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I have a set of small questions about rational curves.

A rational curve is a curve birationally equivalent to $\mathbb{P}^1$. But I've also seen it said that "$X$ contains a rational curve" when there is a nonconstant (rational?) morphism $f$ from $\mathbb{P}^1$ to $X$. Does this mean that the image of $f$ is a rational curve? If so, why is that true? Are there easy conditions that $f$ must satisfy for its image to be isomorphic to $\mathbb{P}^1$?

user64480
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1 Answers1

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1) If $X,Y$ are smooth projective curve (over an algebraically closed field of any characteristic) and if $f\colon X\to Y$ is a non-constant morphism, then we have for their genera $g(X)\geq g(Y)$ (Hartshorne, Chapter IV, Example 2.5.4) so that, indeed, if $X=\mathbb P^1$ then we must have $Y=\mathbb P^1$ too .
However note carefully that $f$ needn't be an isomorphism: it could be a ramified covering of any positive degree $d$, as examplified by $$f\colon \mathbb P^1\to \mathbb P^1\colon[x:y]\mapsto [x^d:y^d] .$$

2) If you have a non constant morphism $f\colon \mathbb P^1\to Y$ where $Y$ is a projective curve which is not supposed smooth, then $Y$ is still rational but of course not isomorphic to $\mathbb P^1$ if $Y$ does have a singularity.
The simplest example is to take for $Y$ the singular plane curve (called cusp) $Y\subset \mathbb P^2$ given by the equation $y^2z=x^3$ and for $f$ its normalization (=desingularization) $$f\colon \mathbb P^1\to Y\colon [u:v]\mapsto [x:y:z]=[u^2v:u^3:v^3]$$ which is a bijective morphism but not an isomorphism.

Georges Elencwajg
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  • Thanks! This is helpful. I was also wondering, if you have a nonconstant morphism $f:\mathbb{P}^1\to X$, where $X$ is a projective variety, why are we guaranteed that the image is a curve? Since in general the image of a morphism may not be a variety. – user64480 Apr 03 '13 at 21:09
  • Dear user, a projective variety $P$ is complete, which implies that the the image of any morphism $P\to W$ (where $W$ is a completely arbitrary algebraic variety) is an algebraic subvariety of $W$. – Georges Elencwajg Apr 03 '13 at 21:30
  • Ah, yes, thanks! It's taking me some time to get used to these things. – user64480 Apr 03 '13 at 21:54
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    Dear user, it takes everybody time to get used to these things! By the way, I should have added in my comment that the completeness of projective varieties is a fundamental but quite difficult result. – Georges Elencwajg Apr 04 '13 at 09:25
  • Also, I still don't see why a nonconstant map $f:\mathbb{P}^1\to Y$ necessarily implies $Y$ is birationally equaivalent to $\mathbb{P}^1$ in the case that $Y$ is singular – user64480 Apr 14 '13 at 01:13
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    Dear user64480, there is a factorization of your $f$ as a composition $\mathbb P^1\to Y^{nor} \to Y$ of two morphisms ( $Y^{nor}\to Y$ being the normalization of $Y$) . You can then apply the argument in the answer to $\mathbb P^1 \to Y^{nor}$ (since $Y^{nor}$ is smooth) and deduce that $ Y^{nor} \cong \mathbb P^1$. Conclude by recalling that normalization is birational. – Georges Elencwajg Apr 14 '13 at 07:29