MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Pardon me if this question is trivial as I am just starting to learn the subject. Let $k$ be a number field. Given a smooth projective variety $X \subset \mathbb{P}_k^n $, a positive divisor $D$ in $\text{Div} _\bar{k} (X)$ and a prime divisor $Y$ in $\text{Div} _\bar{k} (X)$, can we say anything about $\text{deg}(D \cdot Y)$ besides being less than $\text{deg}(D)\cdot\text{deg}(Y)$? Any nontrivial lower bound? Under what condition (on $D$ and $Y$) do we have equality? (If $X = \mathbb{P}_k^n $, then we always have equality, so say we have a general $X$). Any good reference will be appreciated. Thank you.

share|cite|improve this question
P.S. If $Y$ intersects each component of $D$ transversely, then we will have $\text{deg}(D \cdot Y)=\text{deg}(D) \cdot \text{deg}(Y)$. Is there an easier sufficient condition for $Y$ intersects each component of $D$ transversely? For eg: $\text{deg}(Y)$ large enough? or $Y$ is not rationally equivalent to all component of $D$? – user34880 Jun 14 '13 at 20:04
How do you define the degree of a divisor on $X$? If you only consider the degree given by the inclusion in $\mathbb{P}^n$, then equality holds (it is Bezout Theorem). Otherwise, you should say what you mean by the degree here. – Jérémy Blanc Jun 14 '13 at 20:07
I should say everything happens in Chow ring. For $D \in \text{Div}(X)$, $\langle D \rangle_X $ will denote the class of $D$ in the Chow ring of $X$, A(X). Let $i : X \rightarrow \mathbb{P}^n_k$ be the inclusion, then '$\text{deg}(\langle D \rangle_X):=\text{deg}(i_*(\langle D \rangle_X) \cdot \langle H\rangle_{\mathbb{P}^n} ^{\text{dim}(D)})$', where $H$ is any hyperplane in $\mathbb{P}^n_k$. – user34880 Jun 14 '13 at 21:32
So your definition of degree is the classical degree in $\mathbb{P}^n$: intersect with a linear subspace of the good dimension. Hence, $deg(X\cdot Y)=deg(X)\cdot deg(Y)$ is just Bezout Theorem. – Jérémy Blanc Jun 15 '13 at 7:53
I think I have been abusing the notation, which causes the confusion. From the definition, we have '$\text{deg}(\langle D \rangle_X)=\text{deg}(\langle D \rangle_{\mathbb{P}^n}) $' for $D \in \text{Div}(X)$. – user34880 Jun 15 '13 at 16:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.