Let \(Y, Z \subseteq \AA^2\) be two distinct curves, given by equations \(f = 0\) and \(g = 0\). For \(P \in Y \intersect Z\), define the intersection multiplicity \((Y \cdot Z)_P\) of \(Y\) and \(Z\) at \(P\) to be the length of the \(\mco_P\da\)module \(\mco_P / \gens{f, g}\).
Show that \((Y \cdot Z)_P\) is finite, and that \((Y \cdot Z)_P \geq \mu_P(Y) \cdot \mu_P(Z)\).
If \(P \in Y\), show that for almost all lines \(L\) through \(P\), that is, all but a finite number, one has \((L \cdot Y)_P = \mu_P(Y)\).
If \(Y\) is a curve of degree \(d\) in \(\PP^2\), and \(L\) is a line in \(\PP^2\) with \(L \neq Y\), show that \((L \cdot Y) = d\). Here \((L \cdot Y) = \sum (L \cdot Y)_P\), summed over all points \(P \in L \intersect Y\), where \((L \cdot Y)_P\) is computed using a suitable affine cover of \(\PP^2\).
Solution
Because \(k\) is algebraically closed, every closed point considered below has residue field \(k\). Translate coordinates in parts (1) and (2) so that \(P=(0,0)\), and put \[ R=k[x,y]_{(x,y)},\qquad \mathfrak m=(x,y)R. \] Write \[ r=\mu_P(Y),\qquad s=\mu_P(Z). \]
1. The \(R\)-module \(R/(f,g)\) has finite length.
The two irreducible plane curves are distinct, so their irreducible equations \(f\) and \(g\) are not associates and have no common irreducible factor. The polynomial ring \(k[x,y]\) is a UFD. If a height-one prime contained both \(f\) and \(g\), that prime would be generated by an irreducible common factor. Hence every prime containing \((f,g)\) has height two. After localizing at \((x,y)\), the quotient \(R/(f,g)\) is therefore a noetherian local ring of dimension zero. Such a ring is Artinian, so it has finite length. This proves the finiteness assertion in part (1).
2. One has\[ \boxed{(Y\cdot Z)_P\ge rs=\mu_P(Y)\mu_P(Z).} \]
Completion preserves the length of a finite-length module, so \[ (Y\cdot Z)_P =\dim_k k[[x,y]]/(f,g). \] Choose a linear coordinate change for which neither tangent cone vanishes identically on the \(y\)-axis. This is possible because the two nonzero homogeneous tangent cones have only finitely many zeros in \(\PP^1\) and \(k\) is infinite. Then \(f(0,y)\) has order exactly \(r\) and \(g(0,y)\) has order exactly \(s\).
By Weierstrass preparation in \(k[[x,y]]\), multiplication by units replaces \(f,g\) by monic polynomials \[ F=y^r+a_{r-1}(x)y^{r-1}+\cdots+a_0(x), \qquad G=y^s+b_{s-1}(x)y^{s-1}+\cdots+b_0(x) \] in \(k[[x]][y]\) without changing the quotient. Their \(\mathfrak m\)-orders are still \(r\) and \(s\). Consequently \[ \operatorname{ord}_x a_i\ge r-i,\qquad \operatorname{ord}_x b_j\ge s-j, \] because otherwise \(a_i(x)y^i\) or \(b_j(x)y^j\) would contain a term of total degree below the relevant multiplicity.
Put \(A=k[[x]]\). The \(A\)-module \(B=A[y]/(F)\) is free with basis \(1,y,\ldots,y^{r-1}\). Multiplication by \(G\) defines an \(A\)-linear endomorphism \[ m_G:B\longrightarrow B \] whose cokernel is \(A[y]/(F,G)\). Step 1 implies that this cokernel has finite length, so \(\det(m_G)\ne0\). The determinant is, up to sign, the resultant \(\operatorname{Res}_y(F,G)\). Smith normal form over the DVR \(A\) therefore gives \[ \dim_k A[y]/(F,G) =\operatorname{ord}_x\operatorname{Res}_y(F,G). \]
Now set \[ \widetilde F(x,z)=x^{-r}F(x,xz),\qquad \widetilde G(x,z)=x^{-s}G(x,xz). \] The coefficient inequalities above show that both belong to \(A[z]\) and are monic of degrees \(r\) and \(s\) in \(z\). The standard homogeneity of the resultant under scaling of the variable and of the two polynomials gives \[ \operatorname{Res}_y(F,G) =x^{rs}\operatorname{Res}_z(\widetilde F,\widetilde G). \] For completeness, this identity follows from the two equalities \[ \operatorname{Res}_z(F(x,xz),G(x,xz)) =x^{rs}\operatorname{Res}_y(F,G) \] and \[ \operatorname{Res}_z(x^r\widetilde F,x^s\widetilde G) =x^{2rs}\operatorname{Res}_z(\widetilde F,\widetilde G). \] Thus the \(x\)-adic order of the resultant is at least \(rs\). Combining the preceding equalities proves the desired inequality.
3. For all but finitely many lines \(L\) through \(P\),\[ \boxed{(L\cdot Y)_P=\mu_P(Y).} \]
Write the translated equation as \[ f=f_r+f_{r+1}+\cdots, \] with \(f_r\ne0\) homogeneous of degree \(r\). Apart from the vertical line, every line through the origin is \[ L_\lambda: y=\lambda x. \] Its intersection multiplicity with \(Y\) is \[ \operatorname{length}k[x]_{(x)}/(f(x,\lambda x)), \] which is the order of vanishing at \(x=0\) of the one-variable polynomial \(f(x,\lambda x)\). One has \[ f(x,\lambda x) =x^r\bigl(f_r(1,\lambda)+x f_{r+1}(1,\lambda)+\cdots\bigr). \] Therefore the order is exactly \(r\) whenever \(f_r(1,\lambda)\ne0\). The polynomial \(f_r(1,T)\) is nonzero, since dehomogenization of a homogeneous polynomial is injective, so it has only finitely many roots. Adding the one omitted vertical line, the exceptional lines are precisely the finitely many tangent directions, i.e. the linear factors of the tangent cone \(f_r\). All other lines give intersection multiplicity \(r\).
4. Let \(Y\subseteq\PP^2\) have degree \(d\) and let \(L\ne Y\) be a line.Then\[ \boxed{(L\cdot Y)=d.} \]
Choose projective coordinates so that \(L=V(z)\). Let \(F(x,y,z)\) be the irreducible homogeneous equation of \(Y\), of degree \(d\). Since \(L\ne Y\), the binary form \[ F(x,y,0) \] is nonzero. Over the algebraically closed field it factors as \[ F(x,y,0)=c\prod_{i=1}^m\ell_i(x,y)^{e_i}, \qquad e_1+\cdots+e_m=d, \] where the \(\ell_i\) are pairwise nonproportional linear forms. Their zeros are exactly the points \(P_1,\ldots,P_m\) of \(L\cap Y\).
Fix \(P_i\) and choose an affine chart and a local coordinate \(t\) on \(L\) vanishing at \(P_i\). The restriction of \(F\) to \(L\) is a unit times \(t^{e_i}\) in the local ring of \(L\) at that point. Quotienting the two-dimensional regular local ring of the affine plane first by the equation of \(L\) therefore gives \[ \frac{\OO_{\PP^2,P_i}}{(F,z)} \cong \frac{\OO_{L,P_i}}{(t^{e_i})}, \] whose length is \(e_i\). Thus \[ (L\cdot Y)_{P_i}=e_i. \] Summing over all intersection points gives \[ (L\cdot Y)=\sum_i e_i=d. \] This proves part (3) directly from the degree of the restricted binary form.
5. Q.E.D.
Steps <1>1–<1>2 prove part (1), step 3 proves part (2), and step 4 proves part (3).