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?
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 $M=0$.