Index Laws/Product of Indices/Field

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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:

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


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 Product of Powers of Group Elements:

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

$\Box$

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:

\(\ds \paren{\paren{0_F}^m}^n\) \(=\) \(\ds \paren{\paren{0_F}^0}^n\)
\(\ds \) \(=\) \(\ds \paren{1_F}^n\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds 1_F\)
\(\ds \) \(=\) \(\ds \paren{0_F}^0\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds \paren{0_F}^\paren{0 \cdot n}\)
\(\ds \) \(=\) \(\ds \paren{0_F}^\paren{m \cdot n}\)

$\Box$

Case 2: $m \ne 0, n = 0$

Let $m \ne 0$ and $n = 0$.

Hence:

$m n = 0$

We have:

\(\ds \paren{\paren{0_F}^m}^n\) \(=\) \(\ds \paren{0_F}^0\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds 1_F\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds \paren{0_F}^0\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds \paren{0_F}^\paren{m \cdot n}\)

$\Box$

Case 3: $m \ne 0, n = 0$

Let $m \ne 0$ and $n \ne 0$.

Hence:

$m n \ne 0$

We have:

\(\ds \paren{\paren{0_F}^m}^n\) \(=\) \(\ds \paren{0_F}^n\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds 0_F\) Definition of Power of Field Element
\(\ds \) \(=\) \(\ds \paren{0_F}^\paren{mn}\) Definition of Power of Field Element

$\Box$

In all cases:

$\paren{\paren{0_F}^m}^n = \paren{0_F}^\paren{mn}$

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

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

$\blacksquare$

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