Can we construct rational functions with prescribed ramification on an algebraic curve over \Qbar - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T08:39:20Z http://mathoverflow.net/feeds/question/68360 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68360/can-we-construct-rational-functions-with-prescribed-ramification-on-an-algebraic Can we construct rational functions with prescribed ramification on an algebraic curve over \Qbar Ariyan Javanpeykar 2011-06-21T10:22:27Z 2011-06-21T11:13:00Z <p>Let $C$ be a smooth projective connected curve of genus $g$ over $\bar{\mathbf{Q}}$. Fix a finite <strong>non-empty</strong> (Edit) set of closed points $S$ in $C$ and let $U$ be the complement of $S$ in $C$.</p> <p><strong>Q1.</strong> <em>(Algebraic formulation)</em> Does there exist a finite (surjective) morphism <code>$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$</code> such that $\pi|_{U}$ is etale?</p> <p>Equivalently, let $X$ be a compact connected Riemann surface of genus $g$ which can be defined over $\bar{\mathbf{Q}}$ and let $B$ be a finite set of of closed points in $X$ with complement $Y$. </p> <p><strong>Q1.</strong> <em>(Analytic formulation )</em> Does there exist a finite topological cover <code>$Y\longrightarrow \mathbf{P}^1(\mathbf{C})-\{0,1,\infty\}$</code> ?</p> <p>The equivalence of these two questions follows from the proof of Belyi's theorem and Riemann's existence Theorem.</p> <p>If the answer to Question 1 is positive, I would be very interested in knowing if the degree of $\pi$ can be bounded effectively. </p> <p><strong>Q2.</strong> Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1$ such that $\pi|_{U}$ is etale and $\deg \pi \leq c$, where $c$ is a constant depending only on $S$ and $g$?</p> <p><strong>Example.</strong> Suppose that $g=0$. Then, following Belyi's proof of his theorem, the answer to Question 1 is yes. The answer to Question 2 is also positive and an explicit upper bound for such a rational function is given by Khadjavi in <em>An effective version of Belyi's Theorem</em>.</p> <p>I don't expect the answer to Question 1 to be easy. In fact, what I'm asking is to prove the existence of a Belyi morphism <code>$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$</code> with <strong>prescribed ramification</strong>. Now, that's probably very hard but definitely very interesting to find out.</p> <p><strong>Trivial Remark.</strong> Suppose that $g>1$. Then the automorphism group of $C$ is finite. Choose a Belyi morphism <code>$\pi:C\longrightarrow \mathbf{P}^1_{\bar{\mathbf{Q}}}$</code> and let $U_0\subset C$ be the complement of the ramification points of $\pi$. Then we see that Question 1 has a positive answer if we take $U$ to be $\sigma(U_0)$ with $\sigma$ an automorphism of $C$. But that's only finitely many examples.</p> http://mathoverflow.net/questions/68360/can-we-construct-rational-functions-with-prescribed-ramification-on-an-algebraic/68363#68363 Answer by ulrich for Can we construct rational functions with prescribed ramification on an algebraic curve over \Qbar ulrich 2011-06-21T10:40:15Z 2011-06-21T10:40:15Z <p>No, it is easy to construct examples where this is not possible (aside from trivial ones with $|S| &lt; 3$). For example, if $g(C)>0$ one can find $S$ arbitrarily large so that the points of $S$ give linearly independent elements in $Pic(C)$. For such an $S$ there can be no map of the kind you want since the elements of $S$ must be mapped to at least $2$ distinct points of $\mathbb{P}^1$ which would give a non-trivial relation on the classes of elements of $S$ in $Pic(C)$.</p>