A question about $K[x_1,x_2]$ Let $K$ be a field. Let  $p(x_1,x_2),q(x_1,x_2)\in K[x_1,x_2]$ such that $p(x_1,x_2),q(x_1,x_2)$ are   linearly independent over $K[x_1,x_2]$ (considering $K[x_1,x_2]$ as a $K[x_1,x_2]$-module). Must the set $S:=\{(u_1,u_2)\in K\times K\mid p(u_1,u_2)=q(u_1,u_2)=0\}$ be finite ?
If the answer is yes, is it possible to bound the size of $S$ by $\deg(p),\deg(q)$ ?
 A: Any two $p,q$ will be linearly dependent over $k[x_1,x_2]$ so what I think you want is if $p,q$ have no common factors. Now if $(p,q) = (1)$ then their vanishing locus will be empty so we may suppose that $p,q$ have no common factors and $(p,q) \neq (1)$. Now let $F = k(x_1)$. Then Gauss' Lemma says that $p,q$ have no common factors (relatively prime) in $k[x_1,x_2]$ iff they have no common factors in $F[x_2] = k(x_1)[x_2]$. This is a polynomial ring over a field, so having no common factors says that $(p,q)F[x_2] = 1$, i.e. there is $f_1$ and $f_2$ in $F[x_2]$ so that 
$$f_1p + f_2q = 1.$$
Clearing denominators we get that $(p,q)$ contains a polynomial $r \in k[x_1]$. But now $V(p,q) \subseteq V(r)$ the latter being a finite set. This shows that your set is finite.
If you are looking for a bound on your set, you should look at Corollary 4 Section 1.7 of Fulton's Algebraic Curves which says that the number of points in $V(p,q)$ is at most the dimension of $k[x_1,x_2]/(p,q)$ (which is necessarily finite).
