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# 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 similarity between this proof and geometric series: The goal was to produce a multiplicative inverse of $1\!+\!x$, and geometric series yields that

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

Related:

DivisibilityInRings

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## Mathematics Subject Classification

13A10*no label found*16U60

*no label found*

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## Corrections

proof by CWoo ✓

change display by Mathprof ✓

take a look at by Mathprof ✓

delete commutative by Mathprof ✓

proof supplied by Wkbj79 ✓

change display by Mathprof ✓

take a look at by Mathprof ✓

delete commutative by Mathprof ✓

proof supplied by Wkbj79 ✓