Let $p:Y\to X$ be a birational proper surjective morphism of regular surfaces, and let $D$ be a divisor on $Y$ such that $p(D)$ is a point. Then $p_\ast D =0$ by definition. Is there an easy way to show that $p_\ast \mathcal{O}_Y(D) = \mathcal{O}_X$?

More generally, does the equality $c_1(p_\ast L) = p_\ast c_1(L)$ hold, where $L$ is a line bundle on $L$? Or equivalently, does the equality $\mathcal{O}_X(p_\ast D) = p_\ast \mathcal{O}_Y(D)$ hold, where $D$ is a divisor on $Y$?

I just found out the answer is positive, but my proof is a bit complicated I think. So the question remains, but now I'm looking for an easy proof.

  • $\begingroup$ If $D<0$ and is mapped to one point of $X$, then $p_*O_Y(D)=0$. $\endgroup$ – user18119 Mar 13 '13 at 23:45

Here is an idea:

The proper birational morphism $p$ can be factored as a finite sequence of blow-downs of $(-1)$-curves, according to Beauville's Complex Algebraic Surfaces Theorem II.11. Thus, saying that $p(D)$ is a point is equivalent to saying that $D$ is a union of exceptional divisors of the blow-downs. By Proposition II.3 of the same book, we know that $\operatorname{Pic}(Y) = \operatorname{Pic}(X)\oplus\mathbb Z^n$ where $n$ is the number of blow-downs of a single $(-1)$-curve (and the obvious generators of the $\mathbb Z^n$ part are the $(-1)$-curves blown down at each step). Since $D$ is a union of exceptional divisors, we know that $\mathcal O_Y(D) = (0,a_1,\ldots,a_n)\in\operatorname{Pic}(Y),$ i.e., the $\operatorname{Pic}(X)$ component of $D$ in $\operatorname{Pic}(Y)$ must be zero. This tells us that $p_*\mathcal O_Y(D)=\mathcal O_X.$

I believe the same method will work for the more general case, using the definition of $p_*D$ and the fact that $p_*\mathcal O_Y(D)$ will be just the $\operatorname{Pic}(X)$ component of $\mathcal O_Y(D).$


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