This exercise is in the book Ideals, Varieties, and Algorithms.
Exercise.
Given a rational parametrization as in (7), there is one case where the naive ideal $I = \langle g_1x_1 − f_1, \dots , g_nx_n − f_n\rangle$ obtained by “clearing denominators” gives the right answer. Suppose that $x_i = f_i(t)/g_i(t)$ where there is only one parameter $t$. We can assume that for each $i$, $f_i(t)$ and $g_i(t)$ are relatively prime in $k[t]$ (so in particular, they have no common roots). If $I ⊆ k[t, x_1, \dots , x_n]$ is as above, then prove that $\mathcal{V}(I_1)$ is the smallest variety containing $F(k \setminus W)$, where as usual $g = g_1 \cdots g_n ∈ k[t]$ and $W = \mathcal{V}(g) ⊆ k$. Hint: In diagram (8), show that $i(k \setminus W) = \mathcal{V}(I)$
Attempt.
We start proving $i(k\setminus W)=\mathcal{V}(I)$. For the non-trivial inclusion let be $(t,x_1,\dotsc,x_n)\in \mathcal{V}(I)$. We have to check $t\in k \setminus W$. We know $g_i(t)x_i = f_i(t)$. If $g_i(t) = 0$ then $f_i(t)=0$ so the have a common root. Thus, we can clear $x_i$ en $g_i(t)x_i = f_i(t)$.
Since $F(k\setminus W)=\pi_1(i(k\setminus W))=\pi_1(\mathcal{V}(I)) \subset \mathcal{V}(I_1)$, $\mathcal{V}(I_1)$ is a variety containing $F(k \setminus W)$. We need to check is the smaller. Let be $h\in k[x_1,\dotsc,x_n]$ vanishing in $F(k/W)$, we'll show $h\in I_1$.
But I can't show it even following the proof of the theorem in the book.