Nilpotent ring

1. Jan 20, 2010

tom.young84

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:

$$\sum$$i=0m+n$$\binom{m+n}{i}$$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. Jan 20, 2010

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?

3. Jan 21, 2010

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:

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$$\leq$$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?

4. Jan 21, 2010

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.