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Thread: [SOLVED] Seeking a better, more accurate and more elegant proof

  1. #1
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    [SOLVED] Seeking a better, more accurate and more elegant proof

    Let N be the set of all nilpotent elements of a commutative ring R, and show that N\subseteq P for each prime ideal P of R.
    Proof: Let p\in P and a\in N. Then there is a positive integer m\geq 2 with a^m=0. Assume towards a contradiction that a\notin P. Since a^{m-1}a=a^m=0=0p\in P and a\notin P, then by definition of prime ideals we have a^{m-1}\in P. However, the same can be said for any integer n\geq 2, that is, if a^n\in P then a^{n-1}a\in P and therefore a^{n-1}\in P. By induction, since a^m\in P, then a\in P, which contradicts our assumption that a\notin P. So a\in P whenever a\in N, that is, N\subseteq P. \blacksquare

    The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

    Thanks!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    Proof: Let p\in P and a\in N. Then there is a positive integer m\geq 2 with a^m=0. Assume towards a contradiction that a\notin P. Since a^{m-1}a=a^m=0=0p\in P and a\notin P, then by definition of prime ideals we have a^{m-1}\in P. However, the same can be said for any integer n\geq 2, that is, if a^n\in P then a^{n-1}a\in P and therefore a^{n-1}\in P. By induction, since a^m\in P, then a\in P, which contradicts our assumption that a\notin P. So a\in P whenever a\in N, that is, N\subseteq P. \blacksquare
    The reasoning is sound, but I'm not quite sure how to best articulate it. I don't think "by induction" is completely appropriate, here. Also, I like to avoid by-contradiction proofs if possible. Does anyone have an idea for how I can more elegantly and accurately explain what's going on in this exercise?

    Thanks!

    Why did you have to use contradiction at all? Let a be any nilpotent, then a^m=0\in P\Longrightarrow a\in P (as you showed this is fine)...and voila! No contradiction and short.

    Tonio
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  3. #3
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    If you don't like that "finite induction" argument, you can derive the result about nilpotent elements as a corollary of a more general result:
    If a product a_1a_2\cdots a_n lies in a prime ideal P, then at least one of the elements a_i belongs to P.
    That result is easily proved, for all n\geqslant2, by a (standard) induction argument, the base case n=2 being just the definition of primality. The result about nilpotents follows by taking all the a_i to be equal to a.
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