Dimension of irreducible components of single homogeneous polynomial

Question

I'm currently working my way through Algebraic Geometry by Andreas Gathmann and could use a hint to exercise 7.9:

Let $ f : \mathbb{P}^n \to \mathbb{P}^m$ be a morphism. Prove: If $ X \subset \mathbb{P}^m $ is the zero locus of a single homogeneous polynomial in $K[x_0,...,x_m]$ then every irreducible component of $f^{-1}(X)$ has dimension at least $n-1$.

I know that the irreducible components of $X$ in $\mathbb{P}^m$ have dimension $m-1$, but how do i get the needed dimension in $\mathbb{P}^n$.

Help appreciated

Answer 1

Hint: Use the following theorem (together with Krull's principal ideal theorem) to conclude

Theorem: Let $U\subseteq \mathbb{P}^n$ be an open set and $f:U\rightarrow \mathbb{P^m}$ a morphism. Then there are unique (except for a constant) homogeneous polynomials $f_0,\dots,f_m\in k[x_0,\dots,x_n]$ of the same degree such that $f$ is globally defined by them, i.e, $$f(x_0:\dots:x_n)=(f_1(x_0:\dots:x_n):\dots:f_m(x_0:\dots:x_n)) \; \; \forall \,(x_0:\dots:x_n)\in U$$

Hint for the theorem: First prove the uniqueness, then as there is a cover $U=\bigcup_{i=1}^n U_i$ such that $f$ is defined by homogeneous polynomials over each $U_i$ (this should be either your definition of morphism or something already proved) by the uniqueness over $U_i\cap U_j$ you can glue all this polynomials.