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Let $K$ be a field with respect to a valuation and $L$ be a finite extension. For simplicity, assume $K$ is complete. From theory, valuation on $K$ extends to $L$ and the extended valuation on $L$ is also complete. Let $\kappa_K, \pi_K$ and $\kappa_L, \pi_L$ denotes the residue class field and prime element of $K$ and $L$ respectively.

We know that $\kappa_L$ is finite extension of $\kappa_K$, say $\kappa_L = \kappa_K(\overline{\omega_1},...,\overline{\omega_f})$. Further assume that the $\omega_i$ are integral in $\kappa_L$. Let $e = \frac{[L:K]}{f}$ be the ramification index. I know that the collection $$\{\omega_i \pi_L^j : 1 \leq i \leq f, 0 \leq j \leq e - 1\}$$ is a basis for $L|K$. That means $$E = K(\omega_1, ..., \omega_f)$$ is an intermediate subfield. Is it true that the residue class field $\kappa_E$ of $E$ is the same as $\kappa_L$ as it contains all representatives for $\kappa_L$? In case it is not, is there situation where this can be guaranteed?

EDIT: I mistakenly believed that degree of $E/K$ to be $f$.

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Yes it does. See for example Neukirch's proof in his book on Algebraic number theory. You are missing a step where you use Hensel's lemma to lift the $\omega_i$ to $\overline{\omega_i}\in L$, but that's a minor issue. In fact you have rediscovered the classical bit of ramification theory which deals with writing each extension as first an unramified extension then a totally ramified extension. In the case where you have number fields, this is also used in a further decomposition of the extension into a tower of extensions including the base field, the decomposition field, the inertia field, and then the full field extension.

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