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Units and Nilpotents

If $a$ is a unit and $x$ is nilpotent, I'm trying to show that $a+x$ is a unit.

Pf.: If $a$ is a unit, there exists a non-zero invertible element $a^{-1}$ s.t. $a\cdot a^{-1} = 1$, and if $x$ is nilpotent then $x^n$ for some finite $n > 0$ is equal to zero.

So if $a+x$ is a unit, there exists a non-zero invertible element $(a+x)^{-1} = 1/(a+x)$ and dividing through by $a$, I get: $(1/a)/(1+x/a)$ but I am having some trouble expanding this power series it looks like.

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James R.
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2 Answers2

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Let's be "naive": what about $$\frac{1}{a+x}=\frac{1}{a}\frac{1}{1+\frac{x}{a}}=a^{-1}\left(1-\frac{x}{a}+\frac{x^2}{a^2}-\frac{x^3}{a^3}+...\right)$$ Well, now we just need to observe that the above series is...a finite one (can you see why?), and all the denominators are well defined as $\,a\,$ is a unit, so...

DonAntonio
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We can write $a + x$ as $a( 1+ xa^{-1})$ since $a$ is a unit. Notice that $(xa^{-1})$ is nilpotent. By the result in the link above we have that $1 + xa^{-1}$ is a unit. Now the product of two units is a unit so you are done.

You can see the following link too: Let $R$ be a commutative ring with 1 then why does $a\in N(R) \Rightarrow 1+a\in U(R)$?

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