Assume we are given two smooth curves $C_1$ and $C_2$ over an algebraically closed field $k$. It is known that divisorial correspondences between them correspond to homomorphisms between their Jacobians. In particular there exist pairs of curves without nontrivial correspondences between them. Still my construction appears to yield such a correspondence for two arbitrary curves. So I am wondering if my construction is wrong or if all my correspondences are trivial. Could you please help me? I proceed as follows:

First I embed the function fields $K_1,K_2$ of the curves into a finite galois subextension of $k(t)^{alg}/k(t)$. This extension corresponds to a curve $C$ with nonconstant morphisms $f:C\rightarrow C_1$ and $g:C\rightarrow C_2$. The Galois group G of $C/C_1$ operates on C making $f$ into the quotient morphism. By composition with $g$ the operation induces several morphisms $g_i:C\rightarrow C_2$.

Now, I consider the sum of the graphs of the $g_i$ in $C\times C_2$. This is a G-stable Divisor descending to $C_1\times C_2$. Correcting with vertical divisors yields the nontrivial divisorial correspondence.


Clearly, this is just the curve on $C_1 \times C_2$ that consists of those points $(P, Q)$ such that $f^{-1} P \cap g^{-1} Q \neq 0$. If $G_1$ is the Galois group of $f$ and $G_2$ is the Galois group of $g$, and $G_1$ or $G_2$ is normal, then this is just the set of points that map to the same point in $C / G_1G_2$. So the correspondence is a pullback of the diagonal correspondence of $C \times C$. If $C= \mathbb P^1$, which is very common, then this diagonal correspondence is trivial after correcting with vertical divisors.

So this shows that your construction can be trivial.

If you'd like to see explicitly, why not try it with a pair of non-isogenous elliptic curves like $y^2=x^3-x$ and $y^2=x^3-1$.

  • $\begingroup$ Thanks. Your explicit description of the points helped me a lot. $\endgroup$ – Andreas Mihatsch Mar 27 '13 at 17:39
  • $\begingroup$ I find it hard to reconcile $C=P^1$ being "common" with the fact that $C$ always admits a non-constant map to an arbitrary curve $C_1$. $\endgroup$ – user30035 Mar 28 '13 at 7:37
  • 1
    $\begingroup$ Sorry, I meant $C/G_1G_2$. $\endgroup$ – Will Sawin Mar 28 '13 at 14:07

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