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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?

EinStone
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4 Answers4

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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 $N=0$

Lukas Heger
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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.

Qiaochu Yuan
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  • Do you know whether it is true in case $M$ is finitely generated? It seems to me that is should be true (using a minimal generating systen), but I am unable to write down a proof. – Severin Schraven Dec 13 '20 at 18:44
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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.

Severin Schraven
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None. For example, let $M = \oplus_{n\in\mathbb N}\mathbb Z$ and $N = \mathbb Z$.

Hydrogen
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  • Thanks for this. What if we add the assumption that M is finitely generated? – EinStone Dec 13 '20 at 17:35
  • The answer is no in general. For noncommutative ring, the invariance of basis number theorem fails to hold. There are some funny noncommutative ring $R$ such that $R = R\oplus R$. – Hydrogen Dec 13 '20 at 17:55
  • @Hydrogen Surely it is false in the noncommutative setting. Do you also know an example in the commutative setting where $M$ is finitely generated? I cannot come up with one and would be curious to see one :) – Severin Schraven Dec 13 '20 at 18:08
  • @SeverinSchraven Okay, I think I overlook the condition 'commutative'. For commutative case, I have no idea.... – Hydrogen Dec 13 '20 at 18:13
  • @Hydrogen: In case something comes to your mind, I'd be really eager to learn about it! – Severin Schraven Dec 13 '20 at 18:16