= Solution
\b[False.] This is an example of <integral trace need not imply integrality>. Take $K=\mathbb Q_p(\theta)$ with $\theta^2=p$, the <splitting field> of $X^2-p$. Its two conjugates are $\theta,-\theta$, and its normalized <discrete valuation> satisfies $v_K(\theta)=1$, $v_K(p)=2$, by the <Eisenstein polynomial> criterion. For
$$
x=\frac{\theta}{p}
$$
the <field trace> is zero, but $v_K(x)=-1$. Therefore $x$ is not in the <valuation ring>, despite having integral <field trace>. Cancellation in a sum of conjugates explains why a <field trace> test fails.
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