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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.

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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 $f_*$.

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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

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