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Nilpotent ring

  • #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:

Answers and Replies

  • #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?
 
  • #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?
 
  • #4
Dick
Science Advisor
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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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