Group of Units of Field
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Theorem
Let $k$ be a field.
Then $k^\times = k \setminus \set 0$
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 $a \in k^\times$.
$\blacksquare$