Let $f:X\to Y$ be a dominant morphism of integral schemes of finite type over a field $k$.
Let $e=\dim(X)-\dim(Y)$. For any point $y\in f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $\geq e$. [Hint: Let $Y'=\overline{\{y\}}$ and use (a) and Ex. 3.20b]
I think there is a typo and we have to use Ex 3.20d, which says that if $Y$ is a closed subset of $X$ then $\dim(Y)+\mathrm{codim}(Y,X)=\dim(X)$. In addition, (a) says that if $Y'$ is a closed irreducible subset of $Y$ whose generic point $\eta'$ is contained in $f(X)$ and $Z$ is any irreducible componet of $f^{-1}(Y')$ such that $\eta'\in f(Z)$, then $\mathrm{codim}(Z,X)\leq\mathrm{codim}(Y',Y)$.
What I have done
By exercise 3.10a, $X_y$ is homeomorphic to $f^{-1}(y)$ with the induced topology, so I'm trying to show the result for $f^{-1}(y)$ since it looks easier, but using the hint I've only managed to prove it for $f^{-1}(Y')$. This is my proof:
Since $y$ is obviously the generic point of $Y'$, if we take $Z$ to be an irreducible component of $f^{-1}(Y')$ containing $y$, by (a) we have
$$\operatorname{codim}(Z,X) \leq \operatorname{codim}(Y',Y)$$
Now Ex 3.20d implies
$$\dim (X) – \dim (Z) \leq \dim (Y) – \dim (Y')$$
and thus
$$e = \dim (X) – \dim (Y) \leq \dim (Z) – \dim (Y') \leq \dim (Z)$$
But this only shows that any irreducible component of $f^{-1}(Y')$ has dimension $\geq e$. How can I conclude that this is true also for the irreducible components of the fibre $f^{-1}(y)$?
Best Answer
Let $W$ be an irreducible component of $X_y$, and let $k$, $k'$ be the codimensions of $\overline{W}$ (closure in $X$) in $X$ and $Y'$ in $Y$, respectively.
One argues that $\overline{W}$ is an irreducible component of $f^{-1}(Y’)$. Thus $k \leq k'$ by part (a).
Applying Ex. 3.20b and using the additivity of transcendence degrees in towers, we see that $$\text{dim}(W)=e+k'-k \geq e. $$