[proofplan] The argument is a bookkeeping consequence of projective Bezout. First we check that the homogenized equations $F$ and $G$ have no common projective component, so Bezout applies to $\overline C$ and $\overline D$. Then we split the projective intersection points into the affine chart $Z\ne 0$ and the line at infinity $Z=0$. On the affine chart, dehomogenization identifies the relevant projective local [quotient ring](/page/Quotient%20Ring) with the original affine local quotient ring, so the local intersection multiplicities agree. The remaining projective summands are exactly the summands at infinity. [/proofplan] [step:Handle the case where one curve is empty] If $m=0$, then $f\in k^\times$, so $C=V(f)=\varnothing$ and $F=f\in k^\times$, hence $\overline C=V_+(F)=\varnothing$. Both sums in the displayed formula are empty, and the right-hand side is $mn=0\cdot n=0$. The same argument applies if $n=0$. Thus it remains to prove the theorem when $m,n\ge 1$. [/step] [step:Show that the projective closures have no common component] Let $S=k[X,Y,Z]$ and let $A=k[x,y]$. We claim that $F$ and $G$ have no common irreducible homogeneous factor in $S$. Suppose, toward a contradiction, that $H\in S$ is a nonconstant irreducible [homogeneous polynomial](/page/Homogeneous%20Polynomial) dividing both $F$ and $G$. Since $F$ is the degree-$m$ homogenization of the nonzero polynomial $f$, some degree-$m$ term of $f$ occurs in $F$ without a factor of $Z$; hence $Z\nmid F$. Similarly, $Z\nmid G$. Therefore $H$ is not associated to $Z$. Define the dehomogenization of $H$ on the chart $Z\ne 0$ by \begin{align*} h:=H(x,y,1)\in A. \end{align*} Because $H$ is homogeneous, nonconstant, and not divisible by $Z$, the polynomial $h$ is nonconstant. Since $H\mid F$ and $H\mid G$ in $S$, evaluating at $Z=1$ gives $h\mid f$ and $h\mid g$ in $A$. The polynomial $h$ has an irreducible factor in $A$, and that irreducible factor divides both $f$ and $g$, contradicting the hypothesis. Thus $F$ and $G$ have no common projective component. [/step] [step:Apply projective Bezout to the homogenized curves] Since $m,n\ge 1$, the projective plane curves $\overline C=V_+(F)$ and $\overline D=V_+(G)$ have degrees $m$ and $n$. By the previous step, they have no common irreducible component. Therefore [Bezout's theorem](/theorems/2131) for plane projective curves applies (citing a result not yet in the wiki: Bezout's Theorem for Plane Projective Curves), and gives \begin{align*} \sum_{p\in \overline C(k)\cap \overline D(k)} I_p(\overline C,\overline D)=mn. \end{align*} Here $I_p(\overline C,\overline D)$ is the projective local intersection multiplicity, equivalently the length over the local ring $\mathcal O_{\mathbb P_k^2,p}$ of the quotient by the two local equations $F$ and $G$. [/step] [step:Identify affine and projective local multiplicities on the chart $Z\ne 0$] Let $U_Z\subset\mathbb P_k^2$ denote the standard affine chart $Z\ne 0$. Define the chart isomorphism \begin{align*} \psi:\mathbb A_k^2\to U_Z \end{align*} by \begin{align*} (a,b)\mapsto [a:b:1]. \end{align*} On coordinate rings this isomorphism is induced by the $k$-algebra isomorphism \begin{align*} \theta:A\to (S_Z)_0 \end{align*} defined by $\theta(x)=X/Z$ and $\theta(y)=Y/Z$, where $(S_Z)_0$ denotes the degree-zero part of the localization $S_Z$. Let $q=(a,b)\in C(k)\cap D(k)$ and set $p=\psi(q)=[a:b:1]$. Let $\mathfrak m_q=(x-a,y-b)\subset A$ and let $\mathfrak m_p\subset (S_Z)_0$ be the corresponding maximal ideal under $\theta$. The dehomogenized local equations satisfy \begin{align*} \theta(f)=F/Z^m \end{align*} and \begin{align*} \theta(g)=G/Z^n. \end{align*} Since $Z$ is invertible on $U_Z$, replacing $F$ by $F/Z^m$ and $G$ by $G/Z^n$ does not change the generated ideal in the local ring at $p$. Localizing $\theta$ at $\mathfrak m_q$ and $\mathfrak m_p$ gives an isomorphism of local $k$-algebras \begin{align*} A_{\mathfrak m_q}/(f,g)A_{\mathfrak m_q}\cong \mathcal O_{\mathbb P_k^2,p}/(F,G)\mathcal O_{\mathbb P_k^2,p}. \end{align*} Taking lengths over these isomorphic local rings gives \begin{align*} I_q(C,D)=I_p(\overline C,\overline D). \end{align*} [guided] The point of this step is to justify that no intersection multiplicity is lost or changed when we pass from the affine plane to the projective closure on the chart $Z\ne 0$. Let $U_Z\subset\mathbb P_k^2$ be the standard affine chart where $Z$ is nonzero. The map \begin{align*} \psi:\mathbb A_k^2\to U_Z \end{align*} is defined by \begin{align*} (a,b)\mapsto [a:b:1]. \end{align*} Its coordinate-ring form is the isomorphism \begin{align*} \theta:k[x,y]\to (k[X,Y,Z]_Z)_0 \end{align*} with $\theta(x)=X/Z$ and $\theta(y)=Y/Z$. This says algebraically that functions on the affine chart are obtained by setting the projective coordinate $Z$ equal to $1$. Now fix $q=(a,b)\in C(k)\cap D(k)$, and let $p=[a:b:1]$ be the corresponding projective point. The affine local ring at $q$ is $k[x,y]_{\mathfrak m_q}$, where \begin{align*} \mathfrak m_q=(x-a,y-b). \end{align*} The projective local ring at $p$ may be computed inside the chart $U_Z$, so it is the localization of $(k[X,Y,Z]_Z)_0$ at the maximal ideal corresponding to $p$. Under the chart isomorphism, these two local rings are isomorphic. It remains to check that the two equations match under this local-ring identification. Since $F$ is the degree-$m$ homogenization of $f$, dehomogenizing on $Z\ne 0$ gives \begin{align*} \theta(f)=F/Z^m. \end{align*} Similarly, \begin{align*} \theta(g)=G/Z^n. \end{align*} The elements $Z^m$ and $Z^n$ are units in the chart local ring because $Z\ne 0$ on $U_Z$. Therefore the ideals generated by $F$ and $G$ are the same as the ideals generated by $F/Z^m$ and $G/Z^n$ in the local ring at $p$. Consequently the chart isomorphism induces an isomorphism \begin{align*} k[x,y]_{\mathfrak m_q}/(f,g)k[x,y]_{\mathfrak m_q}\cong \mathcal O_{\mathbb P_k^2,p}/(F,G)\mathcal O_{\mathbb P_k^2,p}. \end{align*} Local intersection multiplicity is the length of precisely this local quotient. Isomorphic local quotient rings have the same length, so \begin{align*} I_q(C,D)=I_p(\overline C,\overline D). \end{align*} [/guided] [/step] [step:Split the Bezout sum into affine and infinite parts] The projective plane is the disjoint union of the affine chart $U_Z=\{Z\ne 0\}$ and the line at infinity $V_+(Z)=\{Z=0\}$. Hence the finite set $\overline C(k)\cap\overline D(k)$ is the disjoint union \begin{align*} \overline C(k)\cap\overline D(k)=\bigl(\overline C(k)\cap\overline D(k)\cap U_Z\bigr)\cup\bigl(\overline C(k)\cap\overline D(k)\cap V_+(Z)\bigr). \end{align*} Under the chart isomorphism $\psi:\mathbb A_k^2\to U_Z$, the first subset is identified with $C(k)\cap D(k)$, because the equations $F=0$ and $G=0$ dehomogenize to $f=0$ and $g=0$. Therefore the Bezout sum decomposes as \begin{align*} \sum_{p\in \overline C(k)\cap \overline D(k)} I_p(\overline C,\overline D)=\sum_{q\in C(k)\cap D(k)} I_q(C,D)+\sum_{p\in \overline C(k)\cap\overline D(k)\cap V_+(Z)} I_p(\overline C,\overline D). \end{align*} Combining this decomposition with the Bezout equality \begin{align*} \sum_{p\in \overline C(k)\cap \overline D(k)} I_p(\overline C,\overline D)=mn \end{align*} gives the claimed identity. [/step]