1. A tricky ring problem

[Note: when I pressed "preview post," I got the error "LaTeX ERROR: Unknown error" all over, and most of the text didn't appear. However, it seems to be a problem with my own computer because the code is correct. Please let me know if I did something wrong, and how to fix this display problem. Thanks]

This is a seemingly-innocuous problem from my abstract algebra book that turns out to be getting the better of me. Let $R$ be a ring such that $x = x^3$ for all $x \in R$. Then $R$ is a commutative ring.

I've already proven that in a ring where $x = x^2$, we have commutativity. This was done by first proving that $x + x = 0$, and then I had:

$a + b = (a + b)^2 = aa + ab + ba + bb = a + ab + ba + b$

whence

$ab + ba = 0$

But $0 = ab + ab$ by the lemma discussed above, so we had $ab + ba = ab + ab$, and so $ba = ab$.

I figured I would do something similar for the present problem. In fact, I have already shown that $a + a + a = 0$. Proceeding as before, I then evaluated $(a + b)^3$, but this didn't yield anything too helpful. What did work out nicely was to evaluate $(a + b)^3 + (a - b)^3$ which led quickly to a nice result:

$abb + bab + bba = 0$

In fact, it seems like I'm almost done, because if I can only show that

$abb = bab = bba$

then I can prove commutativity from there. It sounds right, and it is right, but regrettably I don't see the proof of the above fact. In spite of the progress I seem to have made, I'm stuck. Any ideas?

2. You need to use tex /tex instead of math tags. They still need the square braces. []

3. Isn't $x^3 = xxx = x$ in this ring? You seem to be showing $x^3 = x+x+x$. This is $3x$, where the '3' is interpreted as "3 summands".

4. see the problem in here.

5. ok, so if x^3 = x, then R is reduced because x^2 = 0 --> x = x^3 = (x^2)x = 0x = 0. therefore (x^2)^2 = x^4 = (x^3)x = x^2, so any square is idempotent, and thus central.

so x+x = (x^2 + x)^2 - x^2 - x^2, and since this is a sum of squares, is central.

finally, x+x^2 = x^2 + x^2 + x^2 + x^2 + x + x + x + x, so x+x+x = -(x^2 + x^2 + x^2), so x+x+x is central, thus x = (x+x+x) - (x+x) is central, thus R is commutative?

(and, yeah, i know, i skipped a couple steps).

6. First Deveno, try to use Latex; it helps. E.g., $x+x = (x^2 + x)^2 - x^2 - x^2$ is $$x+x = (x^2 + x)^2 - x^2 - x^2$$.

Second, you seem to have it, but I would say that $2x \in Z(R)$ instead of "is a sum of squares." The idea is the same, no doubt, but the idea is that Z(R) is a subring and therefore closed under addition. We expressed 2x in terms of elements in Z(R) and therefore 2x must also be in Z(R), i.e., 2x is central. That link was great. I completely forgot how to approach this problem since I haven't done algebra in a few years, but I knew I did it before.

7. i am still getting the hang of latex. sometimes i use it, sometimes i don't. it's slower.

yeah, Z(R) is a subring, which is why a sum of squares is central (subrings are closed under ring addition).

8. Originally Posted by Deveno
ok, so if x^3 = x, then R is reduced because x^2 = 0 --> x = x^3 = (x^2)x = 0x = 0. therefore (x^2)^2 = x^4 = (x^3)x = x^2, so any square is idempotent, and thus central.

so x+x = (x^2 + x)^2 - x^2 - x^2, and since this is a sum of squares, is central.

finally, x+x^2 = x^2 + x^2 + x^2 + x^2 + x + x + x + x, so x+x+x = -(x^2 + x^2 + x^2), so x+x+x is central, thus x = (x+x+x) - (x+x) is central, thus R is commutative?
maybe i should have explained where $x+x^2$ came from. well, i started with expanding $(x+y)^2$ and $(x+y)^3$ and then i realized that with $y=x^2$ we will get what we wanted.
i started with expanding $(x+y)^2$ and $(x+y)^3$ ...