# 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 $x\neq 0$.

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

 $\displaystyle(1\!+\!x)\cdot\sum_{j=0}^{n-1}(-1)^{j}x^{j}$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{n-1}(-1)^{j}x^{j}\!+\!\sum_{k=0}^{n-1}(-1)^{k}x^{k+1}$ $\displaystyle=$ $\displaystyle\sum_{j=0}^{n-1}(-1)^{j}x^{j}\!-\!\sum_{k=1}^{n}(-1)^{k}x^{k}$ $\displaystyle=$ $\displaystyle 1\!+\!\sum_{j=1}^{n-1}(-1)^{j}x^{j}\!-\!\sum_{k=1}^{n-1}(-1)^{k}% x^{k}\!-\!(-1)^{n}x^{n}$ $\displaystyle=$ $\displaystyle 1\!+\!0\!+\!0$ $\displaystyle=$ $\displaystyle 1$

(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

 $\displaystyle\frac{1}{1\!+\!x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n},$

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

Title unity plus nilpotent is unit UnityPlusNilpotentIsUnit 2013-03-22 15:11:54 2013-03-22 15:11:54 Wkbj79 (1863) Wkbj79 (1863) 21 Wkbj79 (1863) Theorem msc 13A10 msc 16U60 DivisibilityInRings