The many principles of conservation of number March 4, 2014

Posted by David Speyer in Uncategorized.

In algebraic geometry, we like to make statements like: “two conics meet at ${4}$ points”, “a degree four plane curve has ${28}$ bitangents”, “given four lines in three space, there are ${2}$ lines that meet all of them”. In each of these, we are saying that, as some parameter (the conics, the degree four curve, the lines) changes, the number of solutions to some equation stays constant. The “principle of conservation of number” refers to various theorems which make this precise.

In my experience, students in algebraic geometry tend to pick up the rough idea but remain hazy on the details, most likely because there are many different ways to make these details precise. I decided to try and write down all the basic results I could think of along these lines.

Let ${B}$ be some parameter space such as the space of pairs ${(C_1, C_2)}$ of two conics. Let ${X}$ be some space of solutions, such as the space of triples ${(C_1, C_2, x)}$ where ${x}$ is a point on ${C_1 \cap C_2}$. Let ${\pi: X \rightarrow B}$ be a map, such as projection onto the ${(C_1, C_2)}$ components. We want theorems which will discuss the size of the fibers of ${\pi}$, in terms of some global degree of the map ${\pi}$.

We work over some field ${k}$. For simplicity of presentation, we’ll assume that ${B}$ is affine, meaning that it is a subset of ${k^n}$ defined by polynomial equations

$\displaystyle f_1(x_1, \ldots, x_n) = f_2(x_1, \ldots, x_n) = \cdots = f_m(x_1, \ldots, x_n)=0.$

We’ll write ${R}$ for the ring ${k[x_1, x_2, \ldots, x_n]/(f_1, f_2, \ldots, f_m)}$.

It would be silly to ask for any such results if ${B}$ were disconnected. A very basic observation of algebraic geometry is that ${B}$ is connected if and only if ${R}$ has no nontrivial idempotents. In fact, we will ask for something stronger: That ${R}$ is an integral domain. The terminology for this is that ${B}$ is irreducible. From now on, we will make:

Assumption ${B}$ is irreducible. (${R}$ is an integral domain.)

If ${X}$ is also affine, with corresponding ring ${S}$, then ${S}$ is an ${R}$ module. We define the degree of ${\pi}$ in this case to be the dimension of ${S \otimes_R \mathrm{Frac}(R)}$ as a ${\mathrm{Frac}(R)}$ vector space. Degree can be defined in much greater generality; we will feel free to refer to it in greater generality without giving the definition. We will denote the degree of ${X \rightarrow B}$ by ${d}$. Roughly, we want theorems which say that the fibers of ${\pi}$ have size ${d}$.

Here is our first result.

Theorem (Shafarevich, II.6.3, Theorem 4) If ${k}$ has characteristic zero and is algebraically closed then ${|\pi^{-1}(b)|=d}$ for almost all ${b}$ in ${B}$. More precisely, there is some polynomial ${g}$, not identically zero on ${B}$, so that ${g(b) \neq 0}$ implies ${|\pi^{-1}(b)|=d}$.

Warning This isn’t true if ${k}$ is not algebraically closed: Consider the map ${x \mapsto x^2}$ from ${\mathbb{Q} \rightarrow \mathbb{Q}}$.

Warning This isn’t true in characteristic ${p}$: Consider ${x \mapsto x^p}$.

We now want results which let us say something, not just about almost all ${b}$, but about all ${b}$.

Naive size

We will at first focus on counting the size of ${\pi^{-1}(b)}$ in a naive sense: We think of ${X}$ as sitting in ${k^m}$ (or in ${\mathbb{P}^m_k}$) and we literally count points of the fiber. We can’t hope for the fibers to always be of full size ${d}$ because even the nicest map, ${x \mapsto x^2}$, has fiber of size ${1}$, not ${2}$, over the point ${0}$. So, using the naive size, we can only hope for upper bounds.

There are two additional problems. The first one is if we have something like ${\{ (x,y) : xy=x(x-1)=0 \}}$ projecting onto the ${x}$ coordinate. In this case, the degree is ${1}$ but the fiber over ${0}$ has size ${2}$. When ${X}$ is affine, with corresponding ring ${S}$, we can fix this by requiring that ${S}$ is torsion free as an ${R}$-module. In general, the right condition is that no irreducible component of ${X}$ maps to a proper subvariety of ${B}$.

More subtly, suppose that ${B}$ is a nodal curve, such as ${u^2 = v^2 (v+1)}$, and ${X}$ is its desingularization. (In this case, the line with ${t \mapsto (t^3-t, t^2-1)}$ as the map ${X \rightarrow B}$.) Then the degree of the map is ${1}$, but the fiber over ${(u,v)=0}$ is ${t = \pm 1}$, of size ${2}$. The hypothesis to rule this out is that ${R}$ is integrally closed in its fraction field. By definition, this is the same as saying that ${B}$ is normal.

Once we rule out these possibilities, we have

Theorem (Shafarevich, II.6.3, Theorem 3) If ${B}$ is normal, and no irreducible component of ${X}$ maps to a proper subvariety of ${B}$, then every fiber of ${\pi}$ has naive size ${\leq d}$.

I can’t resist mentioning a result which far harder than these:

Theorem (A consequence of Zariski’s Main Theorem) Let $B$ be normal and let $X \to B$ have degree $d$. Assume that no irreducible component of ${X}$ maps to a proper subvariety of ${B}$. For any $b$ in $B$, the number of connected components of $\pi^{-1}(b)$ is at most $d$.

Scheme theoretic size

We now consider counting size in a less naive way. Again, for simplicity, suppose that ${X}$ is affine, with corresponding ring ${S}$. Let ${(b_1, b_2, \ldots, b_n)}$ be a point of ${B}$, so there is a map of rings ${R \rightarrow k}$ by ${x_i \mapsto b_i}$. Consider the ring ${S \otimes_R k}$, where ${R}$ acts on ${k}$ by the above map. The maps from this ring to ${k}$ are the point in ${\pi^{-1}(b)}$. Thus, ${\dim_k S \otimes_R k}$ is an upper bound for the number of points of ${X}$ above ${b}$. We will call this dimension the scheme theoretic size of the fiber. Once again, it can be defined when ${X}$ is not affine as well.

We have the following cautionary example: Let ${X = \{ (x,y) : xy=1 \}}$ mapping onto the ${x}$ coordinate. Then the degree is ${1}$, but the fiber above ${0}$ has size ${0}$, either scheme theoretically or naively. To rule this out, we impose that ${X}$ is finite over ${B}$. By definition, this means that ${X}$ is affine, and ${S}$ is a finitely generated ${R}$ module.

You might worry about how we could ever prove that ${X}$ is affine if it is not given to us as a closed subset of ${k^m}$. Fortunately, we have:

Theorem (Hartshorne, Exercise III.11.2) If ${X \rightarrow B}$ is projective with finite fibers, then it is a finite map. Here projective means that ${X}$ is a closed subset of ${\mathbb{P}^n \times B}$, projecting onto ${B}$. (This is not the morally right definition of a projective map, but if you are ready for the right definition, then you should be working with “proper” rather than “projective” anyway.)

We then have

Theorem (Hartshorne, Exercise II.5.8) If ${X}$ is finite over ${B}$, and no irreducible component of ${X}$ maps to a proper subvariety of ${B}$, then every fiber of ${\pi}$ has scheme theoretic size ${\geq d}$.

Flatness

Theorem Let ${X \rightarrow B}$ be a finite map. Then all fibers have scheme theoretic size ${d}$ if and only if ${X}$ is flat over ${B}$.

Unfortunately, flat is a rather technical condition. The first thing to understand is that some nice looking maps can fail to be flat:

Warning Let ${X}$ be ${\{ (w,x,y,z) : wy=wz=xy=xz=0 \}}$, let ${B = k^2}$ and let the map ${X \rightarrow B}$ be ${(w,x,y,z) \mapsto (w-y, x-z)}$. This is a finite map. (We can alternately describe ${X}$ as ${\{ (w,x,0,0) \} \cup \{ (0,0,y,z) \}}$.) This map is degree ${2}$, but the fiber over ${(0,0)}$ has scheme theoretic size ${3}$ (and naive size ${1}$).

If your eye is well enough trained that this doesn’t look nice to you, try the examples here.

There are two good conditions that imply flatness:

Theorem (Hartshorne III.9.7) If ${B}$ is normal and one dimensional, and no irreducible component of ${X}$ maps to a proper subvariety of ${B}$, then ${X}$ is flat over ${B}$.

Theorem (The miracle flatness theorem) If ${X}$ is Cohen-Macaulay, ${B}$ is smooth of the same dimension as ${X}$, and ${X \rightarrow B}$ is finite, then ${X \rightarrow B}$ is flat.