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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?

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  • $\begingroup$ The intuition that is usually given is a "continuous family of schemes". I know that this is difficult to connect to the formal algebraic definition. $\endgroup$ – Rene Schipperus Feb 23 '17 at 12:56
  • $\begingroup$ Does the "continuous family of schemes" intuition work even when $Y$ is of dimension greater than one? And is there any way at all to be more precise about the words "continuous family"? Perhaps there are some theorems constraining what these "continuous families" may or may not look like? Or am I being too optimistic? $\endgroup$ – Kenny Wong Feb 23 '17 at 13:19
  • $\begingroup$ Flatness is more complicated than what you say. The map $\mathbb{A}^1\to X=\{(t^2, t^3)\in\mathbb{A}^2\}$, given by $t\mapsto (t^2,t^3)$ has the property you state, but not flat. So, even in one dimensional case, it is more complicated. While many intuitive and geometric interpretations of flatness can be given, none of them truly capture the strength of its algebraic nature-why some of the great theorems of Italian geometers were not quite correct. $\endgroup$ – Mohan Feb 23 '17 at 14:19
  • $\begingroup$ @Mohan Interesting comment, I would gladly hear more about it. An answer maybe ? $\endgroup$ – Rene Schipperus Feb 23 '17 at 14:37
  • $\begingroup$ @Mohan I see, $Y$ needs to be smooth for the property I stated to imply flatness. I didn't appreciate this - thanks! $\endgroup$ – Kenny Wong Feb 23 '17 at 14:42
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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.

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  • $\begingroup$ Thank you. These are all very helpful examples. $\endgroup$ – Kenny Wong Feb 23 '17 at 22:56
  • $\begingroup$ You are welcome, dear Kenny. $\endgroup$ – Georges Elencwajg Feb 23 '17 at 23:04

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