Unramified generator criterion with separable reduction

A finite extension of complete discretely valued fields is unramified exactly when it has an integral generator whose reduced minimal polynomial of an algebraic element is separable. In one direction, lift a residue-field primitive element. In the other, the factorization form of Hensel's lemma makes the separable reduction irreducible, forcing residue degree to equal the entire extension degree.

For an unramified extension of local fields $L/K$, an integral element whose residue generates $k_L/k_K$ generates $L/K$: its degree is at least the residue degree $f=[L:K]$. The reduced powers $1,\bar{\alpha },\dots ,{\bar{\alpha }}^{f-1}$ are a basis of the residue field extension. If $x=\sum c_j{\alpha }^j$, factor out a coefficient of least valuation. Reduction of the remaining sum is nonzero by that residue basis, so $v_L(x)={\min }_j{v}_K(c_j)$. Consequently $x$ is integral exactly when all $c_j$ are integral.

Let $L/K$ be a finite unramified extension of complete fields with a Non-Archimedean absolute value, with separable residue extension and $[L:K]=[k_L:k_K]=n$. If $x$ is integral and its residue generates $k_L/k_K$, then $1,x,\dots ,x^{n-1}$ form a $K$-basis. Divide any nonzero combination by a coefficient of largest absolute value. Its remaining coefficients are integral, at least one is a unit, and the residue-basis independence makes the reduction nonzero. The displayed norm formula follows. In particular, the combination is integral exactly when every coefficient is integral, proving ${\mathcal{O}}_L={\mathcal{O}}_K[x]$ without requiring a discrete valuation.