Timeline for For R a ring, if $x^n=x$ for all x∈R then R is commutative
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2021 at 12:12 | comment | added | Benjamin Steinberg | Notice this contains as a special case Wedderburn's theorem that a finite division ring is a field so you shouldn't expect it to be trivial | |
| Jun 18, 2021 at 9:12 | comment | added | YCor | I don't think it's pedantic to say this. Without quantifier on $n$, this is unclear, and it's part of the (easy) efforts to make a post clear, to quantify so as to make a question clear. (Maybe it should be said that $R$ is supposed associative, since a large number of people use "ring" in a broader sense, and even for $n=2$ the conclusion fails without associativity.) | |
| Jun 18, 2021 at 7:33 | comment | added | Derek Holt | To be pedantic, the symbol $n$ in this post is undefined. In fact the correct statement of Jacobson's Theorem is "if $R$ is a ring, and for all $x \in R$ there exists an integer $n>1$ with $x^n=x$, then $R$ is commutative". It is difficult to prove and I don't advise attempting it. Try proving that $\forall x \in X\, x^2=x \Rightarrow R$ commutative, and if you find that easy, try proving $\forall x \in X\, x^3=x \Rightarrow R$ commutative, which is a starred exercise in Herstein's Topics in Algebra. | |
| Jun 18, 2021 at 7:05 | comment | added | Emil Jeřábek | And for $n=0$ it’s trivial. | |
| Jun 18, 2021 at 3:42 | comment | added | Gerry Myerson | And for $n=1$ it's false. | |
| Jun 18, 2021 at 2:06 | comment | added | KConrad | For $n > 1$ this is an old theorem of Jacobson. See mathoverflow.net/questions/207757/…. | |
| Jun 18, 2021 at 2:06 | review | First posts | |||
| Jun 18, 2021 at 5:47 | |||||
| Jun 18, 2021 at 2:01 | history | asked | Learner | CC BY-SA 4.0 |