# Unit + nilpotent = a unit [duplicate]

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.

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.

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

$u$ 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.

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 $u+n$.

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