# Units and Nilpotents

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
Bill's link, cleaned up. – t.b. May 4 '12 at 9:59

If $u=1$, then you could do it via the identity $$(1+a)(1-a+a^2-a^3+\cdots + (-1)^{n}a^n) = 1 + (-1)^{n}a^{n+1}$$ 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
This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises. – Derek Elkins Jan 21 at 4:01

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

Here's a rather different argument. First, suppose that $R$ is commutative. Suppose $u+a$ is not a unit. Then it is contained in some maximal ideal $M\subset R$ (unless $R$ is the zero ring, but I'll leave that case to you). Since $a$ is nilpotent, $a\in M$ (since $R/M$ is a field, and any nilpotent element of a field is $0$). Thus $u=(u+a)-a\in M$ as well. But $u$ is a unit, so it can't be in any maximal ideal, and this is a contradiction.

If you don't know that $R$ is commutative, let $S\subseteq R$ be the subring generated by $a$, $u$, and $u^{-1}$. Then $S$ is commutative: the only thing that isn't immediate is that $u^{-1}$ commutes with $a$, and this this can be proven as follows: $$u^{-1}a=u^{-1}auu^{-1}=u^{-1}uau^{-1}=au^{-1}.$$

The argument of the first paragraph now shows that $u+a$ is a unit in $S$, and hence also in $R$.

[The use of a maximal ideal (and hence the axiom of choice) in this answer can be avoided as follows. First, show that if $I$ is a proper ideal in a commutative ring and $a$ is nilpotent, then $I+(a)$ is still a proper ideal. Now take $I=(u+a)$ and let $M=I+(a)$, and argue as above. Note also that while this argument appears nonconstructive, it can be made constructive, and it is a fun and somewhat tricky exercise to chase through the proof and see how it can give an explicit inverse to $u+a$.]

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