Division field of an elliptic curve ($K(E[n])$)

The $n$-division field is obtained by adjoining to $K$ the coordinates of every point of $E[n]$. Over a finite field its degree is the order of Frobenius acting on $E[n]$.

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 $P$.

The absolute Galois group acts on $E[3]\simeq {\mathbb{F}}_3^2$, giving the mod-three Galois representation. Its invariant vectors are precisely the rational points of order dividing three.