# Rational curves in ${\mathbb P}^n$ and immersion

In the paper of Herbert Clemens

the author shows that for a generic hypersurface $V$ of ${\mathbb P}^n$ of sufficiently high degree there is no rational curve on $V$.

The main theorem is a general statement about immersed genus $g$ curves in $V$, and it seems rather elementary to remove the immersed condition for rational curves, and derive the above statement about rational curves. But what is the argument? Is it because every rational curve in ${\mathbb P}^n$ is a multiple cover of an immersed one (if so why is this true)? Or is it because of the property of $V$?

• There are plenty of rational curves that are not multiple covers of immersed curves. A cuspidal cubic curve in the projective plane is not an immersion from $\mathbb{P}^1$ to $\mathbb{P}^2$. Commented Jan 25, 2018 at 16:50
• To Jason: then is it because of the property of the hypersurface V?
– UVIR
Commented Jan 25, 2018 at 17:57
• What property? Your question is unclear.
– abx
Commented Jan 25, 2018 at 18:36
• To clarify, Theorem 1.1 in the paper implies that a general hypersurface contains no immersed rational curves once its degree exceeds roughly half the dimension. The OP is asking: why we can remove the immersed assumption (which is remarked right below the theorem)?
– DCT
Commented Jan 25, 2018 at 21:19
• once the degree exceeds twice the dimension of the hyperaurface? Commented Jan 25, 2018 at 22:25

The OP clarified that the question is not merely about what is stated and proved in the article of Clemens; the OP would like to know what has been proved after the article of Clemens. There is important work by Lawrence Ein, Gianluca Pacienza, Claire Voisin, and Geng Xu. To the best of my knowledge, the current state of the art is the following theorem of Riedl and Woolf that applies in arbitrary characteristic (not merely characteristic $0$).
Theorem 1.2. [Riedl and Woolf] For every field $K$, for every pair of integers $n,d\geq 1$ with $d\geq 2n-1$, every "very general" degree $d$ hypersurface in $\mathbb{P}^n_K$ contains no rational curves, i.e., every morphism from $\mathbb{P}^1$ to the hypersurface is constant.
To be precise, in the projective $K$-scheme $\mathbb{P} H^0(\mathbb{P}^n_K,\mathcal{O}(d)) \cong \mathbb{P}^{\binom{n+d}{n}-1}_K$ parameterizing degree $d$ hypersurfaces in $\mathbb{P}^n$, there exist countably many irreducible, Zariski closed subschemes $Z_i$ of codimension $\geq d-(2n-2)$ such that for every algebraically closed extension $k/K$ and for every $k$-point of $\mathbb{P} H^0(\mathbb{P}^n_K,\mathcal{O}(d))$ that is contained in no $Z_i$, every $k$-morphism from $\mathbb{P}^1_k$ to the associated hypersurface in $\mathbb{P}^n_k$ is constant.