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For a finite extension of fields complete for a Non-Archimedean absolute value, use the unique extended field absolute value. The extension is unramified when its residue extension is a separable field extension andEquivalently, it has ramification index one, a separable field extension of residue fields, and no defect. For general nondiscrete valued fields, merely saying that the ramification index is one is insufficient; the displayed degree equality is part of the unramified condition.

Put $n=[L:K]=[k_L:k_K]$. Since $\bar{x}$ generates the residue extension, $1,\bar{x},\dots ,{\bar{x}}^{n-1}$ are linearly independent over $k_K$. Their lifts $1,x,\dots ,x^{n-1}$ are linearly independent over $K$: a nonzero relation could be divided by a coefficient of largest field absolute value, giving an integral relation whose reduction has at least one nonzero coefficient, contrary to residue independence. They therefore form a $K$-basis of $L$.

For any $y={\sum }_{j=0}^{n-1}{a}_j{x}^j$, choose $a_r$ of largest field absolute value. All coefficients of $y/a_r$ belong to ${\mathcal{O}}_K$, and at least one is a unit. Its reduction is a nonzero linear combination of the residue basis, so $y/a_r$ is a unit of ${\mathcal{O}}_L$. This proves the residue-basis norm formula for an unramified extensionHence $y\in {\mathcal{O}}_L$ exactly when every $a_j\in {\mathcal{O}}_K$. We concludeThe last equality also uses the fact that every polynomial in the integral element $x$ is integral. The proof requires neither a uniformizer nor discrete valuation, and applies to every residue generator specified in the question.