# If two modules are each isomorphic to direct summands of the other, must they be isomorphic?

Let $R$ be a ring. $M,N$ are left $R$-modules such that $M$ is isomorphic to a direct summand of $N$ and $N$ is isomorphic to a direct summand of $M$. Could we get that $M \cong N$? (This holds if $M,N$ are finitely generated. I think it also holds if $M,N$ are semisimple. But I don't have a proof for the general case.)

• Could you write out what you mean by 'direct summand of', mathematically?
– user403337
Aug 29, 2017 at 2:32
• @ChrisCuster For OP's purposes, $M$ is a direct summand of $N$ if there exists an isomorphism $N\cong A\oplus B$ for some $A$ and $B$, where $B\cong M$ as $R$-modules. (This avoids the internal vs. external distinction.)
– anon
Aug 29, 2017 at 2:40

This is not always true. Here's the simplest example I know of. Let $R=\mathbb{Z}$, and let $M$ be the set of all bounded sequences of elements of $\mathbb{Z}[\sqrt{2}]$, which form an abelian group under pointwise addition. It is easy to see that $M\cong M\oplus\mathbb{Z}[\sqrt{2}]\cong M\oplus\mathbb{Z}^2$. Much less obviously, Jeremy Rickard proved on MO that $M\not\cong M\oplus \mathbb{Z}$. Taking $N=M\oplus \mathbb{Z}$, then, $M$ is a direct summand of $N$ and $N$ is a direct summand of $M\oplus\mathbb{Z}^2\cong M$ but $M$ and $N$ are not isomorphic.
There are many other examples of similar phenomena. For instance, by theorem of Corner, there exists an abelian group $M$ such that $M\cong M^3$ but $M\not\cong M^2$. Taking $N=M^2$, we then have that $M$ is a direct summand of $N$ and $N$ is a direct summand of $M^3\cong M$, but $M$ and $N$ are not isomorphic. For references and some general discussion of results like these, see the answers to this question on MO.
• Thank you for the counterexample. Could we get the result under the condition that $M,N$ are semisimple? Aug 29, 2017 at 7:55
• Yes, because then $M$ and $N$ are direct sums of simple modules and $M$ is a direct summand of $N$ iff each isomorphism class of simple module appears no more times in $M$ than in $N$ (this takes some work to prove). It then follows by Schroder-Bernstein that $M$ and $N$ have the same number of copies of each simple module. Aug 29, 2017 at 15:02