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Let $R$ be a ring and $x \in R$.

Prove that if $x$ is nilpotent then $1-x \in R$ is unit.

May I use the fact that $$ \frac{1}{1-x} = 1 + x + x^2 + \ldots $$ and then say that since $x$ is nilpotent, there exist $n$ such that $x^n=0$, and of course $x^{n+1}=0$, and therefore this sum is finite and $1/(1-x)$ is a well defined element.

will that prove compile ?

gebruiker
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Philip L
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    The other answers have discussed the correctness of your solution, so I'll discuss what is incorrect about it.

    When you write $$1+x+x^2 + \dots,$$ the "$\dots$" implies there is some notion of a limit or distance in your ring, which is generally not the case.

     – user217285 Mar 20 '16 at 23:45
  • I'm aware of the syntax problem with this proof, just seems hard to me to find the multiplicative element of 1-x without thinking of 1/(1-x) as geometric series, which is true only for |X| < 1, how may i find and think of finding this multiplicative from other views ? – Philip L Mar 20 '16 at 23:51
  • @mathftw In a general ring you may not even have absolute value, divisions as you wrote, etc. – DonAntonio Mar 20 '16 at 23:53
  • I find it a useful heuristic to find the right expression to check, but you still need to validate the resulting expression with the definition of a unit. – Derek Elkins left SE Mar 21 '16 at 00:18

2 Answers2

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$(1-x)(1+x..+x^n)=1$ if $x^{n+1}=0$.

Tsemo Aristide
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Better, find the multiplicative element of $\;1-x\;$, which is what must be done when dealing with these things in ring theory. Suppose $\;x^n=0\;$ , then

$$(1-x)(1+x+\ldots+x^{n-1})=1-x^n=1$$

DonAntonio
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