Question on extensions of discrete valuation fields

Question

Let $F$ be a discrete valuation field. Let $L$ be a finite extension of $F$. Let $L=F(\alpha)$ where $\alpha$ belongs to ring of integers of $L$, denoted by $O_L$. Is it always true that $O_L=O_F[\alpha]$? Under what conditions is it true? Is it true if both $L$ and $F$ are complete?

Answer 1

As Hurkyl suggested in the comments, suppose for some $\alpha$ you have $L = F(\alpha)$ and $O_L = O_F[\alpha]$. Let $\nu$ be the valuation on $O_L$. If $t \in O_F \subseteq F$ with $\nu(t) > 0$, then $O_K = O_F[\alpha] \supsetneq O_F[t \alpha']$ (do you see why?). However, $\frac{t \alpha}{\alpha} \in F^{\ast}$, so $F(\alpha) = F(t \alpha)$.

So it is never true that $O_L = O_F[\alpha]$ for all $\alpha$ which generate $L/F$, even when $F$ (and therefore $L$) is complete with respect to $\nu$. There is a criterion for which one can check whether $O_L = O_F[\alpha]$. If $g(X)$ is the minimal polynomial of $\alpha$ over $F$, then $O_L = O_F[\alpha]$ if and only if $\mathfrak D = g'(\alpha)O_L$. Here $\mathfrak D$ is the different of $O_L/O_F$, which is defined to be the inverse of the fractional ideal $\{ x \in L : Tr_{L/F}(x O_L) \subseteq O_F \}$. (4, Proposition 6, Frohlich's section of A.N.T.)