Let $Y$ be a affine variety of dimension $d$ in affine space $\mathbb{A}^n$. Let the vanishing ideal of $Y$ be generated by $f_1,\dots,f_n \in A=k[y_1,\dots,y_n]$. Now suppose that $Y$ is rational. Then $K(Y) \cong k(u_1,\dots,u_d)$, where $u_i$ is indeterminate over $k$. Let $y_i$ be sent to $g_i \in k(u_1,\dots,u_d)$ under the above isomorphism. Then $f_j(g_1,\dots,g_n)=0$ which means that the variety $Y'$ of $\mathbb{A}^n$ parametrized by $y_i = g_i(u_1,\dots,u_d), \, \forall i$, is inside the variety $Y$.
Question: How can we actually see that $Y' = Y$?
PS: I believe this is suggested by the second paragraph of this wikipedia article:http://en.wikipedia.org/wiki/Rational_variety