Assignment 3, Problem 107
Let $ R $ be a commutative ring with identity such that the identity map is the only ring automorphism of $ R $. Prove that the set $ N $ of all nilpotent elements of $ R $ is an ideal of $ R $
Proposed solution (Avi Steiner)
Let $ N $ be the set of nilpotent elements of $ R $. Note that conjugation by a unit $ u $ is an automorphism: $ u(x+y)u^{-1}=uxu^{-1} + uyu^{-1} $, $ uxyu^{-1}=uxu^{-1}uyu^{-1} $, and the map $ x\mapsto uxu^{-1} $ is clearly a bijection (with inverse given by conjugation by $ u^{-1} $). So, since there are no non-trivial automorphisms of $ R $, all such conjugations must be trivial. Therefore, $ R^\times $ commutes pointwise with all of $ R $ and in particular is abelian.
Now, $ 1-a $ is a unit for any nilpotent $ a $: If $ a^n=0 $, then
- $ (1+a+a^2+\cdots+a^{n-1})(1-a) = 1-a^n=1. $
So, letting $ r\in R $, $ a\in N $, we have
- $ r-ar=(1-a)r=r(1-a)=r-ra $,
meaning that $ r $ and $ a $ commute. Therefore, if $ a^n=0 $, we have that $ (ra)^n=(ar)^n=a^nr^n=0 $; i.e. $ ra=ar\in N $. Finally, suppose $ a^n=0 $ and $ b^m=0 $. Since $ a $ and $ b $ commute, $ (a+b)^{n+m} $ is a </math>\mathbb{Z}</math>-linear combination of terms of the form $ a^{n+m-k}b^k $ for $ 0\leq k\leq n+m $. If $ k\geq m $, then $ b^m=0 $, making the corresponding term vanish. If $ k<m $, then $ n+m-k>n $, so that $ a^{n+m-k}=0 $, again making the corresponding term vanish. Thus, $ (a+b)^{n+m}=0 $. Hence, $ a+b\in N $. Thus, $ N $ is an ideal.
