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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.

The degree of the residue field extension induced by a finite extension of complete discretely valued fields. Along with the ramification index, it satisfies $[L:K]=ef$. An unramified extension additionally has separable residue extension; a totally ramified extension has residue degree one.

For a finite unramified extension of non-Archimedean local fields, the field norm on residue units is surjective: on a cyclic group of order $q^n-1$ it is exponentiation by $(q^n-1)/(q-1)$ and has image of order $q-1$. For a common uniformizer $\pi$ and $r\ge 1$, expansion of the conjugate product givesThe field trace of a finite finite field extension is onto: its polynomial $X+X^q+\cdots +X^{q^{n-1}}$ is nonzero of degree below $q^n$, so defines a nonzero linear map to the base field. Lift a residue norm and correct successively with elements of $1+{\pi }^r{\mathcal{O}}_L$. The infinite product converges by completeness, and its norm is the prescribed unit. This argument also works in positive characteristic and when $p$ divides $n$.

Here $a\ge 0$ and $p\nmid m$. The prime-to-$p$ roots of unity generate an unramified extension of degree equal to the multiplicative order of $p$ modulo $m$; for $m=1$ this degree is one. Indeed, their distinct residues lift uniquely by Hensel lemma, and the Frobenius automorphism acts by raising them to $p$. The $p$-power roots of unity generate a totally ramified extension, because the translated cyclotomic polynomial ${\Phi }_{p^a}(1+T)$ is Eisenstein. The two fields have trivial intersection and their compositum has the same residue degree as the prime-to-$p$ field, by unramified base change preserves total ramification. Thus total ramification is equivalent to $p\equiv 1(\mathrm{mod}m)$.

Let $E/K$ have good reduction over a local field of residue characteristic $p$, with $p\nmid n$. All solutions of $nQ=P$, for all $P\in E(K)$ together, lie in a single finite unramified extension of degree at most $nf$, where $f=[K(E[n]):K]$. First lift a solution over the algebraic closure of the residue field to a finite unramified extension; the multiplication isomorphism of a formal group law uniquely corrects the error in $E_1$. This also puts $E[n]$ in the maximal unramified extension. If $\sigma$ is its Frobenius generator, ${\sigma }^f$ fixes $E[n]$. For $nQ=P$, set $T={\sigma }^{f}Q-Q\in E[n]$. Then ${\sigma }^{nf}Q=Q+nT=Q$, uniformly in $P$.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.

If $L/K$ is an unramified extension and $M/K$ is a totally ramified extension of non-Archimedean local fields inside a fixed algebraic closure, then $LM/L$ is totally ramified of degree $[M:K]$. A uniformizer of $M$ has an Eisenstein polynomial over $K$. Since a uniformizer of $K$ remains a uniformizer in $L$, that polynomial remains Eisenstein over $L$ and generates $LM$.

At a prime of good reduction of an elliptic curve with residue characteristic not dividing $m$, multiplication by $m$ extends to a finite étale map of the smooth proper elliptic group scheme over the valuation ring. A local rational point extends to a section by properness. Its division fibre is therefore a finite étale torsor, so the Kummer map of an elliptic curve gives an unramified cohomology class. This bounds ramification in the Weak Mordell-Weil theorem.

A direct proof describes the coordinates, not only the cohomology class. Reduce $P$ by properness and choose a point $\overline{Q}$ with $[m]\overline{Q}=\tilde{P}$ over a finite extension of the residue field. Over the corresponding unramified extension of the complete local field, lift $\overline{Q}$ by the Hensel lemma. The error $[m]Q_0-P$ is in the kernel of reduction of an elliptic curve; prime-to-residue-characteristic multiplication on a formal group corrects it uniquely. There is no $m$-torsion in that kernel, even after a ramified field extension, so every division point with that reduction equals this unramified lift. Consequently all division points of an elliptic curve above $P$ have unramified coordinate fields. The torsion case is discussed in J. S. Milne's Elliptic Curves, Chapter V, Proposition 8.1.

For non-Archimedean local fields, let $L/K$ be an unramified extension of degree $f$, and let $E/K$ be finite. The degree-$f$ unramified extension of $E$ has a residue field containing the degree-$f$ residue extension of $K$. Lift an integral residue generator of $L$ into that field by Hensel lemma. This embeds $L$ into an unramified extension of $E$, so $LE/E$ is an unramified subextension. Applying this argument to each completion of a number field at a prime ideal proves the finite-prime assertion for number fields, even when $E/K$ is ramified.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

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.