1

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.

user26857
  • 53,190
Zee
  • 331

2 Answers2

2

$u$ 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.

Tsemo Aristide
  • 89,587
1

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 $u+n$.

lanskey
  • 1,106