max. ideal). Then we know by Nakayama's lemma that $$\mu(M)=\dim_{R/\mathfrak{m}}(M/\mathfrak{m}M),$$ $$\mu(M\oplus M)=\dim_{R/\mathfrak{m}}((M\oplus M)/\mathfrak{m}(M\oplus M))=\dim_{R/\mathfrak{m}}((M/\mathfrak{m}M)\oplus (M/\mathfrak{m}M))=2\mu(M),$$

where $\mu(M)$ denotes the least possible cardinality of a set of generators of $M$. So if $M \simeq M \oplus M,$ we have $2 \mu(M)=\mu(M)$, which implies that $\mu(M)=0$, so $M=0$.