Do two rational parametric curves intersect only finitely many times? Suppose there are two  rational parametric curves $f = (f_1, \ldots, f_n)$ and $g = (g_1, \ldots, g_n)$ in $\mathbb{R}^n$. I read somewhere that such a parametric expression can always be transformed into an implicit form. I am not sure of this because I didn't find a proof (and I'm yet to do an algebraic geometry course). If it is true, then $f$ can be written in the  form $F(x_1, \ldots, x_n) = 0$. Then the points of intersection of $f$ and $g$ corresponds to the real roots of $F(g_1(t), \ldots, g_n(t)) = 0$, which are only finitely many, if not identically zero.
Thank you in advance.
 A: $$\newcommand{\spec}[1]{\mathrm{Spec}({#1})}$$
Let 
$$
\Phi: t \mapsto (\psi_1(t),\ldots, \psi_n(t))
$$
be a map from $k - \{P_1,\ldots, P_s\}$ to $k^n$ where $k$ is an arbitrary field, $\psi_i(t) = p_i(t)/q_i(t)$ with polynomials $p_i, q_i \in k[t]$ and $P_i$ are the zeroes of all the $q_i(t)$. Furthermore let $\Delta(t) = \prod_i q_i(t)$.
Now consider the map $f:A = k[x_1,\ldots,x_n] \to k[t]_{\Delta(t)} \subset k(t)$ with $f(x_i) = \psi_i(t)$. Call $I = \ker f$, the kernel-ideal of $f$ in $A$.
Then we have a sequence of integral domains and $k$-algebras
$$(**)\quad A \twoheadrightarrow A/I \hookrightarrow k[t]_{\Delta(t)} \hookrightarrow k(t)$$
Now citing a theorem of algebraic geometry, every finitely generated integral $k$--algebra $R$ corresponds to a variety (a scheme) $X = \spec{R}$, which has as its closed points the maximal ideals of $R$. Furthermore $X$ is irreducible (can not be decomposed into the union of two proper closed subsets) and reduced ($R$ has no nilpotent elements).
The ring $A$ from above corresponds to the variety $\mathbb{A}^n_k$, the $n$-dimensional affine space over $k$. A surjection $R \twoheadrightarrow S$ corresponds to a "closed immersion" of varieties $\spec{S} \hookrightarrow \spec{R}$ which is a "good embedding" of $\spec{S}$ as a closed subvariety in $\spec{R}$. A localization $R_r$ with $r \in R$ corresponds to the variety $\spec{R_r} = D(r)$ which is the complement of the zero-set $V(r)$ in $\spec{R}$.
So we can read the sequence $(**)$ as saying that it gives the map $\Phi$ of the affine $1$-space minus the points $P_1,\ldots,P_s$ (which corresponds to $k[t]_{\Delta(t)}$) into $\mathbb{A}^n_k$. The image $\Phi(\spec{k[t]_{\Delta(t)}})$ is dense in the closed subvariety $\spec{A/I}$ of $\mathbb{A}^n_K = \spec{A}$ (the denseness follows from the injectivity of $A/I \hookrightarrow k[t]_{\Delta(t)}$.
Now in case of a general field $k$ the ideal $I$ has for $n > 2$ more than one generator for reasons of dimension. But in case $k = \mathbb{R}$ we can
take generators $F_1,\ldots,F_r \in I$ and form the single polynomial
$F_1^2 + \cdots + F_r^2 = F$. Then $V(F)$ has the same real zeros in $\mathbb{R}^n$ as $F_1,\ldots,F_r$ together.
Computing $I$ is easy with a system like Macaulay 2 (try it online):
i13 : R=QQ[x,y,z]

o13 = R

o13 : PolynomialRing

i14 : S = QQ[t]

o14 = S

o14 : PolynomialRing

i15 : KS = frac S

o15 = KS

o15 : FractionField

i16 : phi = map(KS, R, {t^2/(t^2+1),(1-t^2)/(1+t^2),2*t/(1+t^2)})

                   2       2
                  t     - t  + 1    2t
o16 = map(KS,R,{------, --------, ------})
                 2        2        2
                t  + 1   t  + 1   t  + 1

o16 : RingMap KS <--- R

i17 : ker phi

                          2    2
o17 = ideal (2x + y - 1, y  + z  - 1)

o17 : Ideal of R

So we get the intersection of a plane and a cylinder as the image of $\Phi$ in this case.
