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In $L={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the Eisenstein polynomial $X^2-p$ gives $v_L(\theta )=1$ and $v_L(p)=2$. The element $x=\theta /p$ has field trace zero, but valuation $-1$, so is not in the valuation ring. Cancellation in a field trace contrasts with the multiplicative test in unit norm detects units in a local extension.

True. Each element of the Galois group preserves the extending absolute value on a field, because composing it with that absolute value on a field gives another extension, equal to the original by part (a). Thus all conjugates of $x$ have the same absolute value on a field. Writing $d=[K:{\mathbb{Q}}_p]$, the field norm givesA unit field norm has absolute value on a field one, forcing $|x|=1$. This means both $x$ and $x^{-1}$ are in the valuation ring, so $x$ is a unit.

Equivalently, with normalized discrete valuations and residue degree $f$,This is unit norm detects units in a local extension.