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Suppose $C_0$ is a dense open subvariety of $C$ and $C_0 \rightarrow D_0$ is a finite morphism. Does there exist a finite morphism $C \rightarrow D$ to some other variety $D$ that agrees with $C_0 \rightarrow D_0$ on $C_0$?

I know this is possible if $C_0$ is closed, and that there are counterexamples in general. I'm looking for certain situations where such a morphism might exist.

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  • $\begingroup$ Denote the degree of $C_0\to D_0$ by $d$. Form the fiber product $C_0\times_{D_0} C_0$ as a locally closed subscheme of $C\times_{\text{Spec}\ k} C$ (I assume that we are working over a field $k$). If the closure $Z$ of $C_0\times_{D_0} C_0$ is finite and flat with respect to the projection $\text{pr}_1:Z\to C$, then you can construct the morphism $C\to D$ as the associated morphism to the Hilbert scheme of $C$, $\zeta_Z:C\to \text{Hilb}^d_{C/k}$, where $D$ is the image of $\zeta_Z$. If $Z$ is just a "well-formed zero cycle", use the Chow variety instead of the Hilbert scheme. $\endgroup$ Jan 5, 2017 at 9:49

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I am just posting my comment as an answer. I will assume that we are working over a specified field $k$. I will also assume that $C$ is reduced, irreducible, normal, and proper. In the general case, you can first ask your question for the normalizations of the irreducible components of a proper model, and then you can try to descend from that normal, proper scheme to your original scheme.

The fiber product $Z_0 =C_0\times_{D_0} C_0$, considered as a closed subscheme of $C_0\times_{\text{Spec}\ k} C$, is a well-formed zero cycle of some degree $d$. There is an induced morphism $\zeta_{Z_0}:C_0\to \text{Sym}^d_k(C)$. There is an extension of $C_0\to D_0$ to a finite morphism with domain equal to $C$ if and only if $\zeta_{Z_0}$ extends to all of $C$. In turn, this holds if and only if the closure $Z$ of $Z_0$ in $C\times_{\text{Spec}\ k} C$ is finite with respect to the projection $\text{pr}_1:Z\to C$.

The articles of David Rydh are a great source for Chow varieties.

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  • $\begingroup$ This is very helpful. Thank you. How though can one arrange that the projection is finite on $Z$? It seems to be the same problem as at the start, where the morphism on some open subvariety is finite. $\endgroup$ Jan 5, 2017 at 21:02
  • $\begingroup$ I should mention that I include a variety $D'$ such that $C \rightarrow D'$ is projective but not finite, and that $D_0$ is open in $D'$. $\endgroup$ Jan 5, 2017 at 21:27
  • $\begingroup$ @JeremyBerquist. "How though can one arrange that the projection is finite on $Z$?" I am not certain what you are asking. Do you want an example? Do you want a counterexample? Do you want sufficient conditions so that $Z$ is finite over $C$? If you assume that $C$ is locally $\mathbb{Q}$-factorial and that the complement of $C_0$ in $C$ has codimension $2$ (as can be always arranged), then if there exists an ample $\mathcal{A}$ on $\text{Sym}^d_k(C)$ with $\zeta_{Z_0}^*\mathcal{A}$ globally generated, then $Z\to C$ is finite. Is that useful to you? $\endgroup$ Jan 6, 2017 at 12:40
  • $\begingroup$ @JeremyBerquist. "I should mention that I include a variety $D'$ such that $C\to D'$ is projective but not finite, and that $D_0$ is open in $D'$." On its own, that does not help. There are such examples where $Z$ is finite over $C$, but there are also examples where $Z$ is not finite over $C$. For instance, let $D'$ be the affine plane, let $Y\to D'$ be a finite, flat morphism of degree $2$ that is etale over the origin, and let $X\to Y$ be the blowing up of one of the two preimages of the origin. $\endgroup$ Jan 6, 2017 at 12:49
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    $\begingroup$ @JeremyBerquist. "Can you give a nontrivial example in which the induced morphism is finite?" Yes, yes I can. However, since you are the one who proposed the original example, you should specify the example you would like to understand rather than asking me to list examples that (likely) have nothing to do with your examples. $\endgroup$ Jan 7, 2017 at 22:46

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