# X^3=x => R is a commutative ring

1. May 2, 2006

### StatusX

If R is a ring, and has the property that x^3=x for all x in R, then show R is commutative. Generalize for x^k=x, k an integer >3.

I can do this for x^2=x. First use x^2+x^2=x+x=(x+x)^2 = 4x^2 (abusing notation a little) which means x^2+x^2=x+x=0, ie, x=-x for all x in R. Then x^2+y^2=x+y=(x+y)^2 = x^2+xy+yx+y^2, which gives xy+yx=0. But since xy=-xy, we have xy=yx, and R is commutative.

I try to generalize using the same basic idea, but there are a lot of extra terms, and it seems impossible to do anything with R and its elements being so general (ie, I can't cancel anything because I don't know if anything is a zero divisor). Can anyone give me a hint?

2. May 2, 2006

### Hurkyl

Staff Emeritus
When you have the identity x² = x, there's a shorter way to do that first part:

4 = 2² = 2

and therefore 2 = 0.

Anyways, why don't you show the rest of what you've done? Maybe you're on the right track and just need a slight nudge.

3. May 2, 2006

### StatusX

Well, if x^3=x, then by the same method we get 6x=0, which isn't very helpful (and this gets even worse as k is increased), and xxy+xyx+xyy+yxx+yxy+yyx=0, which can be factored to (x+y)(xy+yx)+xy^2+yx^2=0. Now, if there were no zero divisors and a 1, we'd have x(x^2-1)=0 => x^2=1, and I could get a little further, but I can't assume that. I'm stuck.

Last edited: May 2, 2006
4. May 2, 2006

### Hurkyl

Staff Emeritus
With that assumption you can't get any further: problem's done! x=0, x=1, or x=-1, and so your ring has at most 3 elements.

(Okay, I guess that is a little further. I just like being dramatic!)

It's certainly information. And it suggests stuff, like maybe trying (x+2y)³, (x+3y)³, (x+4y)³. amd (x+5y)³. Actually, that last one sounds very interesting to me, since it's the same as (x-y)³.

Have you tried any other manipulations with the leftover terms? It strikes me that you can write down interesting identities when you multiply it by stuff, but I don't know how much it helps yet.

5. May 4, 2006

### StatusX

Ok, I got something interesting, but I'm still stuck. Here's what I got:

$$x^3+y^3=(x+y)^3 \Rightarrow x^2y+xyx+yx^2+y^2x+yxy+xy^2=0$$

$$x^3-y^3=(x-y)^3 \Rightarrow x^2y+xyx+yx^2-(y^2x+yxy+xy^2)=0$$

We know 6x=0, and it turns out any other expression like (nx+my)3 does not give anything new when reduced mod 6 (as far as I can tell). Define $\alpha=x^2y+xyx+yx^2$. Now:

$$x \alpha x=x^3yx+x^2yx^2+xyx^3=2xyx+x^2yx^2$$

$$x^2 \alpha x^2=2x^2yx^2+xyx$$

Then since $\alpha=-\alpha$ from above, we have:

$$x^2 \alpha x^2 = x^2 \alpha x^2 -x(\alpha +\alpha)x$$

$$=2x^2yx^2+xyx-4xyx-2x^2yx^2 = -3xyx=3xyx$$

Then:

$$0=x \alpha x + x(x^2 \alpha x^2)x = 2xyx+x^2yx^2 -3x^2yx^2=2xyx-2x^2yx^2$$

So we have:

$$2xyx=2x^2yx^2$$

and similarly with x and y switched. But like I said, I'm stuck. The only way I can think to isolate xy and yx, which I'll need to do at some point, is to expand a higher power of x+y (eg, I know x3y=xy). But there are so many bad terms here, and the identities I've found so far do nothing to simplify them. Do you have any other suggestions? Thanks for your help so far.

Last edited: May 4, 2006
6. May 4, 2006

### Hurkyl

Staff Emeritus
I haven't gotten much further. Though I was looking at some other quantities too: (1+x)³ and (xy - yx)³.

I don't have much intuition for noncommutative rings, so I can't divine a nice, clever approach to this!

My best idea at the moment is to notice that anything of the form a² satisfies the equation x² = x, to which I think you can apply your previous method. I.E. (x²)² = x² is satisfied by everything.

7. May 21, 2010

### Smiles

Let R be a ring such that x^3 = x for every x in R, then R is commutative ring.

My steps are:

Given : x^3 = x --------------(a)

Replace x by (x+x)

(x+x)^3 = (x+x)

(x+x) (x+x) (x+x) = (x+x)

( x^2 + x^2 + x^2 + x^2 )(x+x) =x+x
(4x^2 ) ( 2x) = 2x
8x^3 = 2x
8x = 2x by (a)
6x=0 ------------------------(b)

Consider

(x^2 – x)^3 = (x^2 – x)
(x^2 – x) (x^2 – x) (x^2 – x) = (x^2 – x)
(x^4 – x^3 – x^3 + x^2 )(x^2 –x) = ( x^2 – x)
(x.x^3 – 2x^3 + x^2)(x^2 – x) = x^2 – x
(x.x – 2x +x^2)((x^2-x) =x^2 –x by (b)
(x^2 – 2x + x^2)(x^2-x) = x^2 – x
(2x^2 – 2x) ( x^2 – x) = x^2 –x
2( x^2 – x) ( x^2 – x)= x^2 – x
2(x^4 – x^3 – x^3 + x^2) = x^2 – x
2(x.x^3 – 2x^3 + x^2) = x^2 – x
2(2x^2 – 2x) = x^2 – x
4x^2 – 4x = x^2 – x
3x^2 = 3 x ---------------------(c)

Consider S = { 3x | x in R }

It can be easily checked that S is a sub ring of R.

For y in R,

y^2 = (3x)^2 = 9x^2 = 6x^2 + 3x^2 = 6x. x + 3x^2 = 0 + 3x = 3x = y

 y^2 = y for all y in S implies that S is commutative.

So ( 3x) (3y) = (3y)(3x)

9xy = 9yx

Now Consider

(x+y)^3 = x+y

(x+y) (x+y) (x+y) = (x+y)
(x^2 + xy + yx + y^2) (x+y) = (x+y)
x^3 + xyx + yx^2 + y^2 .x + x^2. y + xy^2 + yxy + y^3 = x + y

x + xyx + yx^2 + y^2 .x + x^2. y + xy^2 + yxy + y^3 = x + y by (a)

xyx + yx^2 + y^2 .x + x^2. y + xy^2 + yxy = 0 ------------(g)

Now again

(x-y)^3 = x-y

implies

xy^2 – x^2.y – xyx – yx^2 + yxy + y^2. x = 0 ----------(h)

adding (g) and (h), we get

2xy^2 + 2yxy + 2y^2.x = 0 ---------------(k)

Multiply (k) by y on right

2xy^3 + 2yxy^2 + 2y^2.xy = 0

2xy + 2yx y^2 + 2y^2.xy = 0 ------------------(l)

Multiply (k) by y on left

2yxy^2 + 2y^2. xy + 2y^3.x = 0

2yxy^2 + 2y^2. xy + 2y.x = 0 --------------(m)

Subtract (l) from (m)

-2xy + 2yx = 0

2yx = 2xy

Which shows that R is commutative.

8. May 23, 2010

### Martin Rattigan

$$\text{If R is an associative ring }((\forall x\in R)(\exists n\in\mathbb{N})n>1\wedge x^n=x)\Rightarrow R\text{ is commutative}$$.