Proposition 3

Let A be a K-algebra admitting a finite basis. For an element $a \in A$ to be invertible, it is necessary and sufficient that its $N_{A/\mathbf{K}}(a)$ be invertible in $\mathbf{K}$.

If $a$ admits an inverse $a'$ in $A$, then

by formula (3) of no. 1. Conversely, if $N_{A/\mathbf{K}}(a)$ is invertible, the endomorphism $h : x \mapsto ax$ is bijective (§ 8, no. 2, Theorem 1). Then there exists $a' \in A$ such that $aa' = 1$; then $h(a'a - 1) = aa'a - a = (aa' - 1)a = 0$, whence $aa' = 1$ since $h$ is injective. Hence $a'$ is the inverse of $a$.