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.

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.

I would also be interested in other proofs of this result.

  • 1
    $\begingroup$ You should emphasize your hypotheses more. The reason this would be merely a remark in many birational geometry papers is that it is completely trivial in case X is smooth and the characteristic is zero: there rationally connected is equivalent to being able to join a general pair of points by a rational curve, which is obvious for unirational varieties. $\endgroup$ Dec 6, 2011 at 21:47
  • $\begingroup$ @Jack Why is it completely obvious as you say? For example if my points are outside of the image of rational map from projective space, I don't see why it should be clear I can get a rational curve connecting these points, or even a chain of curves. $\endgroup$
    – HNuer
    Dec 6, 2011 at 21:50
  • 2
    $\begingroup$ You need only join a general pair of points, not every pair. The general pair of points lies in the image of the map from projective space. It's not obvious why the various definitions of rationally connected are equivalent in characteristic zero, however; this typically requires deformation theory. See Debarre, "Higher Dimensional Algebraic Geometry," chapter 4, for this. $\endgroup$ Dec 6, 2011 at 21:54
  • $\begingroup$ I have edited my question to ask why any two points can be joined by a chain of rational curves. I agree if we're using your defition then it's obvious. $\endgroup$
    – HNuer
    Dec 6, 2011 at 22:00

2 Answers 2


In case you are still interested in this question, here is a proof of the explicit statement at the end of the post.

Claim 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 $E$.

Proof: 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 $\widetilde L=\pi^{-1}_*L\subseteq T$ 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 $E$. $\square$


Assume that $X$ is unirational. Then we have a dominant rational map $$\phi:\mathbb{P}^{n}\dashrightarrow X$$ Let $x,y \in X$ be two general points and let us consider two points $p,q$ in $\pi^{-1}(x),\pi^{-1}(y)$ respectively. Take the line $L$ generated by $p$ and $q$. Then $\phi_{|L}:L\rightarrow X$ is a finite morphism and its image $C = \phi(L)$ is a rational curve through $x$ and $y$. So $X$ is rationally connected.

  • $\begingroup$ Why is $\phi|_L$ a morphism instead of a rational map? $\endgroup$
    – S. Carnahan
    Oct 16, 2013 at 15:46
  • $\begingroup$ @Scott: $L$ is a smooth projective curve. There is no room for indeterminacy. (I suppose one may assume that $X$ is projective as well). $\endgroup$ Oct 16, 2013 at 16:54
  • $\begingroup$ Yes, $\phi_{|L}$ is a morphism just because $L$ is a smooth curve. I am assuming $X$ projective. $\endgroup$
    – Puzzled
    Oct 16, 2013 at 17:03
  • $\begingroup$ I acknowledged, after Jack's comments above, that your proof works for the usual definition of rationally connected, namely that two general points can be connected by a rational curve. However, I edited the question to more clearly reflect what I was asking which was why ANY two points can be connected by a chain of rational curves. $\endgroup$
    – HNuer
    Oct 16, 2013 at 21:10
  • $\begingroup$ If X is projective and two general points can be connected by a rational curve then any two points can be connected by a rational curve. $\endgroup$
    – Puzzled
    Oct 23, 2013 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.