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mrw
mrw
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I think so (even in dimension higher than 2, assuming that $F$ and $G$ are still finite sets, and not codimension two subvarieties of course). Let H$H$ be an ample divisor in X$X$ avoiding F$F$ and let H'$H'$ be its strict transform in Y$Y$. Then H'$H'$ is ample in Y$Y$. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.

I think so (even in dimension higher than 2). Let H be an ample divisor in X avoiding F and let H' be its strict transform in Y. Then H' is ample in Y. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.

I think so (even in dimension higher than 2, assuming that $F$ and $G$ are still finite sets, and not codimension two subvarieties of course). Let $H$ be an ample divisor in $X$ avoiding $F$ and let $H'$ be its strict transform in $Y$. Then $H'$ is ample in $Y$. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.

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mrw
mrw
  • 153
  • 6

I think so (even in dimension higher than 2). Let H be an ample divisor in X avoiding F and let H' be its strict transform in Y. Then H' is ample in Y. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.