Group of Units of Field

Theorem
Let $k$ be a field, then $k^\times=k\setminus\{0\}$

Proof
$0$ is not invertible in $k$ since $0a=0$ for all $a\in k$, thus $0\not\in k^\times$

Consider now $0\ne a\in k$, from the axioms of fields it follows that there exists $a^{-1}$ and thus $a\in k^\times$.