Intersection Multiplicity : curves with common component

Question

I am trying to do Exercise 2.7, pg13

Exercise 2.7: Let $F,G$ be two curves through a point $P \in \Bbb A^2$. Show:

(a) If $F,G$ have no common component, the family $(F^n)$ is linearly independent in $O_P/\langle G \rangle $

(b) If $F,G$ have a common component through $P$ then $\mu_P(F,G)=\infty$.

Definition: $O_P$ refers to the stalk of sheaf of regular functions on $\Bbb A^2$ at $P$. The multiplicity $\mu_P(F,G)$ is $$\dim O_p/\langle F,G \rangle $$

I am stuck at (2), and have problem interpreting what (a) means in terms of $\mu_P(F,G)$.

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My attempt:

(a) Suppose $\sum_{ n \ge 0} k_n F^n = f/g \cdot G'H$, where $G'$ is the component of $G=G'H$ thorugh $P$. Then we proceed by induction.

$G'(P)=0$, implies $k_0=0$. Thus, factoring $F$ out, we see that $G'| \sum_{n \ge 1} k_n F^{n-1}$. We repeat the argument.

(b) ??

Answer 1

There's a minor confusion here because you're using $F$ and $G$ for the varieties as well as the equations cutting them out. I'll rephrase this as $F=V(f)$ and $G=V(g)$ for this answer.

Hint for (a): Factor out the highest power of $f$ you can in the LHS of your relation. Look at what's you get when you do this - what conclusions can you draw about each of the factors?

Full solution for (a):

Suppose that there's a linear relation between some powers of $f$ in $O_P/(g)$. Then after factoring out the lowest power of $f$ appearing in the relation, we can write our relation as $f^n\cdot(k_{n+i}f^i+\cdots+k_n)$ with $k_n\neq 0$. But that means that the latter term is a unit, so that means that $f^n=0$ in $O_P/(g)$. This is a contradiction, as $F$ and $G$ having no common components means that $f$ and $g$ are relatively prime and no power of $f$ is in $(g)$.

Hint for (b): If $F,G$ have a common component, then we can write $f=f'h^a$ and $g=g'h^b$ where $h$ is the equation of the common component $H$. Consider the surjective ring homomorphism $O_P/(f,g)\to O_P/(h)$. Do you see how to apply (a) to this?

Full solution for (b):

If you can find a curve which has no common component with $H$, then you'll find an infinite linearly independent set in $O_P/(h)$ by part (a). Since $O_P/(f,g)$ surjects on to $O_P/(h)$, this means that $\dim_k O_P/(f,g) \geq \dim_k O_P/(h)$, and this latter quantity is infinite by the presence of an infinite linearly-independent collection. All that's left is to actually find this curve. The recipe is easy: pick a line along which the $h$ is not zero (some care may be required with finite fields, but it can be done via taking a line defined over some algebraic extension and then using the product of the galois conjugates of the equation defining this line.)