upper semicontinuity of fiber dimension with target dimension 0

Question

I have problem with Ravi Vakil's FOAG, Theorem 11.4.2. The theorem claims that for a morphism $\pi : X \rightarrow Y$ between finite type k-schemes, the map assigning to a point $p \in X$ the dimension of the largest irreducible component of $\pi^{-1}(\pi(p))$ containing $p$ is a upper semicontinuous function.

He proves this by induction on $\operatorname{dim} Y$, and says it is obvious when the dimension is 0. I couldn't figure out the reason why it is obvious. Am I overlooking something?

Answer 1

In short, this is because, in good situations, the dimension of the fiber is bounded from below by the relative dimension (i.e., $\dim X - \dim Y$). You can prove it whenever you are allowed to use Exercise 11.4.A and Theorem 11.2.9.

Proof of the base case. We may assume that both $X$ and $Y$ are irreducible.

Reason: taking an irreducible component of $Y$ containing $q:=\pi(p)$ does not change the fiber. Each irreducible component of $\pi^{-1}(q)$ containing $p$ is contained in some irreducible component $X_i$ of $X$ containing $p$, so $p\in F_n$ if and only if $$p\in F_{n,i}:=\{x\in X_i| \text{the dimension of the fiber at $x$ of the morphism $X_i\to \pi(X_i)$ is at least $n$.}\}$$ for some $i$. Thus $F_n=\bigcup_{i=1}^n F_{n,i}$, and if $F_{n,i}$ are closed for all $i$, then $F_n$ is closed.

Assuming $\dim Y = 0$, by Exercise 11.4.A and Theorem 11.2.9 we see that $$ \dim X - \dim p = \mathrm{codim}_X p + \underbrace{\mathrm{codim}_Y q}_{\leq \dim(Y)=0} \leq \mathrm{codim}_{\pi^{-1}(q)} p \leq d-\dim p, $$ where $d$ denotes the dimension of the fiber at $p$. Thus, $d \geq \dim X$; the fiber is a subset of $X$, so $d = \dim X$.

Hence (when $X$ is irreducible) we have $$ F_n = \begin{cases} X &\text{if $n\leq \dim X$},\\ \emptyset &\text{if $n > \dim X$}. \end{cases} $$