TITLE: Unirational implies rationally connected
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.