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Why there is not nonconstant morphism from $P^{n}$ to $P^{1}$ when $ n>1$, and how about the case when we replace the $P^{1}$ by an arbitary projective curve $C$?

Another one: Let $C\subset P^{3}$ be a non-singular rational curve and $H$ be a hypersurface of degree two. How many points in the intersedtion set $C\bigcap H$?

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A morphism $P^n\to P^1$ would be given by two homogeneous polynomials $f_0,f_1$, but these have a common zero if $n>1$.. –  J.C. Ottem Jun 22 '13 at 12:40
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For the last question, the number of intersection points could be arbitrarily large, if the degree of $C$ is large. –  J.C. Ottem Jun 22 '13 at 12:43
    
A slight variation on J.C Ottem's answer, for an arbitrary curve C: any fibre of such a morphism would have to have dimension n-1, i.e. be a hypersurface in P^n (see Shafarevich Ch.1 for a proof). But any two hypersurfaces in P^n intersect if n>1. –  Artie Prendergast-Smith Jun 25 '13 at 18:19
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closed as off-topic by Allen Knutson, J.C. Ottem, Mariano Suárez-Alvarez, Andrey Rekalo, Alex B. Jun 26 '13 at 22:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – J.C. Ottem, Mariano Suárez-Alvarez, Andrey Rekalo
If this question can be reworded to fit the rules in the help center, please edit the question.

2 Answers

Geometric proof: For $\mathbf{P}^n \to \mathbf{P}^1$: Do you know the theorem on morphisms to projective space, [Hartshorne], Theorem II.7.1? Alternatively, one could use the "Projective Dimension Theorem", [Hartshorne], Theorem I.7.2.

For $\mathbf{P}^1 \to C$: Use the Hurwitz formula, [Hartshorne], Corollary IV.2.4.

For $\mathbf{P}^n \to C$: There is a $k = 2$ or $3$ such that there exists a closed immersion $C \hookrightarrow \mathbf{P}^k$. Similarly to the above, there are no non-constant morphisms $\mathbf{P}^n \to \mathbf{P}^k$ for $n > k$.

Cohomological proof: Alternatively: If $f: X \to Y$ is surjective, $f^\ast: H^\ast(Y) \hookrightarrow H^\ast(X)$ is injective for a Weil cohomology theory $H^\ast(-)$. For $\ell$-adic cohomology, $H^*_\ell(\mathbf{P}^n)$ can be easily calculated: $H_\ell^{2i}(\mathbf{P}^n_{\bar{k}}) = \mathbf{Q}_\ell$ for $i = 0, 1, \ldots, n$, and the other cohomology groups are $0$; and $H^1_\ell(C_{\bar{k}}) = \mathbf{Q}_\ell^{2g}$, $g$ the genus of $C$.

For the last question: [Hartshorne], Theorem I.7.7.

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Why has this been downvoted? –  Timo Keller Jun 22 '13 at 16:10
    
The following statement is very false: "If $f \colon X \to Y$ is surjective, $f^\ast \colon H^\ast(Y) \hookrightarrow H^\ast(X)$ is injective". Consider $Y$ a nodal cubic and $X$ its normalization. –  Dan Petersen Sep 3 '13 at 11:12
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There are no holomorphic maps from a projective space to a Kahler manifold X of smaller dimension. Indeed, the pullback of a Kahler form $\omega_X$ cannot be exact, because a positive closed form, integrated with an appropriate power of a Kahler form, is non-zero. Since the Picard group of P^n is Z, a pullback of a Kahler class is a Kahler class. However, $\omega_X^{dim X+1}=0$ downstairs, and upstairs it is non-zero because the cohomology algebra of P^n is truncated polynomials.

Such X is automatically Kahler, but it would take some effort to prove.

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