Are mod-$R$ and mod-$M_n(R)$ isomorphic for $n>1$?

Let $R$ be a ring (with 1) and $M_n(R)$ be the ring of $n$ by $n$ matrices with entries in $R$. Let mod-$R$ be the category of right $R$ modules and mod-$M_n(R)$ be the category of right $M_n(R)$ modules.

In Jacobson's Basic Algebra 2, (page 31) it is proved that the above two categories are equivalent. One of the exercises in that section asks whether the two categories are isomorphic.

For $n=1$, they are obviously isomorphic. For $n>1$, I believe that they cannot be isomorphic but I don't see a way to explain this. The examples of isomorphic categories I'm aware of are just the trivial ones, and I'm having a problem showing that the two categories are not isomorphic..

Please Enlighten me.


An equivalence between categories is naturally isomorphic to an isomorphism of categories iff the cardinalities of all the corresponding isomorphism classes of objects in the two categories are equal. That is, all you need to know to promote an equivalence $F$ into an isomorphism is that for each object $X$ in the domain category, there are just as many objects isomorphic to $X$ as there are objects isomorphic to $F(X)$. See this answer of mine for a sketch of the proof and some related discussion.

In this case every object in a module category has a proper class of other objects isomorphic to it and this proper class can in fact be put in bijection with the class $V$ of all sets. So any equivalence between two module categories is naturally isomorphic to an isomorphism.

[Technical note: This argument involves choosing proper classes of representatives, and so requires the axiom of global choice and should be formulated in a set theory with classes like NBG. I don't know whether there is a definable isomorphism of categories in ZFC in your case.]

  • $\begingroup$ Thank you very much! I didn't expect "extending" or "modifying" $F$ to be so easy! $\endgroup$
    – Dilemian
    Apr 8 '17 at 5:15
  • $\begingroup$ I find it kind of surprising that the categories are isomorphic, because I cannot see such an isomorphism. But as you write, it may depend of the choice of set theory ... $\endgroup$
    – HeinrichD
    Apr 10 '17 at 13:03

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