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Nilpotent ring

  1. Jan 20, 2010 #1
    I have this question and its a combination of the binomial theorem and nilpotent elements within a ring.

    Suppose the following, am=bn=0. Is it necessarily true that (a+b)m+n is nilpotent.

    For this question I did the following:

    [tex]\sum[/tex]i=0m+n[tex]\binom{m+n}{i}[/tex]am+n-ibi

    If i=m, then a=0. Additionally, if i>m a=0.

    That's actually as far as I've gotten.
     
    Last edited: Jan 20, 2010
  2. jcsd
  3. Jan 20, 2010 #2

    Dick

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    I think you are actually asking whether (a+b)^(m+n)=0. Is that right? If a^m=0, then a^(m+1)=0, a^(m+2)=0 etc etc. Similar for b. All of the terms in your binomial expansion have the general form i*a^k*b^l where (k+l)=(m+n). Is it possible k<m AND l<n?
     
  4. Jan 21, 2010 #3
    So I was working on this today during a lecture.

    (a+b)m+n

    Now we go to some arbitrary term in the middle:

    am+n-ibi

    From here we can notice the following things:
    i>n and i=n
    If this is true, then we know that b=0 and the whole thing equals zero.
    i<n
    If this is true, then we know that a=0 from the following:
    If i<n then we know i[tex]\leq[/tex]n-1. Then with i subbed in, m+n-(n-1) which equals m+1.

    This means that, by substitution, m+1. That makes a=0.

    Done?
     
  5. Jan 21, 2010 #4

    Dick

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    That's more than a little confusing. I mean, you aren't proving a=0 or b=0, you are proving powers of a and b are zero, right? But yes, I think you've got the right idea. You could just state it a lot more clearly.
     
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