Rational Curves 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 $\mathbb{P}^1$?
 A: 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.
