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Given a morphism $\phi : X\rightarrow Y$ of smooth projective varieties and a non-constant map $f : C\rightarrow Y$, where $C$ is amooth projective curve, how to construct a smooth projective curve $C'$ with non-constant map $g : C'\rightarrow X$ such that $\phi \circ g = f\circ h$, where $h : C'\rightarrow C$.

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up vote 11 down vote accepted

First form the fibre product $Z$ of $X$ and $C$ over $Y$, to get maps $f':Z \to X$ and $\phi':Z \to C$ such that $\phi\circ f' = f \circ \phi'.$

Now if the map $\phi'$ is not surjective (equivalently, dominant, since all varieties in sight are projective) then we can't find $C'$. (This happens when the image of $f$ is not contained in the image of $\phi$.) [EDIT: As Sandor Kovacs and Roy Smith noted, I am taking it as a condition that $h$ is also non-constant, although this is not explicitly stated in the question; otherwise one could allow more degenerate solutions. Also, I am assuming that all curves under discussion are connected.]

Suppose now that $\phi'$ is surjective. Then we can find a one-dimensional subvariety $D$ of $Z$ such that $\phi'(D) = C$. Indeed, since $\phi'$ is surjective, we may find at least one component of $Z$ such that the restriction of $\phi'$ to that component is already surjective. Replace $Z$ by this component, and so assume that $Z$ is irreducible. If $Z$ is also one-dimensional, then we can take $D = Z$. If not, a general hyperplane section of $Z$ will also surject onto $C$, and so we may replace $Z$ by this hyperplane section, thus reducing the dimension. Continuing in this way, we eventually find the desired $D$.

Finally, we can form the normalization $C'$ of $D$. The composite $$C' \longrightarrow D \hookrightarrow Z \buildrel f' \over \longrightarrow X$$ is then the desired map $g$, while the composite $$C' \longrightarrow D \hookrightarrow Z \buildrel \phi' \over \longrightarrow C$$ is the desired map $h$.

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I guess moreover that your construction (for an adequate choice of the irreducible component of Z) minimizes the degree of h. Is this true? – quim Dec 10 '10 at 13:04
I would be surprised: presumably one can find lower degree (say for some fixed choice of projective embedding of $Z$) choices of $D$ which don't happen to be very ample, if not in general then surely in some particular cases. – Emerton Dec 10 '10 at 17:06
Correct - in general taking hyperplane sections won't give you the lowest possible degree. If X = P^1 x P^1 (embedded as a quadric surface), Y = C = P^1... then using the hyperplane section will give us a degree 2 map C' --> C but there is a section of the map X --> Y (which isn't realized as the intersection of X with a hyperplane). – mdeland Dec 10 '10 at 18:29
If we modify the question so that both g and h are non constant, then this seems a perfect answer. I.e. imagine this is jeopardy, and this is the answer in search of a question. – roy smith Dec 10 '10 at 18:56
As noted below, the minimum degree of h is zero when all fibers of phi have positive dimension, so this answer almost never minimizes the degree, if those "trivial" solutions are allowed. – roy smith Dec 10 '10 at 19:07

Assuming $\phi$ surjective, Matt's answer is OK if the ground field $k$ is perfect. Otherwise, one should weaken the conditions on $C'$ from "smooth" to "normal".

For instance, assume $p=\mathrm{char}(k)>0$ and $a\in k$ is not a $p$-th power. Take $\phi:\mathbb{P}^2\to\mathbb{P}^2$ to be the Frobenius map $[x:y:z]\mapsto[u:v:w]=[x^p:y^p:z^p]$, and $C\subset\mathbb{P}^2$ to be the line $u+av=0$. Then $\phi^{-1}(C)$ is reduced and geometrically irreducible and its function field is inseparable over $k$, so there is no nonconstant map from any smooth $k$-curve to $\phi^{-1}(C)$.

On the other hand, it's enough to assume $X$ and $Y$ proper and $\phi$ surjective (no smoothness or projectivity needed): let $z$ be the generic point of $C$, then $\phi^{-1}(z)$ is a nonempty $\kappa(z)$-scheme of finite type, hence has a closed point $z'$. The residue field of $z'$ is a one-variable function field over $k$, hence is the generic point of a unique normal projective irreducible $k$-curve $C'$. By the valuative criterion of properness, the canonical map $z'\to X$ extends to $C'\to Y$, and $C'$ is smooth if $k$ is perfect.

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why there is no nonconstant map from any smooth $k$-curve to $\phi^{-1}(C)$? Please tell me as I am just learnig the subject. – Yashica Dec 13 '10 at 4:42
A nonconstant map would induce an embedding of the function field of $\phi^{-1}(C)$ (inseparable over $k$) into the function field of a smooth curve, which is separable. – Laurent Moret-Bailly Dec 13 '10 at 8:41

I am probably confused. Matt's h seems always to be surjective. What if C = Y, f = id, and C = X, and phi = constant map to point p of C? Then we seem to need h to be constant. E.g. taking h = phi, and g = id, seems to work (and the minimum degree which occurs here is zero). Is it possible your construction (essentially) works as long as the images of f and phi meet? I.e. that it works more widely, but that it is only when the image of f is actually contained in the image of phi that h is surjective? No I guess you need the inverse image of f(C) in X, i.e. the fiber product, to contain a curve, in order for a non constant g to exist. Something like that?

Indeed it seems more complicated than that. What if we take f the inclusion of a plane curve C into P^2 = Y, and X the blow up of P^2 at a point p of C. Then the we seem to be able to find both constant and non constant models for h. I.e. C' = C, and either h = id, and g is the "proper transform" of f, or h is a constant map from C to the point p, and g is any non constant map to the exceptional P^1 in X.

So there seem to be various answers of various degrees corresponding to different positive dimensional components of the fiber product, and whether or not they surject onto f(C). Indeed if there is a single point f(p) on f(C), whose inverse image in X contains a curve D, then we seem to be able to take C' = D, and g= id, and h = the constant map to p. Thus in general the minimum degree of h seems to be zero, e.g. whenever dimX > dimY and phi surjects. Does this seem right?

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Yes it does seem right. But presumably Yashica is interested in how to construct these smooth curves C' in X_C dominating C rather than this degenerate case.... – mdeland Dec 10 '10 at 19:19
Dear Roy, this is a reply to your comment to my post that I just deleted. You are right, it was constant. And I knew that. I just realized that I did not read the question completely, and somehow I missed the requirement of being non-constant. That is an obvious thing to ask for and I did wonder about that. I still did not realize that the question asked for that. I guess I read selectively. :( – Sándor Kovács Dec 10 '10 at 19:27
Dear Roy, I'm not sure whether "Your" in your second sentence applies to me, to the author of the question, to Sandor, or to someone else, but just in case: I misread the question and took it to be asking that $h$ should be non-constant as well as $g$. Thus I required that the image of $f$ be contained in the image of $\phi.$ Regards, Matt – Emerton Dec 10 '10 at 19:35
I figured as much, Matt. And I would guess the case of h non constant is the one most interesting to the original questioner as well. But since you had already given the best possible non trivial answer, I had fun with the trivial ones!, regards, roy. – roy smith Dec 10 '10 at 23:05
Sandor, forgive me for possibly causing you to delete yours, since I thought it the best possible answer to the case where g,h are not necessarily constant. I.e. there are three situations of possible interest, answered by the union of the criteria we gave: Z non empty (g,h arbitrary), Z contains a curve (g non constant), and phi surjective (both g,h non constant). (Subject to my introduced errors.) – roy smith Dec 11 '10 at 4:37

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