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Would this proof work?

Let $N$ be a nilpotent in a commutative ring and let $X$ be a unit.

Let $Y$ be an element in the ring.

Proof by contradiction:

Assume

$$Y(X+N) \neq 1$$

Then

$$YX + YN \neq 1 $$

Then

$$YXN^{n-1} \neq N^{n-1} $$

A contradiction since since $Y$ can be the inverse of $X$ and thus the equation can hold above.

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  • $\begingroup$ I LaTeX-ified your solution and added the proof-verification tag. $\endgroup$ Oct 8, 2016 at 1:38
  • $\begingroup$ You shouldnt use capital letters for elements $\endgroup$ Oct 8, 2016 at 1:39
  • $\begingroup$ Thank you Carl, I have to learn Tex one day am just too technophobic... $\endgroup$
    – Zee
    Oct 8, 2016 at 2:16
  • $\begingroup$ @User375942 you may be satisfied with just knowing the answer to a question , am not. I like to actually solve it myself. $\endgroup$
    – Zee
    Oct 8, 2016 at 6:27
  • $\begingroup$ Take $Y = 0$ and then $Y \, (X+N) = 0$, and there's no contradiction. $\endgroup$
    – user144221
    Oct 8, 2016 at 12:38

2 Answers 2

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$u$ unit, $n$ nilpotent, $u+n=u(1+u^{-1}n)$, suppose $n^m=0$, $(u^{-1}n)^m=0$, so $n'=u^{-1}n$ is nilpotent $(1+n')(1+\sum_{i=1}^{i=m-1}(-1)^i{n'}^i)=1+(-1)^{m-1}{n'}^m)=1$, so $1+n'$ is invertible and $u+n=u(1+n')$ is invertible since it is the product of two invertible elements.

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I'll write out how I understand your reasoning, and then point out the mistake.

Let R be a ring, $u$ be a unit, $n$ nilpotent, with $n^k=0$. Suppose, that $y(u+n)\neq1\forall y\in R$. Then $y(u+n)n^{k-1}\neq n^{k-1} \forall y\in R$. A contradiction since $u$ is unit.

The problem with the above, is that you assume $a\neq b \Rightarrow ac\neq bc$, which is wrong. For example in $\mathbb Z/4\mathbb Z$ we have $1\neq3$, but $1\times2=3\times2$.

Here is an easy direct way. If $u$ is unit, $n$ nilpotent, with $uv=1, n^k = 0$, then it is easy to see that $v+\sum_{i=1}^{k-1}(-1)^i(uv)^i$ is the inverse of $u+n$.

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  • $\begingroup$ so the fault is in having contradictory assumptions? $\endgroup$
    – Zee
    Oct 8, 2016 at 6:32
  • $\begingroup$ I edited my answer, hope it helps. $\endgroup$
    – lanskey
    Oct 8, 2016 at 12:11
  • $\begingroup$ Thank you very much lanskey . Now am gonna put back Atiyah-Macdonald and return to analysis. $\endgroup$
    – Zee
    Oct 8, 2016 at 16:44

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