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June 5th, 2016, 02:21 AM   #1
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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.
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June 9th, 2016, 06:22 AM   #2
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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.
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June 18th, 2016, 05:00 AM   #3
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thank you very much. It's done.
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