My Math Forum An exercise of Rodney Y. Sharp-Steps in Commutative Algebra

 Abstract Algebra Abstract Algebra Math Forum

 June 5th, 2016, 02:21 AM #1 Newbie   Joined: Jun 2016 From: HCM Posts: 2 Thanks: 0 An exercise of Rodney Y. Sharp-Steps 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 non-trivial commutative ring, and assume that, for each P € Spec(R), the localization Rp has no non-zero nilpotent element. Show that R has no non-zero 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: 144 Thanks: 16 I would try this: assume R has a non-zero 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. Thanks from adnb46
 June 18th, 2016, 05:00 AM #3 Newbie   Joined: Jun 2016 From: HCM Posts: 2 Thanks: 0 thank you very much. It's done.

 Tags algebra, commutative, exercise, rodney, sharpsteps

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Toncho1 Algebra 0 September 3rd, 2015 10:34 AM raul21 Abstract Algebra 1 May 23rd, 2014 03:25 AM mrwolf Algebra 2 March 6th, 2013 04:57 PM normvcr Abstract Algebra 2 January 16th, 2013 11:06 PM kimmi23 Algebra 2 July 16th, 2012 06:27 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top