When we say that a morphism $f: E \rightarrow M $ between two algebraic varieties (over $\mathbb{C}$) is a flat morphism, what does it mean? Does it mean that that the "dimension" of every fiber $f^{-1}(x)$ is the same for all $x$? Or do we also have to check some compatibility conditions?

The specific example I have in mind is the following:

Let $\mathcal{D} \approx \mathbb{P}^{\delta_d}$ be the space of non-zero homogeneous degree $d$ polynomials in three variables upto scaling, where $\delta_d = \frac{d(d+3)}{2}$ (basically degree $d$-curves in $\mathbb{P}^2$). Let $$ \mathcal{V} := \{ [f]\in \mathcal{D}: f([1,0,0]) =0, ~~\nabla f|_{[1,0,0] =0} \} $$ ie it is the space of curves having $[1,0,0]$ as a singular point. Define $$ \mathcal{C} := \{ ([f], p) \in \mathcal{V} \times \mathbb{P}^2: f(p) =0 \} $$

I think that the "morphism" $$ \pi_{\mathcal{V}}: \mathcal{C} \rightarrow \mathcal{V} $$ is "flat", although it is not a fiber bundle (in the sense of differential topology). Why is this a "flat morphism"?

Secondly I believe that the morphism $$ \pi_{\mathbb{P^2}}: \mathcal{C} \rightarrow \mathbb{P}^2 $$ is not flat. Why is that? Is it because the fiber over $[1,0,0]$ is of a larger "dimension"?

$\textbf{EDIT:}$ One further question about terminology, using this example. Define $$ \mathcal{V}^* := \{ [f]\in \mathcal{D}: f([1,0,0]) =0, ~~\nabla f|_{[1,0,0]} =0, ~~det \nabla^2 f|_{[1,0,0]} \neq 0 \}, $$ $$ \mathcal{C}^* := \{ ([f], p) \in \mathcal{V}^* \times \mathbb{P}^2: f(p) =0 \} $$ I believe that the morphism $$ \pi_{\mathcal{V}^*}: \mathcal{C}^* \rightarrow \mathcal{V}^* $$ is smooth of relative dimension one. I assume it is of relative dimension one, since the fiber at each point is one dimensional. But why is it "smooth"? I don't think $\mathcal{C}^*$ is even a smooth manifold.

  • $\begingroup$ In response to the edit: I'm not sure why you put words like "smooth" in quotation marks; they are well-defined terms, whose meaning you can find in Hartshorne. But I agree that the morphism in question isn't smooth: its fibre over a point $[f]$ is the set of points $p$ such that $f(p)=0$, which is never a smooth curve by definition of $\mathcal{V}$. Why do you say you believe that morphism is smooth? $\endgroup$ – user64687 Jul 2 '13 at 17:52
  • $\begingroup$ Thank you for the reply. I was simply putting those terms in quotation marks for which I wasn't sure of the definition. I must have been mistaken when I said $\pi_{\mathcal{V}^*} $ is a smooth morphism. $\endgroup$ – Ritwik Jul 2 '13 at 18:32

There is a formal definition of flatness, which you can find for example in Hartshorne section III.9. The definition is maybe unsatisfying, because it is algebraic in nature and its geometric meaning is not so clear. But it has some nice geometric consequences, for example Corollary 9.6 in the same section: if $X \rightarrow Y$ is a surjective flat morphism of irreducible algebraic varieties, then (every component of) every fibre has the same dimension. So your guess at a definition is indeed a consequence of flatness.

In fact, under some additional assumptions --- $Y$ regular and $X$ Cohen--Macaulay --- it is in fact equivalent to flatness: see Hartshorne Exercise III.10.9. (But keep in mind that this is not the general definition.) This should show why one of your morphisms is flat, but not the other.

  • $\begingroup$ Thank you for your answer. I just added one further question about such terminology, in the context of this example. $\endgroup$ – Ritwik Jul 2 '13 at 17:34

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