Zero locus of polynomials in three variables I am working through what seems to be a generalization of Bezout's Theorem for affine varieties. $f(x,y,z)$ and $g(x,y,z)$ are homogeneous non-constant polynomials. I need to show that:
If $\gcd{(f,g)}=1$, then $V(f,g) = \{(a,b,c)\in \mathbb{A}^3 : f(a,b,c) = 0 = f(a,b,c)\}$ is the union of finitely many lines through $(0,0,0)$ in $\mathbb{A}^3$
We are only working with affine geometry, so no projective theory can be used. I have been trying to do something with the resultant which lies in $k[x,y]$.
EDIT: $k$ is algebraically closed.
 A: Let $k$ be the base field, which I do not suppose algebraically closed.  
i) It is clear that $C=V(f,g)=\bigcup_{i\in I} L_i$ is a union  of lines i.e. a cone (or is reduced to the origin: e.g. $f=x^2+y^2,\; g=x^2+z^2$ over $\mathbb R$).     
ii) To prove the finiteness of $I$, let us look at the intersection $F=C\cap P$ of $C$ with the affine plane $P$ given by  $z=1$.
This intersection has as equations   $z=1$ and  $f_*=f(x,y,1)=0,\; g_*=g(x,y,1)=0$ .
The latter two equations have finitely many solutions because$f_*,g_*\in k[x,y]$ are relatively prime (cf. Fulton, Algebraic Curves Corollary, page 24 and Proposition 2, page 9 )
So only finitely many lines of our cone $C$ cut the plane $P$.     
iii) Similarly only finitely many lines of the cone cut the plane $y=1$ resp. the plane $x=1$.
Since any line in $k^3$ must cut at least one of these affine planes, $C$ consists of finitely many lines, as desired. 
Remark
The above proof is affine camouflage for a more natural projective argument.
