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Homework Help: Elementary Number Theory Proof, Integral Ideals

  1. Feb 14, 2010 #1
    The first part of the problem is as follows:
    Any nonempty set of integers J that fulfills the following two conditions is called an integral ideal:
    i) if n and m are in J, then n+m and n-m are in J; and
    ii) if n is in J and r is an integer, then rn is in J.

    Let Jm be the set of all integers that are integral multiples of a particular integer m. Prove that Jm is an integral ideal.

    Second part:
    Prove that every integral ideal J is identical with Jm for some m. (Hint: Prove that if J [tex]\neq[/tex]{0}=J0, then there exists positive integers in J. By the least-integer principles, there is a least positive integer in J, say m. Then prove that J=Jm

    For the first part, I'm not really sure if there is not much to it, or maybe I'm just missing something. Either way, I am not really sure how to put it down on paper. If m is in J, and Jm is simply some integer multiple of m, rm. Then it automatically satisfies the ii. For part i, n and m can be factored into (r1+r2)m or (r1-r2)m. Either way this is still an integer multiple of m, which is in J. Not really sure if this is right though...

    For the second part I am a bit more lost. How do I prove that there are positive integers in J? When you go to prove that J=Jm, I think you have to use Euclid's division lemma and the concept that the gcd will divide every integer multiple of J. Still not really sure how this all ties in though...

    Any help would be greatly appreciated! Thanks!
  2. jcsd
  3. Feb 14, 2010 #2


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    I think your problem with the first part is you're confusing distinct variables by giving them the same letter, and distinguishing between two variables which are supposed to be the same.

    In particular, the statement "Jm is an integral ideal" requires you to substitute Jm for J in that definition. Also, the m in Jm has nothing to do with the m in that definition, so you should probably change the dummy variable to help unconfuse yourself.
  4. Feb 14, 2010 #3
    this is how you should think about this proof. you want to show that for X={mr:r is integer for fixed integer m}. thus if x and y are two elements in X and k is an integer we have ax - y = a(mr) - mr'=m(ar-r') so ax-y is in X. this is a short cut which is equivalent to checking the two independently. the second part i think is refering to the generator of an ideal.
  5. Feb 14, 2010 #4
    use the division algorithm for the second part: if x is in J then x=am +r when 0<=r<m. now r = x-am which obviously belongs to J since both x and am do. if x is the smallest positive element in J which exists (since you can multiply any negative number in J by a negative integer) then that means that r=0. therefore x is in Jm. meaning J is a subset of Jm, the reverse proof is trivial.
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