0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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 $D$? $\endgroup$
    – user34880
    Jun 14, 2013 at 20:04
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 14, 2013 at 20:07
  • $\begingroup$ 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 $\mathbb{P}^n_k$. $\endgroup$
    – user34880
    Jun 14, 2013 at 21:32
  • $\begingroup$ 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. $\endgroup$ Jun 15, 2013 at 7:53
  • $\begingroup$ 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 $D \in \text{Div}(X)$. $\endgroup$
    – user34880
    Jun 15, 2013 at 16:47

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.