# Under what conditions does $M \oplus A \cong M \oplus B$ imply $A \cong B$?

This question is fairly general (I'm actually interested in a more specific setting, which I'll mention later), and I've found similar questions/answers on here but they don't seem to answer the following:

Let $R$ be a ring. Are there any simple conditions on $R$-modules $M, A$ and $B$ to ensure that $M \oplus A \cong M \oplus B$ implies $A \cong B$?

This is obviously not true in general: a simple counterexample is given by $M= \bigoplus_{n \in \mathbb{N}} \mathbb{Z}, A = \mathbb{Z}, B = 0$. In the more specific setting that I'm interested in, $R$ is noetherian, each module is finitely generated, reflexive and satisfies $\text{Ext}_R^n(M,R) = 0$ for $n \geqslant 1$ (or replacing $M$ with $A$ or $B$), and $A$ is projective. In this case, do we have the desired result?

• There are rings $R$ where $M\oplus R \cong R^3$ with $M$ not isomorphic to $R^2$ (see math.stackexchange.com/a/1248693/83337), which satisfy all of your assumptions except possibly the reflexivity. – PVAL-inactive Jun 30 '15 at 16:38
• This follows from the Krull-Remak-Schmidt theorem when $M$, $A$ and $B$ are Noetherian and Artinian (I think; I don't remember the exact conditions). – darij grinberg Jun 30 '15 at 16:40
• @PVAL: Hmm, I didn't see that duplicate. But honesty it's the last part which really interests me. – FrancisW Jun 30 '15 at 16:43
• @darijgrinberg: Well, in my case $M$, $A$ and $B$ are all noetherian, but there's certainly no reason for them to be artinian. – FrancisW Jun 30 '15 at 16:45

Are there any simple conditions on $R$-modules $M,A$ and $B$...
The readiest one is that if $R$ has stable range 1 and $M$ is finitely generated and projective, then it cancels from $M\oplus A\cong M\oplus B$. You can find this, for example, in Lam's First course in noncommutative rings theorem 20.13. Examples of rings with stable range 1 include right Artinian rings (and in increasing order of generality, right perfect, semiprimary, semiperfect, and semilocal rings.)
As for conditions on $M_R$, you can say that $M$ cancels if $End(M_R)$ is a ring with stable range 1.