Let $R$ be a commutative ring with unity of prime characteristic $p$ , if $a\in R$ is nilpotent then does $\exists n \in \mathbb N$ such that $(1+a)^n=1$ ?
$R$ be a commutative ring with unity of prime characteristic $p$ , if $a\in R$ is nilpotent then is $1+a$ unipotent?
1
$\begingroup$
ring-theory
nilpotence
positive-characteristic
-
2I think that the characteristic of $R$ doesn't have to be prime. It just has to be nonzero. – 2017-02-07
-
0Yes, the proof is certainly less technical with prime characteristic, but the only thing you need to generalize it to arbitrary finite characteristic is the following: Given a number $d$ (the characteristic) and a bound $m$ (the nilpotency), one can find $n$ such that all binomial coefficients $\binom{n}{1}, \dotsb, \binom{n}{m}$ are divisible by $d$. This is not hard to see...Just make sure $n$ contains $d$ more often than $m!$. – 2017-02-07
1 Answers
2
Yes, because we have
$$(1+a)^{p^e} = 1+a^{p^e} = 1 \text{ for } e \gg 0.$$