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Suppose $X,Y$ are closed subvarities of $\mathbb{P}^n,\mathbb{P}^m$ Can every morphism $f:X \rightarrow Y$ be written as $(x_0,\ldots,x_n)\rightarrow (f_1(x),\ldots,f_m(x))$ where $f_i$ are homogeneous polynomials? It is the right locally on $X$, can it be written globally in this form?

Michael Hardy
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  • Yes, although you have to be careful if the $f_i$ have a common zero. Basically at "almost" every point you can write your function like that, and at the common zeros $f=(g_0,\ldots,g_m)$ where $g_if_j=g_jf_i$ for all $i,j$. – rfauffar Nov 03 '13 at 03:04
  • But the common zeros are precisely the problem! Consider, for instance, $X = V (X_0 X_3 - X_1 X_2) \subset \mathbb{P}^3$, $Y = \mathbb{P}^1$, and the regular map $f : X \to Y$ defined by $(X_0 : X_1 : X_2 : X_3) \mapsto (X_0 : X_1)$ away from $V (X_0, X_1) \cap X$ and $(X_0 : X_1 : X_2 : X_3) \mapsto (X_2 : X_3)$ away from $V (X_2, X_3) \cap X$. – Zhen Lin Nov 03 '13 at 03:14
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    Exactly, my comment essentially means that you can think of regular functions as a tuple of homogeneous polynomials, but when there is a common zero you have to find another way of expressing it at that point. – rfauffar Nov 03 '13 at 03:26
  • I think my comment actually confuses more than it helps. As Zhen Lin shows and as Georges shows below, you can't always write an equation globally. I just meant to say that you can almost everywhere but you have to be careful with common zeros. – rfauffar Nov 03 '13 at 15:11

1 Answers1

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I) No, in general a morphism $f:X\to Y$ cannot be determined by a single sequence of homogeneous polynomials $(f_0:...:f_m)$.

For a counterexample consider the conic $C\subset \mathbb P^2_{x:y:z}$ given by the equation $xz=y^2$.
There is a morphism (actually an isomorphism) $f:C\to \mathbb P^1$ given by $f(x:y:z)=(x:y)$ if $x\neq0$ or $y\neq0$ and by $f(x:y:z)=(y:z)$ if $y\neq0$ or $z\neq 0$.
This morphism cannot be given by a sequence $(f_0:f_1:f_2)$ of homogeneous polynomials without a common zero.

II) However any morphism $f: \mathbb P^n\to \mathbb P^m$ defined on all of $\mathbb P^n$ can be written as $f=(f_0:...:f_m)$ with the $f_i$'s homogeneous of the same degree $d$.
If $n\gt m$ such a morphism will necessarily be a constant, i.e. $d=0$.

III) Hartshorne's Theorem 7.1 in chapter II, page 150 explains in great generality how morphisms $X\to \mathbb P^n$ from a scheme $X$ to projective space are determined.

Georges Elencwajg
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    ..I am not clear how to duduce II) from III)(or do you mean this)? –  Nov 04 '13 at 06:00