Let $R$ be a Dedekind domain and $A, B$ be finitely generated projective $M\_n(R)$-modules. Is it true that


$A\oplus M\_n(R)\cong B\oplus M\_n(R)\:\:\Rightarrow\:\:A\cong B$?


Here, the isomorphism is over $M\_n(R)$. Here are my thoughts so far:


If $P(R)$ is the set of isomorphism classes of finitely generated projective $R$-modules, then it is well known that $P(R)$ is a commutative monoid under the direct sum, i.e. $(A, B)\mapsto A\oplus B$. In fact, as $A\oplus B$ is always free for some projective module $B$, $P(R)$ is a group known as the projective class group $K\_0(R)$ but I'll stick with monoids.


Next, by Morita's Theorem we have $P(M\_n(R))\cong P(R)$.


Finally, as $R$ is a Dedekind domain, $P(M\_n(R))\cong P(R)\cong \mathbb{N}\oplus \widetilde{K\_0}(R)$.


So, if $\widetilde{K\_0}(R)=0$ then $P(R)\cong\mathbb{N}$ and we are done. What about when $\widetilde{K\_0}(R)\neq 0$? My feeling is the answer is still yes. As $M\_n(R)$ is a free $M\_n(R)$-module, then is represents the identity of $\widetilde{K}\_0(R)$. So really the only 'place' $A, B$ could come from where $A\not\cong B$ is from the $\mathbb{N}$, but this is generated by a single element, so they $A\cong B$. This is very hand-wavy, however.



Does $A\oplus M_n(R)\cong B\oplus M_n(R)$ imply $A\cong B$? $R$ Dedekind domain
Under Morita equivalence, $M\_n(R)$ is associated to $R^n$, and your question is equivalent to: For $A, B$ finitely generated projective $R$-modules, does 
$$A \oplus R^n \cong B \oplus R^n\quad \Rightarrow\quad A \cong B?$$
Then you can simply use the classification of finitely generated $R$-modules, and the answer is yes.


