# 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:

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

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

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

$=$ | $1+0+0$ | |||

$=$ | $1$ |

(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 .

Title | unity plus nilpotent is unit |
---|---|

Canonical name | UnityPlusNilpotentIsUnit |

Date of creation | 2013-03-22 15:11:54 |

Last modified on | 2013-03-22 15:11:54 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 21 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 13A10 |

Classification | msc 16U60 |

Related topic | DivisibilityInRings |