Nilpotent Ring Element plus Unity is Unit

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Theorem

Let $A$ be a ring with unity.

Let $1 \in A$ be its unity.

Let $a \in A$ be nilpotent.


Then $1 + a$ is a unit of $A$.


Proof

Because $a$ is nilpotent, there exists a natural number $n > 0$ with $a^n = 0$.

By Sum of Geometric Sequence in Ring:

$\paren {1 + a} \cdot \ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k = 1 + \paren {-a}^n$
$\paren {\ds \sum_{k \mathop = 0}^{n - 1} \paren {-a}^k} \cdot \paren {1 + a} = 1 + \paren {-a}^n$

where $\sum$ denotes summation.

By Negative of Nilpotent Ring Element:

$\paren {-a}^n = 0$

Thus $1 + a$ is a unit.

$\blacksquare$


Also see