Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 21 / 1

Use the usual irreducible algebraic variety convention here, and write $N=\dim X$, $r=\dim Y$. Fix a closed point $y$ and an affine neighborhood $Y_0$ of it. Noether normalization gives a finite surjective map $\pi :Y_0\to {\mathbb{A}}^r$. Put $a=\pi (y)$. Its fibre is a finite set $\{y_1,\dots ,y_s\}$, one of whose members is $y$, because a finite morphism has finite fibres over closed points.

Choose an affine open $U\subseteq {\varphi }^{-1}(Y_0)$ meeting $X_y$. Since $X$ is irreducible, $U$ still has dimension $N$. In $U$, the fibre of $\pi \varphi$ over $a$ is cut out by the $r$ regular functions ${\pi }_j\varphi -a_j$. Apply the supplied principal hypersurface dimension lemma successively on each irreducible component: if a function vanishes identically the dimension does not drop, if its zero set is empty the component disappears, and otherwise each new component loses exactly one dimension. Hence every surviving component after the $r$ cuts has dimension at least $N-r$.

But that composite fibre is the disjoint union of $U\cap {\varphi }^{-1}(y_i)$. Each member is closed, and since there are finitely many members it is also open in the composite fibre. Thus $U\cap X_y$ is a nonempty union of its irreducible components, givingThe same argument around a point on any fibre component gives the componentwise bound. This is the closed-fibre lower bound by finite normalization; it avoids assuming that $Y$ is smooth or that its point ideal has only $r$ generators.

Irreducibility matters if “variety” is generalized to arbitrary reduced reducible spaces. For example $X={\mathbb{A}}^2\bigsqcup {\mathbb{A}}^1\to {\mathbb{A}}^1$ can map the first component constantly to zero and the second by the identity. It is surjective, but nonzero fibres have dimension zero while $\dim X-\dim Y=1$. For reducible sources one uses the individual component dimensions instead; the preceding proof establishes the stated theorem under its standard irreducible-variety convention.

Suppose the map were nonconstant. Since projective space is proper, its image $Z$ is a closed projective subvariety of ${\mathbb{P}}^n$, of positive dimension $r\le n$. Fix $y\in Z$. Regard the map as surjective onto $Z$; the fiber dimension theorem just proved givesChoose a target hyperplane $H$ not containing $y$. It meets $Z$: the affine cone over $Z$ has dimension $r+1\ge 2$, and the hyperplane equation vanishes at its vertex but not identically. By the principal hypersurface dimension lemma its zero set on the cone has dimension $r\ge 1$, so it contains a nonzero cone point and hence a point of $Z\cap H$.

The inverse image $D={\varphi }^{-1}(H)$ is therefore nonempty and proper. Locally the hyperplane equation is a single regular equation, after trivializing the pulled-back line bundle. On the irreducible smooth source it is not identically zero, so the same hypersurface dimension lemma shows that each component of $D$ has dimension $m-1$. Choose one component $D_0$. Its cone is a codimension-one irreducible closed subset of affine $(m+1)$-space, whose prime ideal is principal because the polynomial ring is a unique factorization domain. Thus $D_0$ is a projective hypersurface defined by a homogeneous polynomial $P$ of positive degree.

Choose a positive-dimensional irreducible component $F$ of ${\varphi }^{-1}(y)$. Since $y\notin H$, $F$ is disjoint from $D_0$. On the cone over $F$, the equation $P$ is consequently nonzero everywhere except at the vertex. This contradicts the hypersurface dimension lemma: the cone has dimension at least two, so its nonempty proper zero set for $P$ has dimension at least one and cannot consist just of the vertex. Therefore $\boxed{\varphi \text{ is constant}}$. The argument proves morphisms from projective space to smaller projective space are constant by combining the fibre bound with the forced intersection of a positive-dimensional projective variety and a hypersurface.