Powers of Field Elements Commute
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Theorem
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.
Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.
Then:
- $(a):\quad \forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^n \circ a^m$
- $(b):\quad \forall a \in F : \forall n, m \in \Z_{\ge 0} : a^m \circ a^n = a^n \circ a^m$
Proof
Statement $(a)$
By Definition of Field:
- $\struct{F^*, \circ}$ is an Abelian group
By Definition of Power of Field Element:
- For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$
From Powers of Group Element Commute:
- $\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^n \circ a^m$
$\Box$
Statement $(b)$
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.
For $a \in F^*$, $(b)$ follows from $(a)$.
It remains to show that $(b)$ holds for $0_F$.
Case 1: $m = 0$
Let $m = 0$.
We have:
| \(\ds \paren{0_F}^m \circ \paren{0_F}^n\) | \(=\) | \(\ds \paren{0_F}^0 \circ \paren{0_F}^n\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1_F \circ \paren{0_F}^n\) | Definition of Power of Field Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n\) | Field Axiom $\text M3$: Identity for Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n \circ 1_F\) | Field Axiom $\text M3$: Identity for Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n \circ \paren{0_F}^0\) | Definition of Power of Field Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n \circ \paren{0_F}^m\) |
$\Box$
Case 2: $m \ne 0$
Let $m \ne 0$.
We have:
| \(\ds \paren{0_F}^m \circ \paren{0_F}^n\) | \(=\) | \(\ds 0_F \circ \paren{0_F}^n\) | Definition of Power of Field Element | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0_F\) | Field Product with Zero | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n \circ 0_F\) | Field Product with Zero | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren{0_F}^n \circ \paren{0_F}^m\) | Definition of Power of Field Element |
$\Box$
In both cases:
- $\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^n \circ \paren{0_F}^m$
Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
- $\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^n \circ \paren{0_F}^m$
$\blacksquare$