Bezout fails on the affine plane because points at infinity are missing

Given: AC The Axiom of Choice, affine coordinates $(x,y)$ on ${\mathbf{A}}^2\subset {\mathbf{P}}^2$ with $x=x_1\mathrm{/}{x}_0$, $y=x_2\mathrm{/}{x}_0$, the affine lines $V(x)$ and $V(x-1)$, and their projective closures $C_1=V(x_1)$, $C_2=V(x_1-x_0)$ in ${\mathbf{P}}^2$.

$x_1$ and $x_1-x_0$ are square-free linear forms, so $C_1$ and $C_2$ are plane projective curves of degree one, with no common component; each is a line Plane projective curves and their components.

The affine parts of $C_1$ and $C_2$ are the parallel lines $x=0$ and $x=1$, which are disjoint in ${\mathbf{A}}^2$; hence the count of affine intersection points counted with multiplicity is $0$: $x=0$ and $x=1$ cannot hold simultaneously since $0\mathrm{\ne }1$.

The projective closures meet in the point $[0:0:1]$: solving $x_1=0$ and $x_1-x_0=0$ gives $x_0=x_1=0$ with $x_2\mathrm{\ne }0$, i.e. $[0:0:1]$, a point at infinity of the affine chart $x_0=1$; in the chart $x_2=1$ the local ideal is $(x_1,x_1-x_0)=(x_0,x_1)$, so its quotient is the residue field $k$, of length one projective space points, Local intersection multiplicity of two plane curves.

Bezout for the two projective lines gives ${\sum }_p{I}_p(C_1,C_2)=1\cdot 1=1$, realised at the single point at infinity Bezout's theorem for plane projective curves.