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Let $f: X' \rightarrow X$ be the blowing up of a nonsingular variety $X$ along a nonsingular subvariety $Z$ of codimension $r \geq 2$, and let $Z' = f^{-1}(Z)$. Denote $\mathrm{Cl}(X')$ the divisor class group of $X'$. Why is the map $\mathbb{Z} \rightarrow \mathrm{Cl}(X')$, $q \mapsto q \cdot Z'$, injective? That is, why is not the divisor $q \cdot Z'$ principal for $q \neq 0$?

Thinking about this problem I rose the following question: let $(f, f^{\#}): X \rightarrow Y$ be a birational morphism of integral schemes and let $\xi$ be the generic point of $X$. Then $f_{\xi}^{\#}: K(Y) \rightarrow K(X)$ is an isomorphism between the function fields. Suppose that $f^{-1}(Z)$ is a prime divisor on $X$ for each prime divisor $Z$ on $Y$. Let $\mathrm{div}(g) = \sum n_{Z} \cdot Z$, $g \in K(Y)$, be a principal divisor. Is this true that $\mathrm{div}(f_{\xi}^{\#}(g)) = \sum n_{Z} \cdot f^{-1}(Z)$ Is this true in the case above?

Thanks.

P.S.: I do not know cohomology of sheaves, not yet.

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If $qZ'$ is principal, then it is the divisor $\mathrm{div}_{X'}(f)$ of some $f\in K(X')=K(X)$. In particular, $\mathrm{div}_{X'}(f)$ is supported in $Z'$. But on $X$, the support of the divisor $\mathrm{div}_X(f)$ is not contained in $Z$ (unless it is 0) because the support of a divisor has always codimension $1$. As both divisors have the same support in $X'\setminus Z'\simeq X\setminus Z$, we have $\mathrm{div}_X(f)=0$, thus $f$ is invertible on $X$, hence on $X'$. So $q=0$.

In the general case, how do you define $f^{-1}(Z)$ as a divisor (especially when $Z$ meets the exceptional locus of $f$) ?

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  • $\begingroup$ Thanks for watching. Suppose $f^{-1}(Z)$ is a prime divisor on $X$ for each prime divisor $Z$ on $Y$. $\endgroup$ – rla Nov 21 '11 at 22:39
  • $\begingroup$ @rafaellucas: When $f$ is a blowing-up as in above, this can never happen, because if $Z$ is a prime divisor containing the center of the blowing-up, then $f^{-1}(Z)$ is the union of the strict transform of $Z$ and of the exceptional divisor. Another issue is that one has to affect multiplicities to the irreducible components of the set $f^{-1}(Z)$. $\endgroup$ – user18119 Nov 21 '11 at 23:12

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