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How can I view, in algebraic geometry, a family of curves over a base curve?

For instance, can the family

$y^2 = x(x-1)(x-t)$

be viewed as a family over

$\mathbb{P}^1$ ?

How can I understand this rigorously? Can someone point me to a reference?

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1  
Sure. First, turn it into a homogeneous equation: $y^2 z = x (x - z)(x - t z)$. Then regard it as an equation in $\mathbb{P}^3$, where $t$ is also a variable. The natural projection $(x : y : z : t) \mapsto (z : t)$ then exhibits this as a family of cubic curves over $\mathbb{P}^1$. [Exercise: fix this so that we get a morphism rather than a rational map.] –  Zhen Lin May 7 '12 at 15:44
    
Thanks! But what do you mean by "fix it"? –  averageman May 7 '12 at 15:55
    
The procedure I described is the a very naïve one and suffers various problems. Hence, "fix" it! –  Zhen Lin May 7 '12 at 16:03

1 Answer 1

up vote 2 down vote accepted

A family $X \to S$ is just a flat morphism of varieties. The "members" of the family are the fibers $X_s$ above the points $s \in S$. Intuitively, the flatness of the morphism says that the members of the family look alike.

This is in Hartshorne, the section on flat morphisms. For an introduction without the language of sheaf cohomology, try Eisenbud and Harris's The Geometry of Schemes. I don't know any reference where you can learn about families in the context of varieties and not schemes, but since flatness is the key condition underlying a family, you really need scheme-theoretic language anyway.

In your example, we have a morphism to $\mathbb{A}^1$ which just takes $(x, y, t)$ satisfying your equations and returns $t$. This morphism is flat because the ring of functions $k[t]$ on $\mathbb{A}^1$ is a Dedekind domain and so a module is flat over it iff it is $k[t]$-torsion-free.

When you say it is a family over $\mathbb{P}^1$, I assume you mean some compactification of your surface admits a flat map to $\mathbb{P}^1$ extending this one. One natural compactification is to view your surface as a subvariety of $\mathbb{P}^2 \times \mathbb{P}^1$ where the variables in the first are $x, y, z$ and the latter are $s, t$. So the bihomogeneous polynomial of your surface is then

$$ y^2zs = x^3s - x^2zt -x^2zs + xz^2t $$

(To compute this, I expanded everything out, then made it homogeneous separately in $x, y, z$ and $t, s$) and the map is just the restriction of the projection to $\mathbb{P}^1$; on the open subvariety where $s, z\neq 0$, this is your old family. You'll need to prove this is flat over the new point $\infty$ we have added to the $t$-line.

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Thanks. What is the difference between considering the family as a variety over $k$ and as a curve over $k(t)$? –  averageman May 7 '12 at 16:02
    
These are very different ways of looking at it. For example, the dimension as a curve over $k(t)$ is 1 and the dimension as a variety over $k$ is 2. In scheme-theoretic language, "thinking of it as a curve over $k(t)$" means writing down the family (over the affine line or the projective line, doesn't matter) and base-changing to the generic point. –  user29743 May 7 '12 at 16:07
    
Does it ever come into play the fact that the curve is singular at $t=0$ and at $t=1$? –  averageman May 7 '12 at 16:15
    
Singular is a funny word here. The abstract surface is not singular as a variety over $k$, however, two of the fibers of the morphism to the affine line are singular. (We say that the morphism is flat but not smooth.) –  user29743 May 7 '12 at 16:19
    
I never studied these things and it seems a little overwhelming to me... :p But thanks for the explanation. –  averageman May 7 '12 at 16:23

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