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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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Is it $\Bbb Z^{n^2}$? –  anon Aug 16 '12 at 10:09
When you wrote "directed sum", did you mean "direct sum"? –  Matt N. Aug 16 '12 at 10:10
Yes, thanks Matt –  zacarias Aug 16 '12 at 10:18

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up vote 1 down vote accepted

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.

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