x^3=x implies ring is commutative
themadh...@yahoo.com
1. If R is a ring with unity and x^3=x for all elements then this ring
is commutative.
2. If R is an integral domain and x^3=x for all elements then this
ring is commutative.
I saw problem #1 without unity and I was wondering if there was a clean
proof with unity. I cannot prove it. I made up #2.
2. Proof: For all x in R suppose x^3=x. Then x^3-x=0 and so
x(x^2-1)=0 since R is an integral domain. Now x^2-1=0 but x does not
equal 0 as R is an integral domain. So x^2-1=0 so x^2=1. Thus x is
its own inverse and since x was any element in R, every element is its
own inverse. Now for any y in R, xy is in R as R is closed under
multiplication. Now (xy)^2=xyxy=1. So (xyxy)y=1y and thus xyx=y. Now
(xyx)x=yx implies xy=yx. Thus R is a commutative ring.
Nevermind. I forgot the definition of integral domain meant the ring
was commutative. I still cannot figure out #1. I tried the approach I
used in #2 and it did not work.
Stephen Montgomery-Smith
satisfies x^n=x for some fixed positive integer n>1, then the ring is
commutative. Presumably this proof uses axiom of choice or ultrafilters
or such like.
Thus one can deduce that for each positive integer n>1 that there exists
an elementay "first order" proof that the ring is commutative.
For n=2 the proof is quite well known:
x+y = (x+y)^2 = x + xy + yx + y
So xy = -yx
And -x = (-x)^2 = x^2 = x. QED
Other than that, all I can tell you is I think I remember that n=4 was
easier than n=3.
If anyone knows more details (especially the high powered proof) I would
love to see it.
Johann Wiesenbauer
> Posted: Jul 11, 2005 7:39 PM Reply
>
>
> http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES
> /herstein
>
> A real pain. Begin with the observation that if $y$
> is any element in $R$ so that $y^2=0$, then $y = y^3
> = y(y^2) = y(0)=0$.
> In other words, if the square of an element gives
> $0$, then the
> element itself is $0$.
>
> Next, notice that if $a$ is any element in $R$, then
> $a^4 =
> a^3a=a^2$, so $(a^2)^2 = a^2$. Therefore, we can
> apply problem
> $1$ with $e=a^2$. We get $(xa^2 - a^2xa^2)^2 = (a^2x
> -
> a^2xa^2)^2=0$. Our observation about squares now
> gives us
> $xa^2-a^2xa^2 = 0$ and $a^2x - a^2xa^2 = 0$.
> Combining those
> two, we see that if $x$ and $a$ are any elements in
> $R$, then
> $a^2x = xa^2$.
>
> Now let $b$ be any element of $R$, and set $b+b^2=a$
> in the previous equation. We get(b+b^2)^2x &=
> x(b+b^2)^2\cr
> (b^2 + 2b^3x + b^4)x &= x(b^2 + 2b^3 + b^4)\cr
> (2b^2 + 2b)x &= x(2b^2 + 2b)\cr
> }
> $$
> Since we already know that $b^2x = xb^2$, we get the
> equation
> $2bx = 2xb$. If we could cancel the factor of $2$, we
> would be
> done, but the problem isn't quite that simple, and we
> need one
> last bit of trickery.
>
> Return to the fact that any element is its cube, and
> apply that
> to $b+b^2$:
> $$
> \eqalignno{%
> (b+b^2)^3 &= b+b^2\cr
> b^3 + 3b^4 + 3b^5 + b^6 &= b+b^2\cr
> b + 3b^2 + 3b^3 + b^4 &= b+b^2\cr
> 4b + 4b^2 &= b + b^2\cr
> 3b + 3b^2 &= 0\cr
> \noalign{\hbox{and so}}
> (3b+3b^2)x &= x(3b+3b^2)\cr
> 3bx + 3b^2x &= 3xb + 3xb^2\cr
> }
> $$
> But $b^2x = xb^2$, and so we get $3bx = 3xb$. Since
> we also have
> $2bx = 2xb$, subtraction finally gives $bx=xb$, and
> so $R$ is
> commutative.
Hm, not really bad, but your proof can be shortened to 4 lines or so. You don't believe it? Ok, let's start then, as you did, by observing that an idempotent u in a ring with the law x^3 = x is always in the center of the ring, i.e. ux=ux holds for all x in R. In fact, both ux and xu are equal to uxu as can be concluded from
ux - uxu = (ux-uxu)^3 = (ux-uxu)^2(ux-uxu) = 0(ux-uxu) = 0
xu - uxu = (xu-uxu)^3 = ... = 0
So far, so good. But the rest is plain sailing! See yourself:
xy = (xy)^3 = x(yx)^2y = xy(yx)^2 = xyyxyx =
= xy^2xyx = y^2 x^2 yx = y^3x^3 = yx.
Johann
G. A. Edgar
> Re: x^3=x implies ring is commutative
I once heard Herstein say that he received more mail (this was in the
days before email) about this one problem that on everything else
combined.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Tonico
Why is it that (ux-uxu)^2(ux-uxu) = 0(ux-uxu)? In other words, why (ux
- uxu)^2 =0? I must have missed that...
Thanx
Tonio
Johann Wiesenbauer
u^2 = u implies
(ux-uxu)^2 = (ux-uxu)(ux-uxu) = uxux-uxuxu-uxux+uxuxu = 0.
Johann
Bill Dubuque
>
> If R is a ring with unity and x^3 = x for all elements
1. ab = 0 => ba = 0 via ba = (ba)^3 = b ab ab a = 0
2. cc = c => c central [i.e. xc = cx for all x]
Proof: c(x-cx)=0 so (x-cx)c=0 by 1, so xc = cxc
(x-xc)c=0 so c(x-xc)=0 by 1, so cx = cxc
3. xx central via c = xx in 2.
4. cc = 2c => c central: c = ccc = 2cc central by 3.
5. x + xx central via c = x + xx in 4.
6. x = (x + xx) - xx central via 3, 5. QED
This is a special case of a famous theorem of Jacobson
which implies that a ring is commutative if for every
element x there is an integer n > 1 such that x^n = x.
The above is from my post in the thread below
which you might also find informative
http://groups-beta.google.com/group/sci.math/browse_frm/thread/5a00720c1f4b960b
-Bill Dubuque
themadh...@yahoo.com
turned out to be more difficult than I imagined.
Johann Wiesenbauer
Yes, this is certainly a way to go. In particular, the points 1-3 are a nice alternative in order to prove that idempotents in the rings at issue are central. On the other hand, as to the rest of the proof, though rather simple again, I slightly prefer my way (demonstrated elsewhere in this thread) using the following statement on semigroups
Theorem: A semigroup H, in which the laws x^3=x and
x y^2=y^2 x hold, is commutative.
In particular, it shows that the additive group of the ring and the distributivity law are no longer needed. This is admittedly a subtlety though.
Johann
mr_mathematician
http://www.math.niu.edu/~rusin/known-math/99/commut_ring
Hope this helps.
mr_mathematician