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Let $C_1,C_2$ be two smooth irreducible curves of genus 4,3 resp. Prove there is no morphism $\phi: C_1 \to C_2$.

Well, the tool of treating genera is Hurwitz's theorem, which says here that $6=4+\sum (n(x)-1)$, for $x$ the ramification points of $\phi$ in $C_1$, and we can replace the $n(x)-1$ with $\deg \phi -k(y)$, where k is the number of points mapped by $\phi$ to the branching point $y$. Perhaps being a morphism somehow constricts the behavior of ramification points, but I should still use the irreducibility, and don't have a clue how.

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    $\begingroup$ You don't have Hurwitz' theorem right: it says $2g(C_1)-2=d \cdot (2g(C_2)-2)+\sum (n(x)-1)$ where $d$ is the degree of $\phi$. Since we must have $d>1$, this is impossible. $\endgroup$ – user64687 Jan 28 '14 at 14:04
  • $\begingroup$ Why is $d=1$ impossible? $\endgroup$ – Emolga Jan 28 '14 at 19:09
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    $\begingroup$ Because a morphism of degree 1 between smooth curves is an isomorphism, but these curves have different genera, so are not isomorphic. $\endgroup$ – user64687 Jan 28 '14 at 19:38
  • $\begingroup$ So where have we use the irreducibility? $\endgroup$ – Emolga Jan 28 '14 at 19:48
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    $\begingroup$ Let me also point out why the original problem is false if you omit the word "irreducible". Any curve $C_1$ has a map to $\mathbf{P}^1$, so if $C_2=\mathbf{P}^1 \cup D$ with $D$ a curve of genus 2 glued to $\mathbf{P}^1$ at 1 point, we get a counterexample to the claim of the question. $\endgroup$ – user64687 Jan 28 '14 at 20:22

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