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If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.

I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?

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Dear Shannon, Try the case $u = 1$ first. Regards, –  Matt E Mar 14 '12 at 2:23
    
See also here. –  Bill Dubuque May 3 '12 at 14:43
    
Bill's link, cleaned up. –  t.b. May 4 '12 at 9:59

2 Answers 2

up vote 4 down vote accepted

If $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\cdots + (-1)^{n}a^n) = 1 + (-1)^{n+1}a^{n+1}$$ by selecting $n$ large enough.

If $uv=vu=1$, does $a$ commute with $v$? Is $va$ nilpotent?

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I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a? –  Shannon Mar 14 '12 at 2:51
    
@Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent? –  Arturo Magidin Mar 14 '12 at 2:56
    
oh, I see. I will work on it. Thank you so much. –  Shannon Mar 14 '12 at 3:10

Let $v$ be the inverse of $u$, and suppose $a^2=0$. Note that $$(u+a)\cdot v(1-va)=(1+va)(1-va)=1-v^2a^2=1-0=1.$$ See if you can generalize this.

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The way you constructed the element is wonderful. I've never thought about this way. Can I ask where did this idea come from? –  Shannon Mar 14 '12 at 2:55

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