Binomial Theorem & Nilpotent Elements in a Ring: Is (a+b)m+n Nilpotent?

[tom.young84]

I have this question and its a combination of the binomial theorem and nilpotent elements within a ring.

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

For this question I did the following:

$$\sum$$_i=0^m+n$$\binom{m+n}{i}$$a^m+n-ibⁱ

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

That's actually as far as I've gotten.

 

[Dick]

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?

 

[tom.young84]

So I was working on this today during a lecture.

(a+b)^m+n

Now we go to some arbitrary term in the middle:

a^m+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$$\leq$$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?

 

[Dick]

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.