Theorem 6.1. Assuming that:

Then

(i) $| ∙ |$ extends uniquely to an absolute value ${| ∙ |}_{L}$ on $L$ defined by
${| y |}_{L} = {| N_{L ∕ L} (y) |}^{\frac{1}{n}} \forall y \in L .$
(ii) $L$ is complete with respect to ${| ∙ |}_{L}$ .