Let $F$ be an infinite field and let $f, g \in F[x, y]$. If $f$ and $g$ have no common factors then $V(f) \cap V(g)$ is finite. Note that for $S \subseteq F[x, y], $$V(S)=\left\{a \in F^2 : f(a)=0 \ \forall \ f \in S \right\}$ is a variety.
Does this hold for the higher dimensions? This property is useful when I want to find the closed ideal of a variety $X$ in $F^2$ are prime (e.g. Let $X=V(f)$, $f$ irreducible and $g \in I(X)$, then if $V(f) \cap V(g)=V(f)$ infinite and hence $f$ and $g$ have a common factor, so $f$ irreducible implies $f$ is a common factor of $g$, i.e., $g \in (f)$).
I want to know if I can use the similar arguments if I'm given varieties in $F^n$.
Here is an example demonstrating Gerry Myerson's comment that $V(f) \cap V(g)$ is not necessarily finite if $f$ and $g$ share no common factor.
Let $\mathbb{F} = \mathbb{C}$, $f(x,y,z) = xyz \in \mathbb{C}[x,y,z]$ and $g(x,y,z) = x^2 + y^2 - 1$. Clearly $f$ and $g$ share no common factors,as the irreducible factors of $f$ are $x$,$y$ and $z$ and none of these divide $x^2 + y^2 - 1$. However the intersection $V(f) \cap V(g)$ contains $\{(x,y,0) \in \mathbb{C} : x^2 + y^2 - 1 = 0\}$ which is infinite.
Or, even easier, take $f(x,y,z) = x$ and $g(x,y,z) = y$. Then $V(f) \cap V(g) = \{(0,0,z) : z \in \mathbb{C}\}$.