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I recently had an algebra exam and one of its questions was:

Let $R$ be a ring with a $1$ and let $x\in R$ be nilpotent, i.e., there exists a positive integer $n$ such that $x^n=0$. Then $1+x$ is a unit.

This is what I did: Observe that

$$\begin{align*}\left(1+x\right)\left(1-x\right)&=1-x^2,\\\left(1-x^2\right)\left(1+x^2\right)&=1-x^4,\\\left(1-x^4\right)\left(1+x^4\right)&=1-x^8,\end{align*}$$

and so on. Eventually, we will reach $1-x^\alpha$, where $\alpha\geqslant n$, which would imply that $1-x^\alpha=1$. Therefore, $1+x$ is a unit.

Is this sound? Is there a simpler way to show this?

user26857
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wjmolina
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    Might be related to http://math.stackexchange.com/questions/119904/units-and-nilpotents?rq=1 – Kevin Quirin Dec 16 '15 at 08:17
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    A unit is a divisor of $1$ and the chain of equations in post shows $1+x$ divides $1-x^2,$ which in turn divides $1-x^4$ etc. and eventually one reaches a term that it divides which is of form $1-x^\alpha$ with $\alpha \ge n$ since $x^n=0$ this last term is in fact $1$ so the chain implies that $1+x$ divides $1$ and so is a unit. So it looks like the argument of post is fine. – coffeemath Dec 16 '15 at 08:23

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Note that if $m > n$, then $x^{m} = x^{m-n} \cdot x^{n} = x^{m-n} \cdot 0 = 0$. Pretend you are in a calculus class, how would you write a series for $\frac{1}{1 + x}$. Usually, this somebody's name series expansion will be possible if $|x| < 1$, here the condition that $x$ is nilpotent will play that role.

Write $b = \sum_{k=0}^{n-1}(-1)^{k}x^{k}$.

Then $b \in R$ and calculations show that $(1 + x)b = 1$, which shows that $1 + x$ is invertible (is a unit).

Daniel Akech Thiong
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