A problem from Herstein [6, Problem 3.4.19]:
Let R be a ring in which x^3 = x for every x in R.
Prove that R is a commutative ring.
This problem is single stared in Herstein.
Several solutions have been given [2,4,5,8,1,9].
I still wonder what the intended the solutions would
look like. The problem is in the section quotion rings.
A more general theorem was found by Jacobson:
If for every element x ot the ring there is an integer n, n>1,
such that x^n = x, then the ring is commutative.
For a proof of Jacobson's theorem, see [1,6].
Solution in the way of [5] with a modification of lemma 3.
Lemma 1: x^2 = 0 => x = 0.
x^2 = 0 => x^3 = x0 = 0 => x = 0.
Lemma 2: xy = 0 => yx = 0.
xy = 0 => yxyx = y0x => (yx)^2 = 0 => (Lemma1) yx = 0.
Note that lemma2 shows that a cyclic permutation is valid:
xyz = 0 => 0 = xyz = yzx = zxy. But nothing is said about xzy.
Lemma 3: xy = x^2 y^2 x y for all x,y.
0 = yx - yx = yx - y^3x = y(x - y^2x) = (x - y^2x)y last step by lemma2.
0 = x^2(x - y^2x)y = x^3y - x^2y^2xy = xy - x^2y^2xy.
Lemma 4: (xy - yx)yx = 0 for all x,y.
With lemma3: 0 = x^2y^2xy - xy = x^2y^2xy - (xy)^3 = x(xy - yx)yxy.
By lemma2: 0 = (xy - yx)(yx)^2. Multiply from the right by xy gives
0 = (xy - yx)(yx)^3 = (xy - yx)(yx).
Theorem: xy = yx for all x,y.
Expand (xy - yx)^2 = (xy - yx)xy + (yx - xy)yx
= - (yx - xy)xy - (xy - yx)yx
By lemma4 both terms in the last expression are zero.
The first one according to the variable substitution x<-y, y<-x.
Therefore (xy - yx)^2 = 0 for all x,y. And lemma1 shows xy - yx = 0.
References:
[1] Raymond Ayoub, Christine Ayoub;
On the commutativity of rings,
American Mathematical Monthly, 71 (1964) 267-271
[2] N. Bourbaki;
Elements of Mathematics Algebra I,
Springer-Verlag London, 1989, Chapters 1-3, 2nd printing
[3] David Gries;
A Proof Concerning Rings,
Technical Report, Computer Science Dept.,
Cornell Univ., Sept. 1987
[4] Rob Gross;
x^n=x => xy = yx for n=3,4
sci.math (Wed, 22 Dec 1993)
[5] Ted Herman;
On a Theorem of Jacobson,
in: Eds. Feijen, van Gasteren, Gries, Misra; Beauty is our Business,
Texts and Monographs in Computer Science, Springer, 1990, Sect. 20, 176-181
[6] I. N. Herstein;
Wedderburn's Theorem and a Theorem of Jacobson,
American Mathematical Monthly, 68 (1961) 249-251
[7] I. N. Herstein;
Topics in Algebra,
Wiley, New York, 1975, 2nd Edition
[8] Jiang Luh;
An Elementary Proof of a Theorem of Herstein,
Mathematical Magazine, 38 (1965) 105-106
[9] Alf van der Poorten;
Concerning Commuting,
letter in Austral. Math. Soc. Gazette 21 (1994) 68
www-centre.mpce.mq.edu.au/alfpapers/a127.pdf
http://www.mathematik.uni-bielefeld.de/~sillke/
mailto:Torsten.Sillke@uni-bielefeld.de
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Proof from [9] provided by Hanna Neumann in the mid-sixties.
2x = (2x)^3 = 8x^3 = 8x so 6x = 0 for all x in R. Also
x ± y = (x ± y)^3 = x^3 ± (x^2y + xyx + yx^2) + (xy^2 + yxy + y^2x) ± y^3,
yielding eventually 2(x^2y + xyx + yx^2) = 0. Multiplying this
on the left by x, and on the right by x yields
2(xy + x^2yx + xyx^2) = 0 2(x^2yx + xyx^2 + yx) = 0,
so, subtracting, we have 2(xy - yx) = 0. But it happens that
(x + x^2) = (x + x^2)^3 = x^3 + 3x^4 + 3x^5 + x^6
= x + 3x^2 + 3x + x^2 = 4(x + x^2),
thus 3(x + x^2) = 0 for all x in R. In particular
3(x + y + (x + y)^2) = 3(x + y + x^2 + xy + yx + y^2) = 0,
yielding eventually 3(xy + yx) = 0.
But 6yx = 0 so 3(xy - yx) = 0. From a while ago also 2(xy - yx) = 0 so,
incredibly, we are done.
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Subject: Wanted: a short proof that x^n=x => xy = yx for n=3,4
From: gross@bcuxs2.bc.edu (Rob Gross) Date: Wed, 22 Dec 1993
Hi,
> Does anyone know a short proof of the following statement:
>
> in a ring if for all x: x^n=x, then the ring is commutative
>
> where n is 3 or 4? I looked at the Robinson's paper that considers
> the general case (n is any positive integer), but the proof is long
> and complicated, and I heard that there is a short proof for n 3 or 4.
I once made the mistake of assigning the problem (it's in
Herstein's "Basic Algebra," and it isn't even a starred problem),
and had to type the answer for my students when no one could get
it. That meant, unfortunately, that I had to work out the
answer, and it was not simple; in the end, I needed to get an
answer key for the text and look it up. I enclose it below, in
TeX format; feel free to post it (with credit) if you don't get
anything simpler.--Rob
---
Rob Gross (617) 552-3758 / 787-9593
Department of Mathematics BITNET: GROSS@BCVMS.BITNET
Boston College Internet: gross@bcuxs2.bc.edu
Chestnut Hill, MA 02167-3806 gross@bcvms.bc.edu
-------------------------------------------------------------------
\nopagenumbers
\def\ans{\smallskip\noindent{\it Answer:\/ }}
\centerline{Mathematics 317}
\bigskip
\noindent
1. If $R$ is a ring and $e\in R$ is such that $e^2=e$, show that
$(ex-exe)^2 = (xe-exe)^2 = 0$ for any $x\in R$.
\ans
This one is just a computation: $(ex-exe)^2 = (ex-exe)(ex-exe) =
exex - exexe - exe^2x + exe^2xe = 0$, and similarly for the other
one.
\bigskip\noindent
2. Let $R$ be a ring in which $x^3=x$ for every $x\in R$. Prove
that $R$ is commutative.
\ans
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
$$
\eqalign{%
(b+b^2)^2x &= x(b+b^2)^2\cr
(b^2 + 2b^3x + b^4)x &= x(b^2 + 2b^3 + b^4)\cr
(2b^2 + 2b)x &= x(2b^2 + 2b)\cr
}
$$
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.
\bigskip\noindent
3. Let $R$ be a ring in which $x^4=x$ for every $x\in R$. Prove
that $R$ is commutative.
\ans
This one makes the previous one look simple. We begin with a few
observations.
Since $(-x)^2 = x^2$, we have $-x = (-x)^4 = (x^4) = x$, so
$-x=x$, or $2x=0$ if $x$ is any element in $R$.
If $a^2=0$, then $a=a^4 = (a^2)^2 = 0^2 = 0$. In other words, if
the square of an element is $0$, then the element itself is $0$.
Finally, we have the simple computation that $(a^3)^2 = a^6 =
a^4a^2 = a^3$. That means that we can apply the first problem,
with $e=a^3$: $(xa^3 - a^3xa^3)^2 = (a^3x - a^3xa^3)^2 = 0$. Our
second observation then gives $xa^3 - a^3xa^3 = 0$, and $a^3x -
a^3xa^3=0$, and therefore $xa^3 = a^3x$.
We now set $a=b+b^3$ in the previous equation, though it helps
to first cube $b+b^3$: $(b+b^3)^3 = b^3 + 3b^5 + 3b^7 + b^9 =
b^3 + b^5 + b^7 + b^9 = b^3 + b^2 + b + b^3 = 2b^3 + b^2 + b =
b^2 + b$. Therefore, $x(b^2 + b) = (b^2 + b)x$.
Next, we take this equation, and substitute in $c+d$ for $b$:
$x((c+d)^2 + (c+d)) = ((c+d)^2 + (c+d))x$. Expanding, we have
$x(c^2 + cd + dc + d^2 + c + d) = (c^2 + cd + dc + d^2 + c +
d)x$. However, we also know that $x(c^2+c)=(c^2+c)x$ and
$x(d^2+d)=(d^2+d)x$. Make those two cancellations, and we have
$x(cd+dc)=(cd+dc)x$.
This last equation is true for any $c$, $d$ and $x$ in $R$. In
particular, we can let $x=c$, and then we get $c^2d + cdc=cdc +
dc^2$, or $c^2d=dc^2$. But we also know that
$d(c^2+c)=(c^2+c)d$, and therefore $dc^2+dc=c^2d+cd$. This,
finally, gives $cd=dc$.
\bye
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Boolean Rings (an easy problem):
Let R be a ring in which x^2 = x for every x in R.
Prove that R is a commutative ring and that x+x = 0.
Solution:
(A) x+y = (x+y)^2 = x^2 + xy + yx + y^2 = x + xy + yx + y
Therefore xy + yx = 0.
(B) x+x = x^2 + x^2 = 0. (The last step is a special case of (A).)
Therefore x = -x.
(C) Now (A) and (B) give xy + yx = xy - yx = 0.
Therefore the ring is commutative.