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Question. Let $k$ be an algebraically closed field, an let $\mathbb{P}^n$ be projective $n$-space over $k$. Why is it true that every regular map $\mathbb{P}^n \to \mathbb{P}^m$ is constant, when $n > m$?

I can't see any obvious obstructions: there are certainly homomorphisms of function fields (giving rise to the dominant rational maps), and we're not demanding the map be injective or anything. While it is clear that $(F_0 : \cdots : F_m)$ cannot define a regular map on its own unless $F_0, \ldots, F_m$ are all constants, I don't see why it should be impossible to extend $(F_0 : \cdots : F_m)$ by choosing some other $(G_0 : \cdots G_m)$ which agrees with $(F_0 : \cdots : F_m)$ on the intersection of their domains. Is there something conceptual I'm missing?

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Hint: To get a morphism on all of $\mathbb{P}^n$ the $F_0,\cdots, F_m$ can't have any common zeroes. If $n>m$ show this is impossible. – Lalit Jain Oct 16 '11 at 16:14
@Lalit: That's obvious for dimension reasons. What's not obvious is that a regular map is globally determined by just one $(F_0 : \cdots : F_m)$. – Zhen Lin Oct 16 '11 at 16:29
@rattle: I don't think it's so simple. Consider the surface $V(X_0 X_3 - X_1 X_2) \subset \mathbb{P}^3$ and the map $(X_0 : X_1 : X_2 : X_3) \mapsto (X_0 : X_1)$. This appears at first to be undefined at $(0 : 0 : 1 : 1)$, but can be extended to a true regular map by patching it together with $(X_0 : X_1 : X_2 : X_3) \mapsto (X_2 : X_3)$. – Zhen Lin Oct 17 '11 at 6:25
The point is that it is not obvious that a regular map, or even a rational map, is given on its entire domain of definition by one tuple of polynomials. – Zhen Lin Oct 17 '11 at 15:19
Dear @Parsa: what you say is true but does not address Zhen's excellent point: look at his example. The subtle problem is that it is not clear whether a tuple of polynomials defines a regular map at a point where the polynomials all vanish. – Georges Elencwajg Oct 10 '13 at 16:44
up vote 3 down vote accepted

I apologize in advance if I am using results that you are, yet, unaware of. I still wanted to give it a shot:

A morphism $\mathbb{P}^n\to\mathbb{P}^m$ corresponds to a way of globally generating a line bundle $\mathcal{O}_{\mathbb{P}^n}(d)$ with $m$ generators. We can safely assume $d\ge 0$ here. Now the global sections of that line bundle are precisely the homogeneous polynomials of degree $d$ in $n+1$ variables, and since $m<n$, this must mean $d=0$, i.e. we have chosen $m$ constants from k.

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I'm afraid I don't understand this correspondence you're citing. $\mathscr{O}_{\mathbb{P}^n}(d)$ is the degree $d$ twisting sheaf, isn't it? What does it mean to globally generate a line bundle with $m$ generators? – Zhen Lin Oct 18 '11 at 18:45
The correspondence is Theorem II.7.1 in Hartshorne. Yes, $\mathcal{O}_{\mathbb{P}^n}(d)$ is the $d$-th tensor power of the twisting sheaf, and I am also using Corollary II.6.17, which says that any line bundle on $\mathbb{P}^n$ is isomorphic to one of these for some $d$. To globally generate a line bundle $\mathcal{L}$ on a variety $X$ with $m$ generators means to choose $s_1,\ldots,s_m\in\mathcal{L}(X)$ such that $\mathcal{L}_P$ is generated by the stalks of the $s_{i,P}$. – Jesko Hüttenhain Oct 19 '11 at 16:05
The latter means that not all of the $s_i$ are mapped to zero under $\mathcal{L}(X)\to\mathcal{L}_P\cong\mathcal{O}_{X,P}\twoheadrightarrow\mathcal{‌​O}_{X,P}/\mathfrak{m}_P\cong k$ – Jesko Hüttenhain Oct 19 '11 at 16:06
Finally, let me quote Proposition II.5.13 which says that the global sections of $\mathcal{O}_{\mathbb{P}^n}(d)$ are the homogeneous polynomials of degree $d$ over $k$. – Jesko Hüttenhain Oct 19 '11 at 16:09
Thanks for the references. It's interesting, but a bit too high-brow for this simple question, I think. – Zhen Lin Oct 20 '11 at 6:39

There's no need to phrase Jesko's solution in this high-brow language. In general, any rational map $\mathbb{P}^n \rightarrow \mathbb{P}^m$ can be given by an $m+1$-tuple of polynomials of the same degree, with no common factor. If $n>m$, then the dimension of the common vanishing locus of these polynomials must be positive, since each hypersurface cuts it down by at most one, so it is non-empty. The rational map cannot be defined on this locus.

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