Does adjoining a $p$-power divisor to an elliptic curve, where $p\neq 2$ is a prime, always result in a Galois group of order greater than $2$? The question says it:
Suppose I have a field $K$, whose characteristic is, for simplicity, zero, and an elliptic curve $E$ over $K$, and $x\in E(K)$.
Suppose that $p$ is a prime different from $2$, and $n\in E(\overline{K})$ with $pn=x$, $n\notin E(K)$. If $L$ is the smallest algebraic extension of $K$ so that $n\in E(L)$, is $\sharp\operatorname{Gal}(L/K)>2$? 
 A: I can’t think of negative answers to your question except when $p^r=3$. Look at the following case:
Fortunately you allowed any field of characteristic zero for your $K$, and I’m going to take $K=\Bbb Q_3$, and $E$ the curve $Y^2=X^3+X^2+1$ defined over $\Bbb Z_3$, with good ordinary reduction modulo $3$. The $3$-torsion points are the inflection points when the point $\Bbb O$ at infinity is taken to be the identity. I make the formula for the second derivative to be
$$
\frac{d^2Y}{dX^2}=\frac{15X^4 + 20X^3+4X^2+24X+8}{8Y^3}\,,
$$
which is zero exactly when the numerator is zero, so giving four values of $X$, each with two $Y$-values. Look at the Newton polygon of the numerator: vertices at $(0,0)$, $(3,0)$, and $(4,1)$, so that there is a $\Bbb Q_3$-rational root of the form $u/3$, where $u$ is a $3$-adic unit. This allows your point “$n$” to be $(u/3,\sqrt{u/3}\,)$, necessitating precisely a quadratic extension.
Nor is this phenomenon peculiar to $3$-torsion points. If you take your point “$x$” to be suitably close to $\Bbb O$ in the $3$-adic topology, you’ll have the same behavior, but the formulas will be a lot messier. It’s actually much easier to do this in the language of formal groups, but that’s another story.
