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Let $X$ be a smooth, complete algebraic variety and suppose I have two projective, birational morphisms

$$f:X \to \mathbb{P}^n$$


$$g:X \to \mathbb{P}^m,$$

such that the image of $f$ is the normalization of the image of $g$. Let $L_f$ and $L_g$ the two line bundles on $X$ whose global sections induce respectively $f$ and $g$, in the standard way. Does the condition I require on the normalization of the image imply that $L_f=L_g$?

PS I guess that actually the birationality of the two morphisms is not needed.

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Even if the two images are equal, for example both morphisms could be embeddings, there is no reason why the two line bundles should coincide. – ulrich Sep 19 '11 at 11:54
If the normalization is induced by a linear projection $P^n\rightarrow P^m$ (rational map) then $L_f=L_g$, but essentially that should be the only case. – quim Sep 19 '11 at 12:44

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