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Suppose $M_n(\mathbb{Z})$ is a matrix ring with integer entries. Prove that $M_n(\mathbb{Z})$ is torsion-free.

My attempt: Let $A \in Tor(M_n(\mathbb{Z}))$. Then there exists a non-zero integer such that $rA=0$, where $o$ here denotes zero matrix. Since $A$ is of integer entries and $r \neq 0$, the only way to obtain zero matrix from this is when $A=0$. Hence, $Tor(M_n(\mathbb{Z})) \subset \lbrace 0 \rbrace$.

Clearly $\lbrace 0\rbrace \subset Tor(M_n(\mathbb{Z}))$. Hence, $M_n(\mathbb{Z})$ is torsion-free.

Is my proof correct?

Remark; Sorry for the confusion made. The question goes like this' Prove that $M_n(\mathbb{Z})$ is torsion-free over the ring $\mathbb{Z}$'

Idonknow
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    Yes it is. In fact this idea shows that $M_n(R)$ is torsion free for any domain $R.$ – Ragib Zaman Sep 14 '13 at 13:28
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    Your question is ambiguous: If you are asking if $M_n(\Bbb{Z})$ is torsion-free as a *module over itself* then certainly not. You can have two non-zero matrices to multiply to give the zero matrix. –  Sep 14 '13 at 13:46
  • it would be better if you can specify the base ring... :) though not a big problem.... –  Sep 14 '13 at 14:07

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Your proof misses the essential argument. You write "the only way to ... is when $A=0$", but why?

Any free module over a domain is torsion-free, and matrices constitute a free module.

Martin Brandenburg
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  • The statement 'Any free module over a domain is torsion-free', can we change the domain to arbitrary ring? Because I know the statement' Any free module is torsion-free'. – Idonknow Sep 15 '13 at 06:02
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    Yes, you are right. But over a general ring the definition of torsion-free is a little bit more complicated, it involves regular elements instead of non-zero elements. But clearly, then every free module is torsion-free. More generally, every ring is torsion-free over itsself, and torsion-free modules are closed under direct sums. – Martin Brandenburg Sep 15 '13 at 11:22