 Solution

False. Use the polynomial $X(X^2-p)$, with splitting field $K={\mathbb{Q}}_p(\theta )$, ${\theta }^2=p$, and label the polynomial roots $0,\theta ,-\theta$. Take $\beta =p\theta$. ThenBoth $p-1$ and $p+1$ are p-adic units, also when $p=2$. Hence zero is the uniquely closest polynomial root, but $\beta \notin {\mathbb{Q}}_p(0)={\mathbb{Q}}_p$, because $\theta \notin {\mathbb{Q}}_p$. The polynomial need not be irreducible in the given setup. The inclusion established by Krasner's lemma goes in the direction in part (iii), not its converse.