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?

  • $\begingroup$ It looks like you want a proof of the theorem attributed to Jacobson (too long to prove here, I think) and an example of a 2x2 matrix $a$ which satisfies $a^3=a$? Is this correct? $\endgroup$ – rschwieb Aug 21 '12 at 16:18
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    $\begingroup$ You’re not going to get a $2\times2$ matrix ring over $\Bbb R$ in which every element satisfies $a^3=a$, if that’s what you want. $\endgroup$ – Brian M. Scott Aug 21 '12 at 16:19
  • $\begingroup$ If your matrix is invertible it satisfies $a^2=I$ or alternatively $a=a^{-1}$. If not, then it is singular ... the options are limited. $\endgroup$ – Mark Bennet Aug 21 '12 at 16:21
  • $\begingroup$ Yes, but I want a non-trivial example of a 2x2 matrix if possible. I don't know of a matrix $a$ where $a^2$ gives the identity matrix. An intuitive proof would be nice but not necessary $\endgroup$ – Student Aug 21 '12 at 16:21
  • $\begingroup$ The only non-trivial example is the one with the identity matrix. Brian points out that I can't get a 2x2 matrix if that is the case then I won't press any further $\endgroup$ – Student Aug 21 '12 at 16:26

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 $\begin{bmatrix}0&1\\1&0\end{bmatrix}$.

  • $\begingroup$ I didn't see that, thanks but is there a nontrivial example? Sorry if I'm putting you on the spot someone else can respond $\endgroup$ – Student Aug 21 '12 at 16:15
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    $\begingroup$ @Shaniqua I listed 3 examples with decreasing triviality. $\endgroup$ – rschwieb Aug 21 '12 at 16:19

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