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Let $K^{\mathrm{ur}}$ be the maximal unramified extension of $K$, with residue field $\overline{k}$. First show that every geometric solution of $nQ=P$ lies in $E(K^{\mathrm{ur}})$.

The multiplication-by-n morphism on the reduced elliptic curve is surjective over $\overline{k}$. Choose $\overline{Q}$ with $n\overline{Q}=\overline{P}$, defined over some finite finite field extension $k^{\prime }/k$. Let $L/K$ be the corresponding finite unramified extension. By the surjectivity of good reduction over a local field, lift $\overline{Q}$ to $Q_0\in E(L)$. Then $P-n{Q}_0\in E_1(L)$. The multiplication isomorphism of a formal group law gives a unique $R\in E_1(L)$ with $nR=P-n{Q}_0$, so $Q_0+R$ is an $n$-division point of $P$ defined over $L$.

Taking $P=O$ gives every element of $E[n]$ over $K^{\mathrm{ur}}$: reduction is injective on this torsion point of an elliptic curve group because $[n]$ is injective on the formal kernel, and each reduced $n$-torsion point has the unique corrected lift just constructed. Every other division point differs from one lift by an element of $E[n]$. Hence the entire field $K([n{]}^{-1}P)$ is unramified, not merely a field containing one choice of $Q$.

To bound the composite uniformly in $P$, put $f=[K(E[n]):K]$. This division field of an elliptic curve is unramified. The Galois group of $K^{\mathrm{ur}}/K$ is procyclic, generated topologically by its Frobenius element $\sigma$, and ${\sigma }^f$ fixes every element of $E[n]$. If $nQ=P\in E(K)$, thenIteration gives ${\sigma }^{jf}Q=Q+jT$. In particular ${\sigma }^{nf}Q=Q$. Thus every such $Q$, for every $P$ together, lies in the unique unramified extension of degree $nf$. If $M$ is the composite of all these fields, the uniform unramified division field over a local field givesThe prime-to-$p$ hypothesis is essential to the inverse-series argument; no assertion about arbitrary $p$-division is being made.

Kummer theory begins with the exact sequenceGalois cohomology and Hilbert theorem 90 identifyso cyclic extensions of exponent dividing $n$ are described by adjoining $n$th roots when $K$ contains ${\mu }_n$.

For an elliptic curve $E/K$, multiplication by $n$ gives the Kummer exact sequence of an elliptic curveIts connecting homomorphism is the injective Kummer map of an elliptic curvewhere $nQ=P$. Passing to the finite division field of an elliptic curve $K(E[n])$ makes $E[n]$ constant. Rational functions whose divisors are $n(T)-n(O)$ then express the classes through finitely many elements of $K(E[n]{)}^{\times }/K(E[n]{)}^{\times n}$.

Let $S$ contain the primes above $n$, the primes of bad reduction of an elliptic curve and the finitely many primes introduced by these functions. The local theory of good reduction shows that every Kummer class coming from $E(K)$ is unramified outside $S$, so it lies in an S-unramified power class group. Such a group is finite: valuations outside $S$ vanish modulo $n$, the ideal class group is finite, and the Dirichlet unit theorem makes the group of $S$-units modulo $n$th powers finite. The kernel created by passing to $K(E[n])$ is finite by finite-group Galois cohomology. HenceThis is the Kummer-theoretic proof of the weak Mordell-Weil theorem. Combining it with the height descent lemma proves the Mordell-Weil theorem: $E(K)$ is a finitely generated abelian group.

For $E:y^2=x^3-x$ over ${\mathbb{F}}_3$, each of $x=0,1,2$ gives the single affine point with $y=0$. Including $O$ gives $\#E({\mathbb{F}}_3)=4$, so the Frobenius trace is $a=3+1-4=0$. Therefore the Frobenius isogeny of an elliptic curve $\pi$ satisfiesOn the $11$-torsion, this reads ${\pi }^2=[-3]=[8]$. The element $8\in {\mathbb{F}}_{11}^{\times }$ has order ten and $8^5=-1$. Henceand no smaller positive power of $\pi$ is the identity on $E[11]$. A division field of an elliptic curve over a finite field has degree equal to the order of Frobenius on the torsion module, so