# Boolean Ring

• Oct 17th 2009, 03:24 PM
knguyen2005
Boolean 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 PM
NonCommAlg
Quote:

Originally Posted by knguyen2005
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?

well, if $a^n = a,$ for all $a \in R,$ then $(2^n -2)a=0, \ \forall a \in R.$ because for any $a \in R: 2^n a = (2a)^n = 2a.$ that's jsut trivial! a non-trivial (even hard probably) question is this:

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

for example if $n=3,$ would it be possible to have $m=2$ or $3$?
• Oct 18th 2009, 03:01 AM
knguyen2005

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 AM
NonCommAlg
Quote:

Originally Posted by knguyen2005

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

how did you get that? also we can't cancel $-a$ out because $-a$ is not necessarily a unit in R.
• Oct 18th 2009, 01:16 PM
knguyen2005
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 PM
NonCommAlg
Quote:

Originally Posted by knguyen2005
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

$a+a=(a+a)^3 = (2a)^3=8a^3=8a.$ so $8a=2a$ and thus $6a=0.$