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Let $M$, $N$ be two modules over ring $A$. If $M\oplus M\cong N\oplus N$, can we conclude $M\cong N$? In the case that $M$, $N$ are completely decomposable (e.g. finite-length module by Krull-Schmidt Theorem), it is easy to show this must be true. Does the general case also hold?

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False, even in the locally free case: if $A$ is a Dedekind domain and $I$ is a nonzero ideal of $A$ which is 2-torsion in the class group then $I \oplus I \simeq A \oplus I^2 = A \oplus A$. So if the class group of $A$ is finite with even order then any $I$ representing an element of exact order 2 in the class group provides a counterexample. –  BCnrd May 15 '10 at 5:15
    
@BCnrd: Why don't you put this as an answer. –  Chandrasekhar Feb 3 '11 at 2:50

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up vote 5 down vote accepted

There are even counterexamples in the case $A = {\mathbb Z}$: at the end of B. Jónsson’s paper “On direct decompositions of torsion-free abelian groups,” Math. Scand. 5 (1957), 230–235, an example is given of torsion-free, finite-rank abelian groups $B \not\cong C$ such that $B \oplus B \cong C \oplus C$.

A further counterexample, which I believe has been pointed out independently by L. S. Levy, R. Wiegand, and R. G. Swan: let $A$ be the coordinate ring of the real 2-sphere and ${}_AM$ the module for the tangent bundle; then $M \oplus M$ is free of rank $4,$ but $M$ is not free of rank $2$.

In the positive direction, K. R. Goodearl has proved (“Direct sum properties of quasi-injective modules,” Bull. Amer. Math. Soc. 82 (1976), no. 1, 108–110, Theorem 3) that if $M$ and $N$ are quasi-injective modules over a ring (commutative or not), then $M^n \cong N^n$ implies $M \cong N$ for any positive integer $n$.

Your question is related to an important open problem in noncommutative ring theory, the “separativity” problem for von Neumann regular rings: if $R$ is a von Neumann regular ring (or more generally an exchange ring), and $A$ and $B$ are finitely generated projective left $R$-modules with the property that $A \oplus A \cong A \oplus B \cong B \oplus B$, must we have $A \cong B$?  An affirmative answer would resolve several major open problems, as explained in P. Ara, K. R. Goodearl, K. C. O’Meara, and E. Pardo’s paper “Separative cancellation for projective modules over exchange rings,” Israel J. Math. 105 (1998), 105–137.

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