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.

  • $\begingroup$ I’m not sure what you mean by the isomorphism being “over $M_n(R)$”. $A$ and $B$ are not $M_n(R)$-modules. $\endgroup$ – Jeremy Rickard Jan 17 '18 at 15:53
  • $\begingroup$ Sorry, I meant $A,B$ are projective $M_n(R)$-modules $\endgroup$ – Sam Williams Jan 17 '18 at 15:55

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.

  • $\begingroup$ Thanks for the answer, Dustan. Is this not what I was doing above? By using Morita equivalence, the questions is reduced to one over a Dedekind domain. However, I am unsure of the last part of your answer. Why is the answer true over Dedekind domains? From my understanding of the classification of fg projective modules over a DD, these are isomorphic to $I\oplus R^n$ where $I$ is a non-zero fractional ideal. So does this not just imply $I\oplus R^m\cong J\oplus R^m$, where $I, J$ are fractional ideals. Isn't this where $P(R)\cong \mathbb{N}\oplus\widetilde{K_0}(R)$ comes from? $\endgroup$ – Sam Williams Jan 17 '18 at 16:35
  • $\begingroup$ @SamWilliams I'm not sure precisely what $\widetilde{K_0}(R)$ means. Under the classification that you can read off of Wikipedia, $I \oplus R^m \cong J \oplus R^m$ is equivalent to $I \cong J$. More generally, for fractional ideals $I_1, \ldots, I_k$ and $J_1, \ldots, J_k$, we have $I_1 \oplus \cdots \oplus I_k \cong J_1 \oplus \cdots \oplus J_k$ if and only if the products of fractional ideals $I_1\cdots I_k \cong J_1 \cdots J_k$ are isomorphic. $\endgroup$ – Dustan Levenstein Jan 17 '18 at 16:48
  • $\begingroup$ It looks like $\widetilde{K_0}(R)$ is maybe the ideal class group is this context? In which case note that the direct sum of two cancellative monoids is cancellative. $\endgroup$ – Dustan Levenstein Jan 17 '18 at 16:49
  • $\begingroup$ Ah yes, of course. Thank you so much! And yes, in this context it is $\endgroup$ – Sam Williams Jan 17 '18 at 16:55

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