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 $f^*\mathbb C(Y)$?
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1$\begingroup$ Think of the case where $Y$ is a point. $\endgroup$– Laurent Moret-BaillyCommented Feb 23, 2016 at 14:22
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$\begingroup$ Sorry, the question was badly formulated, I have rewritten it. $\endgroup$– user88055Commented Feb 23, 2016 at 14:53
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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 $f^{*}\mathbb{C}(Y)$.
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1$\begingroup$ How are you getting the morphism $Y' \to Y$? A map between the function fields gives a dominant rational map, not a morphism. And then some care is needed when talking about the fibers. $\endgroup$– user47305Commented Feb 24, 2016 at 22:32
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1$\begingroup$ Yeah, you're right. I suppose the main thing would be to argue that the rational map $Y' \to Y$ is defined on an entire fiber over some closed point of $Y$. $\endgroup$ Commented Feb 25, 2016 at 15:49
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$\begingroup$ See Example 2.1.12 of Lazarsfeld's "Positivity in Algebraic Geometry". $\endgroup$– pgrafCommented Feb 29, 2016 at 15:19
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$\begingroup$ @Mark: you can take $Y'$ to be the integral closure of $Y$ in $L$. If $X$ is normal, then $X \to Y$ has to factor through $Y'$. In this case, a general fibre will be $\mathbb C^d$ (since $Y' \to Y$ is generically finite étale, hence finite étale on some open). Note however that $Y' \to Y$ need not even be flat, so there could be fibres of different length. $\endgroup$ Commented Mar 26, 2016 at 1:56