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.
$\blacksquare$