Let $d \geq 2$ be an integer, $k$ a number a field containing $d$-th roots of unity and $X$ and $Y$ smooth varieties over $k$.
Let $\pi: Y \to X$ be an unramified cyclic cover of degree $d$. Let $x$ be a point in $X$, with residue field $k(x)$ and let $y$ be a point in $Y$ such that $\pi(x)=y$. Denote by $k(y)$ the residue field of $y$.
Is it true that $[k(y): k]$ divides $d$? If so, can you help me to prove it?
This is not possible because you can simply take $y$ to have huge residue degree. Let me be more precise.
Let $f:X\to Y$ be a finite surjective morphism (unramified is unnecessary) of smooth varieties over a number field $k$ of degree $d$.
Let $L/k$ be a field extension of prime degree $>d$ and $y \in Y(L)\backslash Y(K)$. Note that $[k(y) :k] = [L:K] > d$ does not divide $d$. (Such field extensions $L/K$ exist if $Y$ is a proper curve.)
Nevertheless, by surjectivity there are $d$ points in the pre-image of $y$ (counted with multiplicity). Let $x \in f^{-1}(y)$ so that $f(x) = y$. This contradicts what you are asking for.
