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My question is really simple, I'm beginning to study Algebraic Geometry and I'm still struggling to get the basic concepts.

I would like to know if a Veronese map $v_{n,d}:\mathbb P^n\to \mathbb P^N$ is a morphism.

Thanks

user42912
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1 Answers1

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A morhism $\varphi:X \to Y$ between two projective varieties is a continous map that pulls back regular functions to regular functions. Explicitly, this means that if $U \subseteq Y$ is an open subset and $f:U \to k$ is a regular function, the composite $f \circ \varphi:\varphi^{-1}(U) \to k$ is also regular.

To check if a map of sets $\varphi:X \to \mathbb P^N$ is a morphism, you can do it locally. That is, let $U_i$ we the standard open cover of $\mathbb P^N$. Then each $U_i$ is isomorphic to $\mathbb A^{N}$ and has coordinate functions $\{ x_0,\cdots,x_n\}$. Then it is enough to check that each $x_i \circ \varphi:\varphi^{-1}(U_i) \to k$ is regular. (this is basically Lemma 3.6 in Chapter I of Hartshorne)

Now, lets look at your example in the case $n=1,d=2$ (for notational simplicity). Then the Veronese map takes the form $$v(p_0:p_1)=(p_0^2:p_0p_1:p_1^2)$$ for a point $(p_0:p_1) \in \mathbb P ^1$. The inverse images of the sets $U_0,U_1,U_2$ are $\mathbb P^1 \backslash\{p_0=0\}$, $\mathbb P^1 \backslash \{ (0:1),(1:0) \}$ and $\mathbb P^1 \backslash \{ p_1 = 0 \}$, respectively.

Restricted to $U_0$, the Veronese map takes the form $$ v(1:p_1) = (1:p_1:p_1^2),$$ which is clearly regular, because it is a map defined by polynomials from one affine space to another.

The other charts are handled similarly, and it should not be difficult to generalize this to general $n$ and $d$.

Fredrik Meyer
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  • In the first paragraph did you mean $\varphi^{-1}(U)$, instead of $\varphi^{-1}(V)$? – user42912 Nov 06 '13 at 11:42
  • @user42912: Yes. Thank you for the correction. – Fredrik Meyer Nov 06 '13 at 11:49
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    Dear Frederik, In the last sentence: "it should be difficult" might be missing the word "not". Cheers, – Matt E Nov 06 '13 at 11:55
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    @MattE Thanks! You might be right. – Fredrik Meyer Nov 06 '13 at 12:02
  • In order to prove that $\varphi:X\to Y$ is a morphism, why is it enough to prove $\varphi^{-1}:U_i\to k$ is regular for the basic open subsets of $Y$, instead to prove it for the every open subset $U$ of $Y$? – user42912 Nov 06 '13 at 12:03
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    Excellent explanation: an explicit special case is much clearer that the obscure juggling of indices of the general case. – Georges Elencwajg Nov 06 '13 at 12:05
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    @user42912: Dear user, The answer to the question in your comment is that the property of being a morphism is Zariski local on the source. Consider the following analogous problem: suppose that $f: X \to Y$ is a map between topological spaces. Prove that the following are equivalent: (a) $f$ is continuous; (b) for every open subset of $U$ of $X$, the restriction of $f$ to $U$ is continuous; (c) there exists an open cover $U_i$ of $X$ such that the restriction of $f$ to each $U_i$ is continuous. (Hopefully you can do this.) Now the definition of morphism in algebraic geometry is ... – Matt E Nov 06 '13 at 12:10
  • ... arranged to that the analogous statement is true, and Fredrik is checking that the Veronese map is a morphism by using criterion (c). Regards, – Matt E Nov 06 '13 at 12:11
  • Dear @MattE using the equivalence you mentioned we prove that the pullback is continuous for every $U$, but what about been regular for every $U$? thanks – user42912 Nov 06 '13 at 12:33
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    @user42912: Dear user, You have to strengthen the equivalence I mentioned, not just apply it literally; i.e. you have to prove the analogous statement with "continuous map" replaced by "morphism" or "regular function". This is an exercise just using the definition. Regards, – Matt E Nov 06 '13 at 13:37
  • @MattE I'm doing this right now, thanks – user42912 Nov 06 '13 at 14:10
  • @MattE one last question, I didn't understand why $v$ restricted to $U_0$ is a map defined by polynomials from one affine space to another and where the answerer uses the lemma 3.6 exactly, sorry I'm really new on this subject, regards. – user42912 Nov 06 '13 at 14:14
  • @user42912: It is a fact that $\mathbb P^n \backslash H$ where $H$ is a hyperplane is isomorphic to $\mathbb A^n$. This is for example Prop 2.2 in Hartshorne. – Fredrik Meyer Nov 06 '13 at 15:54