 Integral trace need not imply integrality

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.