TITLE: Why are there no non-trivial regular maps $\mathbb{P}^n \to \mathbb{P}^m$ when $n > m$?
QUESTION [13 upvotes]: 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?

REPLY [4 votes]: 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.