In a class on algebraic curves we were given the problem to show, that the union of the three axes $$C=\lbrace(x,y,z)\in\mathbb{C}^3\vert xy=yz=xz=0\rbrace$$
is not the intersection of two surfaces, so for any $F,\tilde{F}\in C[X,Y,Z]$ we have $$C\neq \lbrace (x,y,z)\in\mathbb{C}^3\vert F(x,y,z)=\tilde{F}(x,y,z)=0\rbrace.$$
We were given the hint to use $\mathbb{CP}^2$ and Bézout's theorem. I tried the following:
The polynomials $xy$, $yz$, $xz$ are homogeneous, so we pass to $\mathbb{CP}^2$ and try to show, that there are no two projective curves, such that their intersection consists of exactly the three points $[1:0:0],[0:1:0],[0:0:1]$ (corresponding to the three axes). The naive approach is to say if two curves intersect in exactly three points, one of them has to be of degree 1 and one of degree 3 and the three points do not lie on a line, however we have to count intersections with multiplicity. So assume $F, \tilde{F}$ are homogeneous of degrees $d$ and $\tilde{d}$ such that $C=\lbrace F=0, \tilde{F}=0\rbrace$. By plugging in the three points we see easily, that the coeffitients of $x^d$, $y^d$, $z^d$, $x^\tilde{d}$, $y^\tilde{d}$, $z^\tilde{d}$ in $F$ respectively $\tilde{F}$ are $0$. So when computing the resultant of $F$ and $\tilde{F}$ with respect to $Z$ we get, that the upper row is zero, so the resultant is zero, so $F$, $\tilde{F}$ share a component and we get more intersections hence a contradiction.
Now I have two questions. First: Although I have shown, that if $F$, $\tilde{F}$ are any two homogeneous poynomials intersecting at least in $[1:0:0],[0:1:0],[0:0:1]$, then they have a common component. I have the feeling, that one should be able to "see" what this common component is. Is this possible?
Second: It os not clear to me why I am done here: Why would any two (affine) polynomials $F,\tilde{F}$ intersecting exacly in the axes have to be homogeneous?