= Solution
\b[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 $\beta=p\theta$. Then
$$
|\beta|=p^{-3/2},\qquad
|\beta-\theta|=|(p-1)\theta|=p^{-1/2},\qquad
|\beta+\theta|=|(p+1)\theta|=p^{-1/2}.
$$
Both $p-1$ 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.
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