Index Laws/Sum of Indices/Field

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:

$(a):\quad \forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$

$(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : a^m \circ a^n = a^\paren{m + n}$

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 Sum of Powers of Group Elements:

$\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$

$\Box$

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

\(\ds \)

\(=\)

\(\ds \paren{0_F}^\paren{0 + n}\)

\(\ds \)

\(=\)

\(\ds \paren{0_F}^\paren{m + n}\)

$\Box$

Case 2: $m \ne 0$

Let $m \ne 0$.

Hence:

$m + n \ne 0$

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

\(\ds \)

\(=\)

\(\ds \paren{0_F}^\paren{m + n}\)

Definition of Power of Field Element

$\Box$

In both cases:

$\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$

Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:

$\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$

$\blacksquare$