Morphisms with connected fibres and rational functions Let $X$ and $Y$ be complex projective varieties and let $f\colon X\to Y$ be a surjective morphism with connected fibres. Is it true that any element of $\mathbb C(X)\setminus f^*\mathbb C(Y)$ is transcendental over $f^*\mathbb C(Y)$?
 A: This is equivalent to asking whether $\mathbb{C}(X)$ has a subfield which is a nontrivial algebraic extension of $f^{*}\mathbb{C}(Y)$.  I think the answer is yes.  Suppose that there is an intermediate field $L \subset \mathbb{C}(X)$ where $L$ is a nontrivial algebraic extension of $f^{*}\mathbb{C}(Y)$.  We know that the extension $L / f^{*}\mathbb{C}(Y)$ is also finite, because $X \to Y$ is a morphism of varieties and therefore of finite type; let $d \geq 2$ be its degree.  The field inclusions $f^{*}\mathbb{C}(Y) \subset L \subset \mathbb{C}(X)$ correspond to a composition of surjective morphisms $X \to Y' \to Y$, where $Y'$ is a variety whose function field is $L$.  Over a $\mathbb{C}$-point of $Y$, the fiber in $Y'$ is isomorphic to $\mathrm{Spec} (\mathbb{C}^{\oplus d})$ and is therefore not connected.  The fact that the fiber in $X$ is also not connected then follows from continuity of the morphisms.  Since that violates your hypothesis, $\mathbb{C}(X) \setminus f^{*}\mathbb{C}(Y)$ has no elements which are algebraic over $f^{*}\mathbb{C}(Y)$.
