Intersection multiplicity Bezout's theorem

Question

I was reading Algebraic Curves by Fulton, where I have doubt in applying Bezout's theorem. Let $$F:=x^2+y^2-z^2$$ $$ G:=x^2+y^2-2 z^2 $$ be two curves in $ \mathbb{P}^2$, $\deg F=\deg G=2$ $\therefore$ by Bezout's Theorem $$\sum_{p \in C \mathbb{P}^2} I(P, F \cap G)=2 \cdot 2=4.$$ Note that $[1, i, 0]$ lies on $F$ and $G$. Then how to find $I(P, F \cap G)$, where $p=[1: i: 0]$? $$ \begin{aligned} & I(p, F \cap G)=\frac{\mathcal{O}_p\left(\mathbb{P}^2\right)}{\left(F_*, G_*\right)} \\ & F_*=x^2+y^2-1 \\ & G_*=x^2+y^2-2 \end{aligned} $$ but I am getting $\left(F_*, G_*\right)=(1)$ which will give me $I(p, F\cap G)=0$ which is not true.

I am not getting how to find $ I(p, F \cap G)=\frac{\mathcal{O}_p\left(\mathbb{P}^2\right)}{\left(F_*, G_*\right)}$ for $p=[1:i:0]$.

I will appreciate any hints. Thanks.

Answer 1

In $\bar{\Bbb Q}[x,y,z]$ (or even ${\Bbb Q}(\sqrt{-1})[x,y,z]$) $$\langle x^2+y^2-z^2,x^2+y^2-2z^2 \rangle=\langle x^2+y^2,z^2 \rangle=\langle (x-\sqrt{-1}y)(x+\sqrt{-1}y),z^2 \rangle=\\\langle x-\sqrt{-1}y,z^2 \rangle\cap \langle x+\sqrt{-1}y,z^2 \rangle.$$

But you want the local ring around $p=[1:\sqrt{-1}:0].$

In the affine $x=1,$ localized at ${\mathfrak m}_p=\langle y-\sqrt{-1},z\rangle,$ that is in the local ring $k[y,z]_{{\mathfrak m}_p}$ the ideal in question is $\langle y-\sqrt{-1},z^2 \rangle.$

The latter corresponds to a line intersecting a double line resulting in a double point supported on $p$. For $q=[1:-\sqrt{-1}:0]$ it goes completely similarly, so there are two double points, and Bezout holds.

Answer 2

Personally, I like to calculate intersection multiplicities using discrete valuations. This may not be the cleverest approach, but I'm generally confident that I can get this approach to work in a reasonable amount of time.

Since the problem is essentially local in nature, let's assume that the curves $C_1$ and $C_2$ are affine curves, living inside $\mathbb A^2$. Let $f$ and $g$ be the polynomials that define $C_1$ and $C_2$ respectively.

Suppose that the point $p$ lies on the intersection of $C_1$ and $C_2$. Some simple commutative algebra shows that $$ \frac{\mathcal O_{p}(\mathbb A^2)}{(f, g) } \cong \frac{\mathcal O_{p}(C_1)}{(g)}.$$ If $C_1$ is smooth at $p$, then $\mathcal O_p(C_1)$ is a discrete valuation ring, and $$ \text{dim}_k \left(\frac{\mathcal O_{p}(C_1)}{(g)} \right) = v_p(g),$$ where $v_p(g)$ is the valuation of $g$ in $\mathcal O_p(C_1)$.

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In your example, we would take $\mathbb A^2$ to be the affine patch $$ \mathbb A^2 \cong \{ [1: y : z] : (y, z) \in \mathbb A^2 \} \subset \mathbb P^2.$$

The curves $C_1$ and $C_2$ are defined by the polynomials $$ f(y, z) = 1 + y^2 - z^2, \ \ \ \ \ g(y, z) = 1 + y^2 - 2z^2.$$

The point $p$ is located at $(y, z) = (i, 0)$.

The gradient of $f$ at $p$ is $$ \nabla f (p) = \left( \frac{\partial f}{\partial y}(p) , \frac{\partial f}{\partial z} (p) \right) = ( 2i , 0) . $$

Since $\nabla f(p)$ is non-zero, the curve $C_1$ is smooth at $p$.

The vector $(0, 1)$ complements the vector $\nabla f(p) = (2i, 0)$ to form a basis for the $2$-dimensional vector space $k^2$. Hence $z$ is a local parameter for $C_1$ at $p$; that is, $z$ generates the unique maximal ideal in the discrete valuation ring $\mathcal O_{p}(C_1)$.

Viewing $g$ as an element of $\mathcal O_{p}(C_1)$, we have $$ g = 1 + y^2 - 2z^2 = -z^2 + (1 + y^2 - z^2) = -z^2 + 0 = -z^2.$$ Notice that we have explicitly written $g$ in the form $uz^2$, where $u := -1$ is a unit in $\mathcal O_{p}(C_1)$. So the valuation of $g$ in $\mathcal O_{p}(C_1)$ is $$v_p(g) = 2.$$

Thus we have shown that the intersection multiplicity of $C_1$ and $C_2$ at $p$ is $2$.