Proposition

Give an example to show that the result is false over $\mathbb{R}$.

Solution

Let $F: \mathbb{R} \rightarrow \mathbb{R}$ be defined such that $F(t) = t^2$.

Let $I = \ev{x - f_1(t)} = \ev{x - t^2}$. Let $V = V(I) = \{ (a, a^2) \mid a \in \mathbb{R} \}$. Then $\pi_1(V) = [0, \infty)$. On the other hand, $I_1 = \ev{0}$, so $V(I_1) = \mathbb{R}$. Suppose there exists a variety $W$ such that $V(I_1) \setminus W \subset F(\mathbb{R}) = [0, \infty)$. Then $(-\infty, 0) \subset W \subsetneq V(I_1) = \mathbb{R}$. $W$ is a vanish set of an ideal of polynomials of 1 variable, and if the ideal is nonzero, $W$ cannot have infinitely many elements. Therefore, $W$ must be a vanish set of $\ev{0}$. However, this implies $W = \mathbb{R}$, so this is a contradiction.