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Ring Theory - a Telescoping Problem

Date: 01/20/2002 at 16:30:59
From: VAH
Subject: Abstract Algebra - Ring Theory

An element a in a ring R is said to be nilpotent if there exists a 
positive integer n such that a^n = 0.

Show that if a is nilpotent, a-1 is a unit. 

a-1 is a unit if there is an element b such that (a-1).b = 1,

I tried various approaches. One was to rewrite a^n = 0 as 
a.(a^(n-1)) = 0, but I don't seem to be able to make progress.
I'm not sure if I should approach it this way:

              a^n = 0, 
 so       a^n - 1 = -1
 and    a^n - 1^n = -1, 

then write it in the form

   (a-1)(a^(n-1) - 1^(n-1)...

I would appreciate any assistance.
Thank you,

Date: 01/20/2002 at 23:49:22
From: Doctor Paul
Subject: Re: Abstract Algebra- Ring Theory

Hi Veronica.  Thanks for writing to Ask Dr. Math.

Your efforts are valiant, but they're not going to lead to a solution.

Sometimes it's best to approach a problem like this by actually 
constructing an inverse manually. It helps to be able to recognize a 
telescoping problem when you see one.

Let's suppose n = 3.

Then we want:

  (a - 1) * (             ) = 1

Surely we must use the fact that a^3 = 0. So what if we 
try this:

  (a - 1) * (a^2             ) = 1

This gives a^3 - a^2 = 1.

This is fine, but it adds an extra -a^2 term, and our desired 1
is nowhere in sight.  Can we get rid of the -a^2?  I think we can.  

  (a - 1) * (a^2 + a            ) = 1

Now we have:

  a^3 + a^2 - a^2 - a = 1

              a^3 - a = 1

This is fine, but it adds an extra -a term, and we still don't have
the 1 we want.  Can we get rid of the extra -a?  Yes:

      (a - 1) * (a^2 + a + 1) = 1

  a^3 + a^2 + a - a^2 - a - 1 = 1

                      a^3 - 1 = 1

Well, we're close. We got -1 instead of 1 so all we need to do is 
multiply everything by negative one and we'll be done:

  (a - 1) * (-a^2 - a - 1) = 1

So this works for n = 3.

This is called a "telescoping" series because all of the terms vanish 
except the first and the last one. And in this special case, the first 
term also vanishes. And this leaves just the desired 1.

Do you see that in the general case, the inverse of (a - 1) will be:

  (-a^[n-1] - a^[n-2] - .... - a^2 - a - 1).

I hope this helps.  Please write back if you'd like to talk about this 

- Doctor Paul, The Math Forum   

Date: 01/21/2002 at 08:51:55
From: VAH
Subject: Abstract Algebra - Ring Theory

Doctor Paul,

Thank you so much for your explanation. I was surprised by the speed 
of the response too. 

Thanks again,
Associated Topics:
College Modern Algebra

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