For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R

What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4?

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- Oct 17th 2009, 03:24 PMknguyen2005Boolean Ring
For the case a^2 = a (a belongs to R), I already showed that a+a = 0 for all a in R

What can you say a ring in which for all a in R, a^3 = a and more generally, for all a in R, a^n = a where n >= 4? - Oct 17th 2009, 05:35 PMNonCommAlg
well, if $\displaystyle a^n = a,$ for all $\displaystyle a \in R,$ then $\displaystyle (2^n -2)a=0, \ \forall a \in R.$ because for any $\displaystyle a \in R: 2^n a = (2a)^n = 2a.$ that's jsut trivial! a non-trivial (even hard probably) question is this:

suppose $\displaystyle R \neq (0)$ is a ring with identity element and $\displaystyle n \geq 3$ is the*smallest*integer for which $\displaystyle a^n = a, \ \forall a \in R.$ what is the*smallest*integer $\displaystyle m$ for which $\displaystyle ma=0, \ \forall a \in R$?

for example if $\displaystyle n=3,$ would it be possible to have $\displaystyle m=2$ or $\displaystyle 3$? - Oct 18th 2009, 03:01 AMknguyen2005
Thanks for a quick reply

If we use the formula http://www.mathhelpforum.com/math-he...e9129494-1.gif

Then for the case n = 2, we have : 2a = 0 implies that a + a = 0. So the formula works

when n = 3, 6a = 0 hence 3a+3a = 0

But when I do the calculation:

Let a in R, a^3 = a ad so (-a)^3 = (-a) by hypothesis

But (-a)^3 = (-a)(-a)(-a) = (a^2)(-a)

Thus, (-a) = (a^2)(-a) hence a^2 = 1.

What am I doing wrong here?

Why did I not get the result 6a = 0 when n = 3?

Thanks again - Oct 18th 2009, 10:08 AMNonCommAlg
- Oct 18th 2009, 01:16 PMknguyen2005
I knew my calculation was wrong, I just want to do another way to get to the answer.

If I dont want to use the formula http://www.mathhelpforum.com/math-he...e9129494-1.gif when n = 3. How am I supposed to get to the answer 6a = 0 ? I mean can we do it another way instead of the formula u gave?

Thank you so much for your valuable time - Oct 18th 2009, 01:19 PMNonCommAlg