A common component makes the intersection sum infinite

Given: AC The Axiom of Choice, the algebraically closed field $k$, the plane projective curves $C=V(x_0)$ of degree one and $D=V(x_0{x}_1)$ of degree two, and an arbitrary point $p$ of the shared line. In a chart $x_i=1$ containing $p$ (where $i=1$ or $2$), write $f=x_0\mathrm{/}{x}_i$ and $g=(x_0\mathrm{/}{x}_i)(x_1\mathrm{/}{x}_i)$ for the local equations.

$x_0$ and $x_0{x}_1$ are nonconstant square-free forms, so $C$ and $D$ are plane projective curves; $V(x_0)$ is an irreducible component of both, so $C$ and $D$ share the component $V(x_0)$ Plane projective curves and their components.

At $p\in V(x_0)$ the local ideal is $(f,g)=(x_0\mathrm{/}{x}_i)$, whose nonunit irreducible generator is a local coordinate vanishing on the line; hence $O\mathrm{/}(f,g)$ is not of finite length and the definition records $I_p(C,D)=\mathrm{\infty }$ Local intersection multiplicity of two plane curves, Finite local length exactly when no common local branch, Prime ideals and maximal ideals in a commutative ring.

The Bezout theorem is stated under the no-common-component hypothesis; it computes a finite sum of finite multiplicities Bezout's theorem for plane projective curves. A module of infinite length is not a finite summand Composition series and length of a module.