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Let $L/K/\mathbb Q_p$ be finite extensions of local fields and let $v_L$ and $v_K$ be normalised discrete valuations on $L$ and $K$ respectively.

My question is quite a general one:

If the valuation rings are given by $\mathcal O_L=\mathcal O_K[\alpha]$ for some $\alpha\in \mathcal O_L$, then is it true that $L=K(\alpha)$ as fields?

Can this also be true in the case where $\mathcal O_K$ is a Dedekind domain, $K$ its quotient field, $L$ a finite separable extension of $K$ and $\mathcal O_L$ the integral closure of $\mathcal O_K$ in $L$?

Thank you for your help!

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Yes it is.

$L$ is the fraction field of $\mathcal{O}_L$, so it certainly contains $\mathcal{O}_K$, so it contains $K$, and it contains $\alpha$, so $L\supset K(\alpha)$.

Conversely, $K(\alpha)$ contains $\mathcal{O}_K$, and it contains $\alpha$, so it contains $\mathcal{O}_L$, so it contains $Frac(\mathcal{O}_L) = L$.

Note that this doesn't use anything about the existence of a discrete valuation. It only relies on the fact that $L = Frac(\mathcal{O}_L)$ and $K = Frac(\mathcal{O}_K)$.

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