Unirational implies rationally connected - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:58:06Z http://mathoverflow.net/feeds/question/82823 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/82823/unirational-implies-rationally-connected Unirational implies rationally connected HNuer 2011-12-06T21:21:49Z 2012-12-08T00:16:25Z <p>It is evidently a well-known fact that a unirational variety $X$ over an algebraic closed field (i.e. there is a dominant rational map from $\mathbb P^n$ to $X$) is rationally connected (by which I mean that any two points can be joined by a chain of rational curves). Numerous authors on birational geometry seem to state this as a remark, but don't indicate how one might prove it. The only proofs I have found of this fact (i.e. Fulton's Intersection Theory book example 10.1.6 and the paper of Samuel he quotes there) use the completion of local rings and power series. I was wondering if there was a purely algebraic (i.e. without completions) proof of this result. </p> <p>In particular, by blowing $\mathbb P^n$ at the indeterminancy locus of the rational map to $X$ we get a commutative diagram involving a birational, projective, surjective morphism from $\tilde{\mathbb P^N}$ to $\mathbb P^n$, our original rational map from $\mathbb P^n$ to $X$, and a projective, surjective morphism $\tilde{\mathbb P^n} \rightarrow X$, so if we can show that the blowup is rationally connected then mapping to $X$ will give us our chain of rational curves connecting any two points of $X$. This reduces to the following affine case: We are then left with the case of showing that if $\pi: T\rightarrow \mathbb A^n$ is the blow-up of $\mathbb A^n$ along a subcscheme Z, with exceptional divisor $E$, and $t\in E$, then there is a morphism $h: \mathbb A^1\rightarrow T$ with $h(0)=t$ but $h(\mathbb A^1)$ not contained in $E$. It is here that I was wondering if people knew of a way to procede without using power series as Fulton and Samuel do. </p> <p>I would also be interested in other proofs of this result. </p> http://mathoverflow.net/questions/82823/unirational-implies-rationally-connected/115759#115759 Answer by Sándor Kovács for Unirational implies rationally connected Sándor Kovács 2012-12-08T00:16:25Z 2012-12-08T00:16:25Z <p>In case you are still interested in this question, here is a proof of the explicit statement at the end of the post.</p> <blockquote> <p><strong>Claim</strong> Let $\pi: T\rightarrow \mathbb A^n$ be the blow-up of $\mathbb A^n$ along a subcscheme $Z$ with exceptional divisor $E$. Then for any $t\in E$, there exists a morphism $h: \mathbb A^1\rightarrow T$ with $h(0)=t$ but $h(\mathbb A^1)$ not contained in $E$.</p> </blockquote> <p><strong>Proof</strong>: Assume that $\pi(t)=0\in \mathbb A^n$. Then the point $t\in E$ corresponds to a normal direction of $Z$ at $0$. Let $L\subseteq \mathbb A^n$ be a line pointing in that direction and let <code>$\widetilde L=\pi^{-1}_*L\subseteq T$</code> be the strict transform of $L$ on $T$. Observe that by choice $L\not\subseteq Z$ and hence $\widetilde L\not\subseteq E$. Also note that $\pi|_{\widetilde L}: \widetilde L\to L$ is the blow up of $L$ along $L\cap Z$, and hence it is an isomorphism. Therefore there exists a morphism $h: \mathbb A^1\rightarrow \widetilde L\subseteq T$ with $h(0)=t$ but $h(\mathbb A^1)=\widetilde L$ not contained in $E$. $\square$</p>