Does $M \oplus N \cong M$ imply that $N = 0$ for modules $M,N$? Let $R$ be a commutative ring and $M,N$ be $R$-modules. Does $M \oplus N \cong M$ imply that $N = 0$?
I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true? And if yes, how would the first isomorphism induce the second one? Sorry if this question has been answered before, I could not find an answer.
Edit: And what if $M$ is assumed to be finitely generated?
 A: Assume that $M$ is finitely generated. Let $f:M \xrightarrow{\cong} M \oplus N$ be an isomorphism and let $p:M \oplus N \to M$ be the projection on the first factor.
Now $p \circ f:M \to M$ is a surjective endomorphism. It is hence an isomorphism, see here. Now as $p\circ f$ and $f$ are both isomorphisms, so is $p = (p \circ f) \circ f^{-1}$. But the kernel of $p$ is $N$, so $N=0$
A: Nope, that is in general not true. Not even if both of them are free and over a field. Consider $M=\mathbb{R}^\mathbb{N}=N$.
Added:
In case you also assume that $M$ is finitely generated, then it becomes true if you have no torsion and $R$ is an integral ring. We can define $n(M)=:L$ to be the maximal cardinality of a linearly indepedent (over $R$) family in $M$ ($L$ is finite as $M$ is finitely generated and $R$ an integral ring). One readily checks that this number is preserved under isomorphism. However, if $N$ was different from $\{0\}$, then we can take $((x_1,0), \dots, (x_L,0), (0,s))$ with $s\neq 0$. As we have no torsion, we get that $n(M)=n(M\oplus N)\geq n(M)+1$, which gives you a contradiction.
A: 
I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true?

As you've already seen from the other answers, this argument doesn't work. Don't let the notation fool you too much; if you just have some arbitrary isomorphism $\varphi : M \oplus N \cong M$, then when you write "$M/M$" what you really mean is $\varphi^{-1}(M)/M$, or $M/\varphi(M)$. There are two different copies of $M$ here which the notation is fooling you into identifying! Your argument requires the much stronger condition that the isomorphism $M \oplus N \cong M$ is the natural projection $M \oplus N \to M$, so that the two copies of $M$ involved are the same. This is a good example of the difference between a natural isomorphism and an "unnatural" isomorphism.
A: None. For example, let $M = \oplus_{n\in\mathbb N}\mathbb Z$ and $N = \mathbb Z$.
