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Is ring of integers of $\mathbb{Q}_p(α) $ is $\mathbb{Z}_p[α] $ ?

If $α$ is primitive root of unity, dependentable reference reads that the titled statement holds. But in general, does the statement hold?

My try: In local fields, rings of integers is the same thing as valuation ring(I don't know hot to prove this). So, there are denominator has no prime factor $p$, so the integer ring is $\mathbb{Z}_p[α] $.

Pont
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  • What if $\alpha=p^{(p+1)/p}$? – Aphelli Oct 05 '21 at 20:49
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    Even easier, note that $\mathbb Q_p(\alpha) = \mathbb Q_p(p^k \alpha)$ for every $k \in \mathbb Z$, but certainly not $\mathbb Z_p[\alpha] = \mathbb Z_p[p^k \alpha]$ in general. (So take $\alpha = p^{-1}\cdot \zeta_n$ for a counterexample. Or maybe most basic, $\alpha= p^{-1}$.) – Torsten Schoeneberg Oct 05 '21 at 21:23

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