unity plus nilpotent is unit

Theorem.

If $x$ is a nilpotent element of a ring with unity 1 (which may be 0), then the sum $1+x$ is a unit of the ring.

Proof.

If $x=0$, then $1+x=1$, which is a unit. Thus, we may assume that $x\ne 0$.

Since $x$ is nilpotent, there is a positive integer $n$ such that $x^n=0$. We multiply $1+x$ by another ring element:

$(1+x)\cdot {\displaystyle \sum _{j=0}^{n-1}}{(-1)}^j{x}^j$$=$${\displaystyle \sum _{j=0}^{n-1}}{(-1)}^j{x}^j+{\displaystyle \sum _{k=0}^{n-1}}{(-1)}^k{x}^{k+1}$

$=$${\displaystyle \sum _{j=0}^{n-1}}{(-1)}^j{x}^j-{\displaystyle \sum _{k=1}^n}{(-1)}^k{x}^k$

$=$$1+{\displaystyle \sum _{j=1}^{n-1}}{(-1)}^j{x}^j-{\displaystyle \sum _{k=1}^{n-1}}{(-1)}^k{x}^k-{(-1)}^n{x}^n$

$=$$1+0+0$

$=$$1$

(Note that the summations include the term  ${(-1)}^0{x}^0$, which is why $x=0$ is excluded from this case.)

The reversed multiplication gives the same result. Therefore, $1+x$ has a multiplicative inverse and thus is a unit.∎

Note that there is a this proof and geometric series: The goal was to produce a multiplicative inverse of $1+x$, and geometric series yields that

$$\frac{1}{1+x}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^n,$$

provided that the summation converges (http://planetmath.org/AbsoluteConvergence). Since $x$ is nilpotent, the summation has a finite number of nonzero terms and thus .