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Prove $1+x$ is a unit in $R: \text{commutative ring}$ where $x$ is nilpotent

do I need to make use of a Taylor series expansion for this?

$(1+x)(1+x)^{-1} = 1 \implies (1+x)^{-1} = \displaystyle\frac{1}{1+x}$

and $x$ being nilpotent then for some positive integer n, $x^n=0$ I need to make use of this.

Or is there another way to do it?

oliverjones
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    In theory, you don't need to directly appeal to taylor series, you can just claim $(1-x+x^2-\ldots+(-1)^{n-1}x^{n-1})$ is an inverse and demonstrate it gives $1$ when multiplied by $(1+x)$. – Adam Hughes Mar 25 '15 at 04:43
  • The linked duplicate is a slight generalization, but you can find more exact duplicates in the dozen or so duplicates linked to that. Please use the search feature before asking a question. Searching for nilpotent unit already brings up several of the duplicates. – rschwieb Mar 25 '15 at 12:25

3 Answers3

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If $x^n=0$, then $$(1+x)\left(\sum_{k=0}^{n-1} (-1)^kx^k\right)=1+(-1)^{n-1}x^n=1.$$

AMPerrine
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Hint: $$(1+x)(1-x+x^2-\cdots\pm x^{n-1})=1\pm x^n=\cdots?$$

David
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You are exactly right about using the geometric series expansion. The point is the terms with high enough power are zero since $x$ is nilpotent, so it is actually a finite sum. Suppose $x^n=0$. Then $(1+x)(1-x+x^2+...+(-1)^{n-1} x^{n-1})=1$

ET93
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  • This is the answer that most helped me understand the problem, especially the phrase "The point is the terms with high enough power are zero since x is nilpotent". – blackbrandt Jul 25 '21 at 15:24