Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given $X$, and a locally free sheaf $\mathcal{F}$ of rank $n$ on it. We have the induced map $f:\mathbb{P}(\mathcal{F})\rightarrow X$. Let $\phi$ be the divisor associated to the line bundle $\mathcal{O}_{\mathbb{P}(\mathcal{F})}(1) $.

Is it true $f_{\ast}(\phi^{n-1})=[X]$ in the Chow ring $A^\ast(X)$? Also, is trace map $A^{top}(.)\rightarrow\mathbb{Q}$ compatible with pushforward $f_{\ast}$?

Thank you.

share|improve this question

1 Answer 1

up vote 0 down vote accepted

The class $\phi^{n-1}$ has degree $1$ when restricted to each fibre of $f$ since $H^{n-1}$ has degree $1$ on $\mathbb{P}^{n-1}$, where $H$ is the class of a hyperplane, i.e. a divisor associated to $\mathcal{O}(1)$. It follows that $f_*(\phi^{n-1})$ is indeed $[X]$ in $A^*(X)$.

I suppose what you call the trace map is just the degree map $A^{top}(X) \to \mathbb{Z}$ (for a proper variety $X$). It follows immediately from the definition of proper pushforward that it is compatible with $f_*$.

share|improve this answer
    
Thank you. The trace map I mentioned is exactly your degree map. However I think the answer depends on the filed $k$ of variety. Let $X$ is spec$\mathbb{Q}[t]$, a closed point $p$ is defined by the ideal $(x^2+1)$. Then the function filed $K(p)$ is $\mathbb{Q}(i)$, not $\mathbb{Q}$. By the definition pushforward, for a proper morphism $f:X \rightarrow Y$, $f_{\ast}(p)=[K(p): K(q)]q$, where $q=f(p)$. The term [K(p): K(q)] will change the degree of 0-cycles. –  Thunder Sep 18 '11 at 21:06
    
I don't see the problem. The degree of your closed point $p$ is $2$; if it gets sent to a $\mathbb{Q}$ rational point by a proper morphism then it will acquire a multiplicity and the degree of the image cycle remains $2$. –  ulrich Sep 19 '11 at 4:07

Your Answer

 
discard

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

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