I'm reading Algebraic Geometry written by Hartshorne, and my question is about proposition 9.5 in chapter III:
Proposition. Let $f:X\to Y$ be a flat morphism of schemes of finite type over a field $k$. For any point $x\in X$, let $y= f(x)$. Then $\dim_x X_y= \dim_x X- \dim_y Y$.
Proof.
First we make a base change $Y'\to Y$ where $Y'= \operatorname{Spec}\mathcal{O}_{y, Y}$, and consider the base extension $f': X'\to Y'$. Then $f'$ is also flat and the three numbers in question are the same. Thus we replace $X, Y, f$ with $X', Y', f'$ and we assume $y$ is a closed point.
Now we use induction on $\dim Y$.
If $\dim Y= 0$, then $X_y$ is defined by a nilpotent ideal in $X$, so we have $\dim_x X_y= \dim_x X$, and $\dim_y Y= 0$… (The proof continues to the inductive step.)
I don't understand the bold sentence.
When I saw the similar expression "$Y$ is defined by an ideal in $X$", that meant $Y$ is a closed immersion of $X$, but the fiber $X_y$ is not always closed.
What does it mean?
And is the proof of $\dim_y Y= 0$ roundabout?
We assume $Y= \operatorname{Spec}\mathcal{O}_{y, Y}$, so $Y$ is just the spectrum of a field.
Therefore we have $X_y= X$, and nothing special happens.
Is this wrong?
Thank you.
Best Answer
The reduction step gives that $y$ is closed in $Y$, and therefore $X_y$ is closed in $X$: the map $X_y\to X$ is the base change of the closed immersion $\operatorname{Spec} k(y) \to Y$ by $X\to Y$, and closed immersions are stable under base change.
This is not correct. All you know is that the reduction is the spectrum of a field, but $Y$ can have nilpotents - for instance, $\operatorname{Spec} k[\varepsilon]/(\varepsilon^n)$.