# Max Noether's AF+BG theorem

I'm looking for an example of the following situation, related to Max Noether's AF+BG Theorem (see Bill Fulton's book on algebraic curves, page 61, at http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf).

Fulton motivates the AF+BG Theorem by saying that if $F, G, H$ are curves in the projective plane with no common components then we could have the inequality of cycles $H \cdot F \geq G \cdot F$ and we might be interested in knowing when there is a curve $B$ so that $H \cdot F = B \cdot F + G \cdot F$. To produce such a curve it is enough to find forms $A$ and $B$ so that $H = AF+BG$ (here I'm using the same letter to denote a curve and its defining form) since then $H \cdot F = BG \cdot F = B \cdot F + G \cdot F$.

Noether's fundamental theorem (the $AF+BG$ theorem) says that the condition $H = AF + BG$ is equivalent to the local conditions that say that for each $P \in F \cap G$, we have $H \in (F,G)\mathcal{O}_P(\mathbb{P}^2)$. Many uses of the theorem rely on being in a situation where the local conditions are obviously met and so we obtain the global fact $H = AF + BG$ and hence $H \cdot F = B \cdot F + G \cdot F$.

My question is to find an example where the local conditions are NOT met but we still have $H \cdot F = B \cdot F + G \cdot F$. If this is impossible, please explain.

## 1 Answer

Define $F$ as $x^2-y^2=0$, $G$ as $x^2+y^2=0$ and $H$ as $xy=0$. (here $(x:y:z)$ are homogeneous coordinates in $\mathbb P^2$.)

It is clear that $H$ can not be expressed as $AF+BG$. At the same time if we take $B=0$, then $H\cdot F=B\cdot F+G\cdot F=4\cdot (0:0:1)$.