I'm studying algebraic geometry from "Introduction to algebraic geometry" by Hassett, and I did not understand a step in his proof of the following result (page 52):
"If $W$ is an affine variety admitting an injection of $k$-algebras $\psi : k[W] → = k(x_1, \cdots , x_n)$ then $W$ is unirational."
The proof presented is:
"Suppose we have an injection $\psi$ as above. By previous results, such a function yields a rational map $\rho : \mathbb{A}^n(k) → W$. If $\overline{\rho(U)} \neq W$ (where $U$ is the open set where $\rho$ is defined) then there must exist some $f \neq 0 \in k[W]$ vanishing on $\rho(U)$, contradicting the assumption that $\psi$ is injective."
My question is in the passage "there must exist some $f \neq 0 \in k[W]$ vanishing on $\rho(U)$, contradicting the assumption that $\psi$ is injective.". Why must such a $f$ exist?
Thank you in advance!
$W$ is affine which means that its closed subsets are in one-to-one correspondence with radical ideals of $k[W]$ (this being the correspondence coming from the Nullstellensatz). Since $W$ is a variety, $k[W]$ has no nilpotent elements, so $(0)$ is radical, and corresponds to the entire closed set $W$ itself. Hence if $\overline{\rho(U)} \not= W$ then, since it is closed, there must be a corresponding radical ideal of $\overline{\rho(U)}$ in $k[W]$ which is not the zero ideal. $f$ is a non-zero element of this ideal.