Given a map $f: \mathbb{P_C^3 \rightarrow P_C^5}$ defined by $(x_0:x_1:x_2:x_3) \mapsto (x_0x_1:x_0x_2:x_0x_3:x_1x_2:x_1x_3:x_2x_3)$ how does one check if it induces a morphism when restricted to a hypersurface $X \subset \mathbb{P_C^3}$ given by $2x^2_0+3x_1^2+x_2^2+x_3^2=0$, or in other words is $f_{|X}:X \rightarrow \mathbb{P^5}$ a morphism? Moreover is $f_{|X}:X \rightarrow f(X)$ an isomorphism?

As far as I know it only needs to be checked if it is defined on the entire $X$ to be a morphism since it is a rational map. But then it is only not defined at points $(1:0:0:0)$, $(0:1:0:0)$, $(0:0:1:0)$ and $(0:0:0:1)$, and they certainly are not on $X$. Am I missing something?

As for the isomorphism part, I don't really know how to show that. Any help? Thank you

  • $\begingroup$ It's a little weird to say you have the map $f$ when $f$ is not defined everywhere. $\endgroup$
    – Jim
    Jan 16, 2015 at 23:53
  • $\begingroup$ @Jim what should I say? $\endgroup$
    – user198182
    Jan 17, 2015 at 0:46
  • 1
    $\begingroup$ I would just define the map from $X \to \mathbb P^5$ and not try to say it's a restriction from something else. You'd have to through the same steps to show that it's a morphism from the something else, so skip the middleman and do it from $X$. $\endgroup$
    – Jim
    Jan 17, 2015 at 17:56

1 Answer 1


You've observed that $X \to \mathbb P^5$ is given by homogeneous polynomials in the coordinates so yes, you have a morphism $X \to \mathbb P^5$. It is not an isomorphism because it is not injective, if $ab \neq 0$ then $(a:b:0:0) \mapsto (1:0:0:0:0:0)$.


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