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[[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]]
 
[[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]][[Category:2013 Summer MA 598A Weigel]]
  
=Assn3Prob107=
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=Assignment 3, Problem 107=
  
  
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Let <math>R</math> be a commutative ring with identity such that the identity map is the only ring automorphism of <math>R</math>. Prove that the set <math>N</math> of all nilpotent elements of <math>R</math> is an ideal of <math>R</math>
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== Proposed solution (Avi Steiner) ==
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Let <math>N</math> be the set of nilpotent elements of <math>R</math>. Note that conjugation by a unit <math>u</math> is an automorphism: <math>u(x+y)u^{-1}=uxu^{-1} + uyu^{-1}</math>, <math>uxyu^{-1}=uxu^{-1}uyu^{-1}</math>, and the map <math>x\mapsto uxu^{-1}</math> is clearly a bijection (with inverse given by conjugation by <math>u^{-1}</math>). So, since there are no non-trivial automorphisms of <math>R</math>, all such conjugations must be trivial. Therefore, <math>R^\times</math> commutes pointwise with all of <math>R</math> and in particular is abelian.
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Now, <math>1-a</math> is a unit for any nilpotent <math>a</math>: If <math>a^n=0</math>, then
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:<math>(1+a+a^2+\cdots+a^{n-1})(1-a) = 1-a^n=1.</math>
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So, letting <math>r\in R</math>, <math>a\in N</math>, we have
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:<math>r-ar=(1-a)r=r(1-a)=r-ra</math>,
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meaning that <math>r</math> and <math>a</math> commute. Therefore, if <math>a^n=0</math>, we have that <math>(ra)^n=(ar)^n=a^nr^n=0</math>; i.e. <math>ra=ar\in N</math>. Finally, suppose <math>a^n=0</math> and <math>b^m=0</math>. Since <math>a</math> and <math>b</math> commute, <math>(a+b)^{n+m}</math> is a </math>\mathbb{Z}</math>-linear combination of terms of the form <math>a^{n+m-k}b^k</math> for <math>0\leq k\leq n+m</math>. If <math>k\geq m</math>, then <math>b^m=0</math>, making the corresponding term vanish. If <math>k<m</math>, then <math>n+m-k>n</math>, so that <math>a^{n+m-k}=0</math>, again making the corresponding term vanish. Thus, <math>(a+b)^{n+m}=0</math>. Hence, <math>a+b\in N</math>. Thus, <math>N</math> is an ideal.
  
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Revision as of 08:24, 18 June 2013


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.



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