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The morphism $\phi : \mathbb P^1 \to \mathbb P^3:(u:v)\mapsto(u^4:u^3v:uv^3:v^4)$ gives an embedding of $\mathbb P^1$ in to $\mathbb P^3$. What is $\phi^*( \mathcal O_{\mathbb P^3}(1))? $.

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    $\begingroup$ This is $\mathcal O_{\Bbb P^1}(4)$. $\endgroup$
    – user171326
    May 24, 2017 at 10:57
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    $\begingroup$ @N.H.: Post as an answer...? :) $\endgroup$ May 24, 2017 at 11:27
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    $\begingroup$ sure, thanks :) $\endgroup$
    – user171326
    May 24, 2017 at 11:31

1 Answer 1

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We have $\phi^* \mathcal O_{\Bbb P^3}(D) \cong \mathcal O_{\Bbb P^1}(\phi^*D)$ for any divisor $D$ on $\Bbb P^3$ such that $\phi(\Bbb P^1) \not \subset D$, as MooS wrote in comments (else the pullback won't be a divisor). Notice that $\mathcal O_{\Bbb P^3}(1) = \mathcal O_{\Bbb P^3}(H)$ where $H$ is an hyperplane, say $x_4 = 0$. Then $\phi^*H = 4 [\infty]$ and so $\phi^*O(1) \cong O_{\mathbb P^1}(4)$.

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    $\begingroup$ Your first isomorphism only holds, when $D$ does not contain the image of $\phi$. $\endgroup$
    – MooS
    May 24, 2017 at 18:21
  • $\begingroup$ Thank you, I edited my answer. $\endgroup$
    – user171326
    May 24, 2017 at 18:26

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