From: jpcolm@aol.com (JPCOLM)
Subject: Re: Question
Date: 21 Jul 1999 02:48:46 GMT
Newsgroups: sci.math
Keywords: proving commutativity in rings from x^n=x
jvoight wrote:
>Let $R$ be a ring in which $x^3=x$ for every $x\in R$. Prove that $R$
>is a commutative ring.
>
>Any takers?
>
The analogous theorem is true for any
n >= 2: if n is a fixed integer and R is
a ring for which x^n = x for all x, then R
is commutative. I think it is due to
Jacobson. The only proof that I know of
relies on fairly involved structural
considerations (any such ring can be
embedded in a product of finite such rings,
which can in turn be seen to be commutative since finite division rings are
commutative.) It would be of some
interest to come up with purely equational
proofs of these theorems (one theorem
for each n). Such proofs must exist by
Birkhoff's completeness theorem for
equational logic but for n greater than 3 or 4 they are still (I think) unkown.
Is anyone
familar with any work in this area?
- John Coleman
==============================================================================
From: Robin Chapman
Subject: Re: Two other questions
Date: Wed, 21 Jul 1999 10:49:54 GMT
Newsgroups: sci.math
To: jvoight34@my-deja.com
In article <7n2bu0$2ks$1@nnrp1.deja.com>,
jvoight34@my-deja.com wrote:
>
> > > And: suppose in a ring that x^3=x for all x. Show that the ring is
> > > commutative.
> >
> > It's a theorem of Jacobson that a ring R with the property that
> > for each x in R there is m(x) > 1 with x^{m(x)} = x is commutative.
> > The proof I've seen is a bit difficult....
>
> This is an early problem in a chapter on ring theory, so the proof
> can't be that difficult if not proven in its full generality.
OK. As (2x)^3 = 2x then 8x^3 = 2x or 8x = 2x for all x in R. Thus
6x = 0 for all x in R. Thus R is the direct sum of the ideals 3R
and 2R. Both of these are rings in their own right, both have
x^3 = x for all x in them and they have characteristics 2 and 3
respectively. Also 3R and 2R commute since the product of
elements from two of them is zero (either way). We can reduce
to two special cases (i) R of characteristic 2 and (ii) R of
characteristic 3.
Expanding out the relations (x+y)^3 - x^3 - y^3 = 0 and
(x+y+z)^3 -(x+y)^3 -(x+z)^3 - (y+z)^3 + x^3 + y^3 + z^3 = 0
gives
x^2y + xyx + yx^2 + xy^2 + yxy + y^2x = 0 (1)
and
xyz + xzy + yxz + yzx + zxy + zyx = 0 (2)
for all x, y and z. Putting z = x in (2) gives
2(x^2y + xyx + yx^2) = 0
and in characteristic 3 we get x^2y + xyx + yx^2 = 0.
Thus in characteristic 3 we have
0 = x(x^2y + xyx + yx^2) - (x^2y + xyx + yx^2)x = x^3y - yx^3 = xy - yx
and so R is commutative in characteristic 3.
Assume now R has characteristic 2. Then x = -x for all x. Putting
z = xy in (2) gives
xyxy + x^2y^2 + yx^2y + yxyx + xyxy + xy^2x = 0
or
x^2y^2 + yxyx + xy^2y + yx^2y = 0. (3)
Swap x and y in (3) and add to 3 to get
x^2y^2 + y^2x^2 + xyxy + yxyx = 0. (4)
From (1) we get
0 = x(x^2y + xyx + yx^2 + xy^2 + yxy + y^2x)
+(x^2y + xyx + yx^2 + xy^2 + yxy + y^2x)x
= x^3y + yx^3 + x^2y^2 + xyxy + yxyx + y^2x^2
= xy + yx
by (4).
Thus xy = yx and R is commutative.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
==============================================================================
From: "G. A. Edgar"
Subject: Re: Question
Date: Wed, 21 Jul 1999 08:04:09 -0400
Newsgroups: sci.math
> > Let $R$ be a ring in which $x^3=x$ for every $x\in R$. Prove that $R$
> > is a commutative ring.
>
> This is (word-for-word) in Herstein, Topics in Algebra, Ch. 3 sec. 4,
> exercise 19.
I once heard Herstein comment that this one problem generated
more mail than the entire remainder of the book.
--
Gerald A. Edgar edgar@math.ohio-state.edu
Department of Mathematics telephone: 614-292-0395 (Office)
The Ohio State University 614-292-4975 (Math. Dept.)
Columbus, OH 43210 614-292-1479 (Dept. Fax)
==============================================================================
From: Robin Chapman
Subject: Re: Two other questions
Date: Wed, 21 Jul 1999 14:33:39 GMT
Newsgroups: sci.math
In article <7n4laf$tds$1@nnrp1.deja.com>,
chri0562@my-deja.com wrote:
>
> > OK. As (2x)^3 = 2x then 8x^3 = 2x or 8x = 2x for all x in R. Thus
> > 6x = 0 for all x in R. Thus R is the direct sum of the ideals 3R
> > and 2R. Both of these are rings in their own right, both have
> > x^3 = x for all x in them and they have characteristics 2 and 3
> > respectively. Also 3R and 2R commute since the product of
> > elements from two of them is zero (either way). We can reduce
> > to two special cases (i) R of characteristic 2 and (ii) R of
> > characteristic 3.
>
> I've just got a little query - what do you mean by 3R and 2R ?
> R doesn't have a 1, right?
I wasn't assuming that R was unital, since the result doesn't need
this. By 3R I mean the set {x+x+x: x in R} etc.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
==============================================================================
From: Robin Chapman
Subject: Re: Two other questions
Date: Wed, 21 Jul 1999 20:27:39 GMT
Newsgroups: sci.math
In article <7n48kg$or4$1@nnrp1.deja.com>,
Robin Chapman wrote:
> In article <7n2bu0$2ks$1@nnrp1.deja.com>,
> jvoight34@my-deja.com wrote:
> >
> > > > And: suppose in a ring that x^3=x for all x. Show that the ring
> is
> > > > commutative.
> > >
> Assume now R has characteristic 2. Then x = -x for all x.
I am being really obtuse today. This case is much better dealt with
as follows. From (x + x^2)^3 = x + x^2 we get 3x + 3x^2 = 0
or x = x^2 for all x. From this it's easy to see that R is
commutative (R is a Boolean ring) as then (x + y)^2 = x + y
gives xy + yx = 0.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
==============================================================================
From: "pascal ORTIZ"
Subject: Re: Two other questions
Date: 22 Jul 1999 08:31:38 GMT
Newsgroups: sci.math
Robin Chapman a écrit dans l'article
<7n48kg$or4$1@nnrp1.deja.com>...
...
>
> Expanding out the relations (x+y)^3 - x^3 - y^3 = 0 and
> (x+y+z)^3 -(x+y)^3 -(x+z)^3 - (y+z)^3 + x^3 + y^3 + z^3 = 0
> gives
> x^2y + xyx + yx^2 + xy^2 + yxy + y^2x = 0 (1)
> and
> xyz + xzy + yxz + yzx + zxy + zyx = 0 (2)
> for all x, y and z. Putting z = x in (2) gives
> 2(x^2y + xyx + yx^2) = 0
> and in characteristic 3 we get x^2y + xyx + yx^2 = 0.
> Thus in characteristic 3 we have
> 0 = x(x^2y + xyx + yx^2) - (x^2y + xyx + yx^2)x = x^3y - yx^3 = xy - yx
> and so R is commutative in characteristic 3.
>
(following hints from a problem in a french textbook) :
by expanding out (x+y)^3 - (x-y)^3 you get xyx + x^2y + y^2x = 0 and
multiplying by x on the left and on the right, you get xy = yx ( = - x^2yx
- xyx^2).
==============================================================================
From: Robin Chapman
Subject: Re: Two other questions
Date: Thu, 22 Jul 1999 07:58:34 GMT
Newsgroups: sci.math
In article <7n2bu0$2ks$1@nnrp1.deja.com>,
jvoight34@my-deja.com wrote:
>
> > > And: suppose in a ring that x^3=x for all x. Show that the ring is
> > > commutative.
Here's a better (I think) argument than the one I posted yesterday.
For all x and y we have
0 = (x + y) - x - y = (x + y)^3 - x^3 - y^3.
Expanding this out gives
x^2 y + xyx + yx^2 + y^2 x + yxy + xy^2 = 0. (1)
Setting y = x in (1) gives 6x^3 = 0 so that 6x = 0 for all x.
Setting y = x^2 in (1) gives 3x^4 + 3x^5 = 0. Using x^3 = x
we get 3x^2 + 3x^3 = 0 = 3x + 3x^2. Hence 3x^2 = - 3x = 3x.
Hence
0 = 3(x + y) - 3x - 3y = 3(x + y)^2 - 3x^2 - 3y^2 = 3xy + 3yx.
Thus 3xy = -3yx = 3xy.
Replace x by -x in (1) to get
x^2 y + xyx + yx^2 - y^2 x - yxy - xy^2 = 0. (2)
Adding (1) and (2) gives
2(x^2y + xyx + yx^2) = 0
and so
0 = 2x(x^2y + xyx + yx^2) -2(x^2 y + xyx + yx^2)x
= 2x^3 y - 2yx^3 = 2xy - 2yx.
Thus 2xy = 2yx. Subtracting this from 3xy = 3yx gives xy = yx.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.
==============================================================================
From: mareg@primrose.csv.warwick.ac.uk (Dr D F Holt)
Subject: Re: Question
Date: 22 Jul 1999 08:47:19 GMT
Newsgroups: sci.math
In article <210719990804094467%edgar@math.ohio-state.edu.nospam>,
"G. A. Edgar" writes:
>
>> > Let $R$ be a ring in which $x^3=x$ for every $x\in R$. Prove that $R$
>> > is a commutative ring.
>>
>> This is (word-for-word) in Herstein, Topics in Algebra, Ch. 3 sec. 4,
>> exercise 19.
>
>I once heard Herstein comment that this one problem generated
>more mail than the entire remainder of the book.
>
I remember spending many hours on this as an undergraduate. Here is the
most concise proof that we managed to come up with:
An element x is called central if xy=yx for all y in R.
Note that the central elements form a subring of R.
1. xy = 0 => yx = 0.
(Proof: yx = (yx)^3 = y (xy)^2 x = 0.)
2. x^2 = x => x central.
(Proof: x(y - xy) = xy - x^2y = xy-xy=0, so (by 1) (y - xy)x = 0,
and yx = xyx.
Similarly, (y - yx)x = 0 => x(y - yx) = 0 => xy = xyx.)
3. x^2 is central for all x in R. (by 2, because (x^2)^2 = x^4 = x^2).
4. If x^2 = nx for an integer n, then x is central.
(Proof: x = x^3 = qx^2, which is central by 3.)
5. x + x^2 is central for all x in R.
(Proof: By 4, because (x + x^2)^2 = 2(x + x^2).)
6. By 3 and 5, x = (x + x^2) - x^2 is central, completing the proof.
Derek Holt.