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I need an example of a ring consisting of 2 by 2 matrices where $a^3=a$ with $a$ belonging to this ring. If someone can list the elements I would be satisfied.

What I'm trying to get at it is conceptualize why a ring $R$ is always commmuative when $a^3=a$. I know of one such example and that is the factor ring $\mathbb{Z}/3\mathbb{Z}.$ Does anyone know how to prove this statement mathematically as well as giving me an example of a ring of 2 by 2 matrices?

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    So you want $a^2$ to be something other than the identity. Not sure that's possible... thinking! Brian said you won't get$a$matrix ring with **every** matrix having that property... because it is impossible due to the theorem we've cited!2012-08-21

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There is a theorem due to Jacobson that says if for every $a\in R$ there exists an $n\in\mathbb{N}$ such that $a^n=a$, then $R$ is commutative. (See this, or this for example).

Obviously the identity matrix cubed is itself... is this the sort of thing you're looking for?!

In general matrix rings are going to have a lot of idempotent elements $e$ such that $e^2=e$, and for all of those $e^3=e$ as well.

For an example where $a^2\neq a$, you could use $\begin{bmatrix}0&1\\1&0\end{bmatrix}$.

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    @Shaniqua I listed 3 examples with decreasing triviality.2012-08-21