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 1


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 $H\cdot F=B\cdot F+G\cdot F=4\cdot (0:0:1)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.