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In Wikipedia it says that "A $n\times n$ matrix of a ring $R$ is commutative if and only if $n=1$ and $R$ is commutative". Could someone please provide me with a proof/reference to a proof of that ?

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Hint: All you need is to find two $2\times 2$ matrices that do not commute. –  Andrew Dec 8 '12 at 18:49
    
groupprops.subwiki.org/wiki/… –  user27126 Dec 8 '12 at 18:50
    
I cannot find your exact quote there. How is a single matrix commutative? –  Hagen von Eitzen Dec 8 '12 at 19:19

1 Answer 1

All this is saying is that if you have the group of invertible $n\times n$ matrices $G = GL_n(R)$ where the matrices have entries from a ring $R$, then for $G$ to be commutative you need

  1. $R$ to be commutative
  2. $n=1$.

You probably already know that the group of $2\times 2$ matrices isn't commutative even over a (non-trivial) field. And if $n=1$ then $G \simeq R^{\times}$ and so $\dots$

Now, you can of course just consider $M_n(R)$ and say that this as a set is commutative if the elements commute, but then you still have the same thing.

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