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Every finitely generated module $M$ over a principal ideal domain $R$ is isomorphic to a direct sum of cyclic modules. If $M=M_n(\mathbb Z)$, the matrices of order $n$ over the integers. What is the decomposition in sum of cyclic modules of $M_n(\mathbb Z)$?

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    Yes, thanks Matt2012-08-16

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As was suggested above, the additive group on the $n\times n$ matrices is just a sum of $n^2$ copies of $\mathbb{Z}$, as isn't hard to see: it's torsion-free, in particular. I could imagine you were thinking about how to make the multiplicative structure work, but the theorem only applies to modules, not algebras. The algebra $M_n$ is not a sum of cyclic $\mathbb{Z}$-algebras, since all of the latter are commutative.