Group of Units of Field

Theorem

Let $k^\times$ denote the group of units of a field $\struct {k, +, \times}$.

Then:

$k^\times = k \setminus \set 0$

where:

$\setminus$ denotes set difference

$\set 0$ denotes the set containing only the zero of $k$.

That is, $0$ is the only element of $k$ which does not have a multiplicative inverse in $k$.

Proof

$0$ is not invertible in $k$ since $0 a = 0$ for all $a \in k$.

Thus $0 \notin k^\times$.

Consider now $a \in k$ such that $a \ne 0$.

From Field Axiom $\text M4$: Inverses for Product it follows that there exists $a^{-1}$ such that:

$a \times a^{-1} = 1 = a^{-1} \times a$

where $1$ is the unity of $k$.

So:

$a \in k^\times$

Hence we have demonstrated that $k^\times$ equals $k$ without $0$, that is:

$k^\times = k \setminus 0$

$\blacksquare$