A condition that implies commutativity - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:26:51Z http://mathoverflow.net/feeds/question/29590 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29590/a-condition-that-implies-commutativity A condition that implies commutativity J. H. S. 2010-06-26T08:21:21Z 2010-07-24T22:09:43Z <p>Let $R$ be a ring. A cute theorem by <strong>N. Jacobson</strong> 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.</p> <p>The proof of the result for the cases $n=2, 3,4$ is the subject matter of several well-known exercises in <strong>Herstein</strong>'s <em>Topics in Algebra</em>. The corresponding proofs rely heavily on "elementary" manipulations. For instance, the proof of the case $n=3$ can be done as follows:</p> <p>1) If $a, b \in R$ are such that $ab= 0$ then $ba=0$.</p> <p>2) $a^{2}$ and $-a^{2}$ belong to $\mathbf{Z}(R)$ for every $a \in R$.</p> <p>3) Since $(a^{2}+a)^{3} = (a^{2}+a)^{2}+(a^{2}+a)^{2}$ it follows that</p> <p>$a=a+a^{2}-a^{2} = (a+a^{2})^{3}-a^{2} = (a^{2}+a)^{2}+(a^{2}+a)^{2}-a^{2}$</p> <p>and whence the result. ▮</p> <p>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: <strong>is an analogous demonstration of the general theorem possible?</strong> The one that appears in [<strong>1</strong>] depends on some non-trivial structure theorems for division rings.</p> <p>As usual, I thank you in advance for your insightful replies, reading suggestions, web links, etc...</p> <p><strong>References</strong></p> <p>[<strong>1</strong>] I. N. Herstein, <em>Noncommutative rings</em>, The Carus Mathematical Monographs, no. 15, Mathematical Association of America, 1968.</p> <p>[<strong>2</strong>] I. N. Herstein, Álgebra Moderna, Ed. Trillas, págs. 112, 119, and 153.</p> http://mathoverflow.net/questions/29590/a-condition-that-implies-commutativity/33234#33234 Answer by Bill Dubuque for A condition that implies commutativity Bill Dubuque 2010-07-24T21:38:30Z 2010-07-24T22:09:43Z <p>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 <a href="http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30273#30273" rel="nofollow">post</a>. 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 <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">work [1]</a></p> <p><a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">1</a> S. Burris and J. Lawrence, Term rewrite rules for finite fields.<br> International J. Algebra and Computation 1 (1991), 353-369. <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf</a> </p>