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June 5th, 2016, 02:21 AM  #1 
Newbie Joined: Jun 2016 From: HCM Posts: 2 Thanks: 0  An exercise of Rodney Y. SharpSteps in Commutative Algebra
I'm reading Rodney Y. Sharp  Steps in Commutative Algebra and I have a problem with this exercise 5.34. Let R be a nontrivial commutative ring, and assume that, for each P € Spec(R), the localization Rp has no nonzero nilpotent element. Show that R has no nonzero nilpotent element. I was trying to prove that P={x∈R∃y∈R,y≠0,xy=0}∈Spec(R) But it looks like P is not a Group with + Can anyone give tell me if I'm going wrong? Or any hint?? And sorry for my bad English. Last edited by skipjack; June 5th, 2016 at 03:39 AM. 
June 9th, 2016, 06:22 AM  #2 
Senior Member Joined: Feb 2012 Posts: 139 Thanks: 15 
I would try this: assume R has a nonzero nilpotent element x. The ideal generated by x is included in a maximal (hence prime) ideal, say P, by Krull's theorem. Then the image of x by the cannonical ring morphism R>R_P is nilpotent in R_P.

June 18th, 2016, 05:00 AM  #3 
Newbie Joined: Jun 2016 From: HCM Posts: 2 Thanks: 0 
thank you very much. It's done.


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algebra, commutative, exercise, rodney, sharpsteps 
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