Geometric intuition for flat morphisms

I'm trying to develop some geometric intuition for what it means for a morphism of schemes to be flat. The definition of flatness in Hartshorne says (if I'm correct) that a morphism $f: X \to Y$ is flat iff pullbacks of SESs of quasicoherent sheaves on $Y$ are exact on $X$. But this is very algebraic, and not at all easy to visualise!

The most helpful thing I've found in Hartshorne is Prop. 9.7: If (for instance) $X$ and $Y$ are varieties and $Y$ is smooth of dimension 1, then $f$ is flat iff the image of every irreducible component of $X$ is dense in $Y$. Thus the irreducible components of $X$ lie "flat" over $Y$, hence the terminology.

But what if $Y$ is of dimension bigger than 1? What is the intuition for flatness now?

And is there another way to think about flat morphisms which is more intuitive altogether?

Solution 1:

Here is a hotch-potch of examples, counterexamples, theorems, ... which I plagiarized adapted from the answer by some guy with a complicated name to the analogous question for complex analytic spaces.
I hope they will give you some intuition for flatness, that "riddle that comes out of algebra, but which technically is the answer to many prayers" (Mumford, Red Book, page 214).

Let $f:X\to Y$ be a scheme morphism, locally of finite presentation. Then:

a) $f$ smooth $\implies$ $f$ flat.

b) $f$ flat $\implies$ $f$ open (i.e. sends open subsets to open subsets).
Beware however that the natural morphism $\operatorname {Spec}\mathbb Q \to \operatorname {Spec} \mathbb Z$ is flat and yet not open: this is because it is not locally of finite presentation.

c) Open immersions are flat.

d) However general open maps need not be flat. A counterexample is: $$\operatorname {Spec k}\to \operatorname {Spec} k[\epsilon]=\operatorname {Spec} \frac {k[T]}{\langle T^2\rangle }$$

e) The normalization $X=Y^{\operatorname {nor}}\to Y$ of a non-normal scheme is NEVER flat.
For example the normalization of the cusp $C=V(y^2-x^3)\subset \mathbb A^2$ :$$\mathbb A^1\to C:t\mapsto (t^2,t^3)$$ is not a flat morphism.

f) A closed immersion it is NEVER flat, unless it is also an open immersion [cf. c)].

g) If $X,Y$ are regular and $f:X\to Y$ is finite and surjective, then $f$ is flat.
for example the projection of the parabola $y=x^2$ onto the $y$-axis is flat, even though one fiber is single point (but a non reduced one!) while the other fibers have two points (both reduced). As another illustration, every non constant morphism between smooth projective curves is flat.

h) If $Y$ is integral and $X\subset Y\times \mathbb P^n$ is a closed subscheme, the projection $X\to Y$ is flat if and only if all fibers $X_y=\operatorname {Spec}\kappa(y)\times X$ ($y$ closed in $Y$) have the same Hilbert polynomial.
In particular the fibers must have the same dimension, so that for example the blow-up morphism $\widetilde {\mathbb P^n}\to \mathbb P^n$ of $\mathbb P^n$ at a point $O$ is not flat, since all fibers are a single point, except the fiber at $O$ which is a $\mathbb P^{n-1}$.
Notice how the morphism $\operatorname {Spec k}\to \operatorname {Spec} k[\epsilon]$ evoked above (for which you have only one fiber!) yields a counterexample to g) if you do not assume $Y$ reduced.
This very general result h) (which is at the heart of the theory of Hilbert schemes) might be the best illustration of what flatness really means.