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I can't tell where I went wrong, am still a beginner.

Let N be a nilpotent in a commutive ring , let X be any element.

I'll will be showing X + N is a unit .

Assume Y(X+N) =1

Then YX(N^n-1) = N^n-1

Multiplying both side by X+N

We get X(N^n-1) = X(N^n-1)
Thus nilpotent summed with any element gives unit.

user26857
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Zee
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    The claim is false and you assumed your conclusion in the first step of the 'proof'... – basket Oct 06 '16 at 22:50
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    Try $X = -N$, $X = 0$, and so on. The statement is true for $X = 1$. –  Oct 06 '16 at 22:51
  • Side question , if I assume something and the conclusions is wrong then the assumption is wrong. What about if the conclusion is right does that imply assumption is correct? – Zee Oct 06 '16 at 23:20
  • If your assumption is wrong, you can deduce anything from it :-) –  Oct 06 '16 at 23:21
  • Ya that's true. Thanks for the answers everyone. – Zee Oct 07 '16 at 00:58
  • It seems to me that people are misunderstanding my question , am not asking for the answers to the question , am asking what is wrong with my proof to make the obviously wrong statement above seem correct to me. – Zee Oct 07 '16 at 19:28
  • Also I came up with this argument which am not sure if it works or not: assume there is no unit Y , X is a unit and N is a nilpotent then YX + YN not= 1 which implies.YX(N^n-1) not= N^n-1 for all Y , a contradiction since X is a unit. – Zee Oct 07 '16 at 19:34

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I assume that you were mistaking the proposition and are really trying the show that the sum of a unit and a nilpotent is a unit. I can start you off with two methods. Let $x$ be nilpotent and let $u$ be a unit.

The first method is basic prealgebra.

Notice that $(x-1)(x^{n-1}+x^{n-2}+ \dots 1) = x^n-1$ in any commutative ring. If $n$ is a natural number such that $x^n = 0$...

The second method involves the characterization of nilpotents in terms of ideals. It is elementary that the set of nilpotents of a ring form an ideal and this ideal is the intersection of all prime ideals of the ring (a proof is in the first chapter of Atiyah-MacDonald for instance). So suppose that $u+x$ is not a unit and so let $\mathfrak{p}$ be a prime ideal that contains $u+x$. Since $x$ is nilpotent it is also contained in $\mathfrak{p}$ and we have [contradiction]...

basket
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  • It was originally about the sum of a unit and nilpotent and I did the above argument but then I realized that X does not need to be a unit so I must done something wrong but I couldn't see it. – Zee Oct 06 '16 at 23:16
  • but it does need to be a unit. – basket Oct 06 '16 at 23:17