Precise general statement of flat morphisms being “equidimensional”

Question

I am given to understand that the main property of flat morphisms is that they give some precise notion of "continuously varying family of fibres". I realise there is a fair bit of literature on this, but I really wanted to nail down what exactly is the most general form of this statement. Most of the introductory texts do this in the case of varieties, and I am finding it hard to distil out exactly what hypotheses are necessary.

I am happy to allow some mildly reasonable assumptions: Suppose $f: X \rightarrow Y$ is a finite type morphism of noetherian schemes. Then we obtain the following: For $p \in X$,
$$
\dim (\mathcal{O}_{X,p}) = \dim( \mathcal{O}_{Y, f(p)} )+ \dim( \mathcal{O}_{X,p} \otimes \kappa(f(p))).
$$
I was surprised when I learned the proof of this to find that it is entirely a claim about local rings: One inequality is obtained by going-down property of flatness, and the other is obtained by lifting generators of an $\mathcal{m}$-primary ideal in the local ring. This most definitely requires noetherianness, but only locally of finite type, but let's stick with finite type.

But this doesn't really seem to tell us the dimension of the fibre. Indeed it may be the case that neither $p$ nor $f(p)$ is a closed point, and so the codimension of the point $p$ in the fibre $f^{-1}(p)$ may not be the dimension of $f^{-1}(p)$ at all.

So to give some kind of precise questions:

1) What exactly is "the dimension of the fiber at $p \in X$"?

2) What extra hypotheses need to be put on $f: X \rightarrow Y$ in order to ensure that the fibre is of pure dimension? Since indeed it may have many irreducible components.

3) What extra hypotheses are needed on $f: X \rightarrow Y$ in order to ensure all fibres have the same dimension, assuming "dimension of the fibre" has been defined.

Answer 1

Here is what I found.

For the irreducible case we have:

Theorem: Let $X,Y$ be Noetherian irreducible schemes and $f:X\rightarrow Y$ a flat morphism of finite type. Then, for each $y\in Y$ the fiber $f^{-1}(y)$ is pure dimensional and its dimension is independent of $y$.

Proof/Source: See these notes of Brian Conrad

For the non-irreducible case we have:

Theorem: Let $X,Y$ be Noetherian schemes and $f:X\rightarrow Y$ be an open morphism of finite type (for example, $f$ flat and of finite type). Moreover, suppose that $Y$ is universally catenary, irreducible and $\dim Y<\infty$, that $X$ is equidimensional and

• For any irreducible component $X'$ of $X$ one have $$\dim Y= \sup_{y\in f(X')} \dim \mathcal{O}_{Y,y}.$$

Then, for all $y\in f(X)$ the fiber $f^{-1}(y)$ is equidimensional and its dimension is independent of $y$.

The property $\bullet$ is satisfied automatically if for example:

• The restriction of $f$ to each irreducible component of $X$ is surjective.
• $f$ is a closed morphism (for example: if $f$ is proper).
• $Y$ is of finite type over a field.
• $Y$ is of finite type over $\mathbb{Z}$.

Proof/Source: Görtz and Wedhorn, Algebraic geometry I. Theorem 14.114 and the remark after.

So as a corollary from the above, you don't have to worry about irreducibility if you are working with varieties.

Now, a counterexample that escape both theorems:

Counterexample: There are schemes $X,Y$ and a morphism $f:X\rightarrow Y$ such that:

1. $X,Y$ are Noetherian.
2. $Y$ is irreducible (and $X$ is not) and universally catenary.
3. $X$ is equidimensional.
4. $f:X\rightarrow Y$ is a (faithfully-)flat morphism of finite type.
5. $f^{-1}(\eta)$ has two components, one of dimension 1 and other of dimension 2.

More concretely:

• $Y=\mathrm{Spec}(R)$ for $R$ a discrete valuation ring with uniformizer $\pi$
• $X=\mathrm{Spec}(A)$ for $A=R[X,Y,Z]/(X(\pi Z-1),Y((\pi Z-1)))$
• $f$ is the morphism corresponding to the ring map $R\rightarrow A$.

Proof/Source: Görtz and Wedhorn, Algebraic geometry I. Exercise 14.24.