If $K/k$ is a one-dimensional function field and $D\in \operatorname{Der}_k(K)\setminus\{0\}$, is it true that $\ker D=k$? Let $K$ be a one-dimensional function field over a characteristic zero field $k$. I.e., $K$ is a finite extension of $k(x)$. We know how to calculate $\Omega_{k[x]/k}\cong k[x]dx$, and the module of Kahler differentials is invariant under localization and finite extensions. This implies that $\Omega_{K/k}\cong Kdx$ and so $\operatorname{Der}_k(K)=\Omega_{K/k}^\vee$ is a one-dimensional $K$-vector space.
If $D\in \operatorname{Der}_k(K)\setminus\{0\}$, then clearly $k\subset \ker D$ for $D$ is $k$-linear. I'm pretty sure that we have an equality there. This is true for $K=k(x)$ since it holds for $D=d/dx$ and any other derivation is a multiple of this one. Nevertheless, I'm a little confused about this maybe being affected by the finite extension.
I would appreciate any clarifications.
 A: I’m turning my comment into an answer with a little bit more detail.
Let $D: K \rightarrow K$ be a nonzero $k$-derivation.
Let $F=\{z \in K,\, Dz=0\}$ be the kernel of $D$. Clearly $k \subset F$ and $F$ is a $k$-vector subspace. Moreover, let $u,v \in F$, then $D(uv)=uD(v)+vD(u)=0$ so $F$ is stable under product. Similarly, if $u=vw$ with $v \neq 0$ and $u,v \in F$, then $vD(w)=D(u)-wD(v)=0$. So $F$ is a $k$-subfield of $K$.
Moreover, let $P \in F[t]$ be an irreducible (thus separable) polynomial. For any $u \in K$, it’s easy to see that $D(P(u))=P’(u)D(u)$. So if $P(u)=0$, then $P’(u)\neq 0$ so $D(u)=0$. Thus $F$ is algebraically closed in $K$.
Now, because $K$ is algebraic over some $k(t)$, it has exactly two $k$-subfields which are algebraically closed in $K$: the algebraic closure of $k$ in $K$ or $K$ itself.
Since $D\neq 0$, $F$ can’t be $K$, so $F$ must be the algebraic closure of $k$ in $K$.
In particular, if $k$ is algebraically closed, or $K$ is the field of functions of a smooth projective geometrically connected curve over $k$, then $F=k$.
