Ring problem: nilpotent elements and units

rachellcb
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Homework Statement


Let R be a ring with multiplicative identity. Let u \in R be a unit and let a1, ..., ak be nilpotent elements that commute with each other and with u. To show: u + a1 + ... + ak is a unit.

The Attempt at a Solution


Need to show that u'(u + a1 + ... + ak)=1 for some u' \in R.
I know that 1 - ai is a unit (i=1,2,...,k) since each ai is nilpotent.
So I'm trying to think of possible candidates for u'.
My attempt so far have been to try and cancel the ai terms by multiplying
u + a1 + ... + ak by a1n-1, where a1n =0 and so on for all all a_i but this leaves me with ua1n-1...akm-1 and doesn't seem to be leading anywhere...
Any suggestions?
 
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Show first that the sum of two nilpotent elements which commute with each other is nilpotent (Hint: Generalize the binomial theorem to an arbitrary ring). Then use induction to show that the sum of k nilpotent elements which commute with each other is nilpotent. Then lastly prove that the sum of a nilpotent element and a unit is a unit (Hint: This is similar to proving 1+x is a unit, where x is nilpotent).
 
Solved, thanks! :smile:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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