Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a ring. A cute theorem by N. Jacobson states that if the identity $x^{n}=x$ holds for every $x \in R$ and a fixed $n \geq 2$ then $R$ is a commutative ring.

The proof of the result for the cases $n=2, 3,4$ is the subject matter of several well-known exercises in Herstein's Topics in Algebra. The corresponding proofs rely heavily on "elementary" manipulations. For instance, the proof of the case $n=3$ can be done as follows:

1) If $a, b \in R$ are such that $ab= 0$ then $ba=0$.

2) $a^{2}$ and $-a^{2}$ belong to $\mathbf{Z}(R)$ for every $a \in R$.

3) Since $(a^{2}+a)^{3} = (a^{2}+a)^{2}+(a^{2}+a)^{2}$ it follows that

$a=a+a^{2}-a^{2} = (a+a^{2})^{3}-a^{2} = (a^{2}+a)^{2}+(a^{2}+a)^{2}-a^{2}$

and whence the result. ▮

Certainly, the mind can't but boggle at the succinctness of the above solution. Actually, it is the conciseness of this argument that has prompted me to pose the present question: is an analogous demonstration of the general theorem possible? The one that appears in [1] depends on some non-trivial structure theorems for division rings.

As usual, I thank you in advance for your insightful replies, reading suggestions, web links, etc...

References

[1] I. N. Herstein, Noncommutative rings, The Carus Mathematical Monographs, no. 15, Mathematical Association of America, 1968.

[2] I. N. Herstein, Álgebra Moderna, Ed. Trillas, págs. 112, 119, and 153.

share|improve this question
2  
As a commutative reader, I'd like to learn what are your (2) and (3) above. What is **Z**$(R)$? Why $(a^2+a)^3=(a^2+a)^2+(a^2+a)^2$? –  Wadim Zudilin Jun 26 '10 at 8:40
    
My guess is that Z(R) is the center of the ring and 3) follows from a^3=a, I think. –  efq Jun 26 '10 at 12:04
1  
The standard notation for the centre of the ring $R$ is $Z(R)$ (the $Z$ not usually bold). My idle speculation is that there should be a "calculational" proof like the given one for any particular $n$, but there being such a proof for all $n$ seems unlikely. Treating general $n$ I suspect needs "second-order" concepts: subrings, ideals, quotients, stuff like that. –  Robin Chapman Jun 26 '10 at 18:55
2  
1. $\mathbf{Z}(R)$ is the center of the ring. 2. I don't see what the problem with #3 is: $(a^2+a)^3=(a^2+a)^2(a^2+a)=(a^2+a+a^2+a)(a^2+a).$ 3. Indeed, there are several stronger results (cf. chapter 3 of Herstein's Noncommutative rings). Yitz expressed therein that the version mentioned by Pete, "as proved has one drawback; true enough, it implies commutativity but only very few commutative rings exist which satisfy its hypothesis." –  J. H. S. Jun 26 '10 at 19:46
1  
@Viktor- 3) gives any element 'a' as the sum/difference of squares of elements and from 2) (and the closure of the centre under addition) we have that 'a' belongs to the centre. –  Tom Boardman Jun 27 '10 at 11:15

1 Answer 1

up vote 15 down vote accepted

For fixed $n \in \mathbb{N}$, Birkhoff's completeness theorem implies that such a proof must exist in the first-order equational theory of rings - as I mentioned here in a recent post. Many years ago Stan Burris told me that John Lawrence discovered such an equational proof that works uniformly for all $n$ (possibly also for Jacobson's form $x^{n(x)} = x$). I don't know if the proof is published yet, but some clues as to how it may proceed might be gleaned from their earlier joint work [1]

1 S. Burris and J. Lawrence, Term rewrite rules for finite fields.
International J. Algebra and Computation 1 (1991), 353-369. http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.