Group of Units of Field

Theorem
Let $k$ be a field.

Then $k^\times = k \setminus \left\{ {0}\right\}$

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

Thus $0 \notin k^\times$.

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

From the axioms of fields it follows that there exists $a^{-1}$.

Thus $a \in k^\times$.