If $|\Phi(A)|=|\Pi|$, then $O_{K}=o_{k}[A]$.

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Here is the problem: K/k is a finite algebraic extension, $\Pi$ is a prime element in K, $A\in O_{K}$, p,P is the maximal ideal of k and K.

We have that $\bar{A}:=A ~mod ~P$ generates the residue class field extension $F:=(O_{K}/P)/(o_{k}/p)$. Let $\phi(x)\in o_{k}/p$ be the minimum polynomial of $\bar{A}$ and $\Phi$ be a lift in $o_{k}$.

a) If $|\Phi(A)|=|\Pi|$, then $O_{K}=o_{k}[A]$.

b)$|\Phi(A)|\neq |\Pi|$, then $|\Phi(A+\Pi)|= |\Pi|$. Deduce that there is a $B\in O_{K}$ s.t. $1,B,...,B^{n-1}$ is a $o_{k}$-basis for $O_{K}$

Any suggestions?

Questions: Does $\bar{A}$ generate F mean $1,\bar{A},\bar{A}^{2},...,\bar{A}^{f}$, where $f=|W|$, is a basis for F?

Isn't $\Phi$ a minimum polynomial for A from Hansel's lemma? Then if yes, then I can show $\Phi$ is separable and so shows $O_{K}=o_{k}[A]$.

Thanks