Why are there no non-trivial regular maps $\mathbb{P}^n \to \mathbb{P}^m$ when $n > m$?

Question

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?

Answer 1

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.

Answer 2

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.