Alterations and resolution of singularities

Lecture 2: towards straightenning up the fibers

0.1. The situation. We now have Z ⊂ X →^{f} P (where so far P = ℙ^{d−1}), such that

1. X is normal projective, Z the support of a Cartier divisor

2. f is equidimensional of relative dimension 1 3. The smooth locus of f is dense in all fibers 4. f|_{Z} is finite and generically étale

5. At least one fiber of f is smooth

6. The fibers are connected.

0.2. The idea. It is easy to make an alteration of P, and replace X and Z by their pullbacks, such that Z becomes the union of sections of X → P. Now we can think of the generic fiber of X → P as a smooth curve with a number of points marked on it. Say the genus of this curve is g, and the number of points is n. By the defining property of moduli spaces, we get a rational map P −−→ \overline{M}_{g,n}. Suppose for a moment that \overline{M}_{g,n} were a fine moduli space. This means that there would be a family of curves C → \overline{M}_{g,n} of genus g, and n sections s_{i} : \overline{M}_{g,n} → C, such that any family Y → B of stable n-pointed curves of genus g is obtained by pulling back along some morphism B → \overline{M}_{g,n}. Well then, in our case, the generic fiber of X → P would be the pullback of U along the map P −−→ \overline{M}_{g,n}. If P_{1} → P resolves indeterminacies of this rational map, then we get a diagram

\begin{matrix}X_{1} & −−→ & X \\ ↓ & & ↓ \\ P_{1} & → & P \end{matrix}

The new morphism f_{1} : X_{1} → P_{1} is much nicer than f, since at least the fibers are as nice as one can expect: they are nodal curves. Moreover, Z was made much nicer: it is replaced by n sections which are mutually disjoint, and pass through the smooth locus of f_{1}. If we can resolve P_{1} (say using induction on dimension), then it is easy to resolve X_{1} as well. There are problems though: First, \overline{M}_{g,n} is not a fine moduli space. This will be one reason for using an alteration in place of the modification P_{1} → P. Second, if we want to repeat this inductively, we cannot allow a rational map X_{1} −−→ X - since we cannot pull back nicely along rational map. So we must find a way to make sure that X_{1} −−→ X is actually a regular map.

We will now make steps to deal with the second problem.

0.3. Auxiliary divisor. De Jong’s idea is, that if a component of a fiber of X → P has at least 3 marked points on it, it must survive after stable reduction. To that end, we have the following lemma (copied from de Jong):

0.4. Lemma. Let X → P be as above. There exists a divisor H ⊂ X such that

1. f|_{H} : H → P is finite and generically étale, and

2. for any irreducible component C of a geometric

fiber of f, we have

#sm(X/P) ∩ C ∩ H ≥ 3. Here we count the points without multiplicities.

0.5. Proof. Let n ≥ 1 be an integer. Given a very ample line bundle L on X, consider the embedding

i : X ↪ ℙ = ℙ(Γ(X, L^{⊗n}))

associated to L^{⊗n}.

0.6. Claim. Given any irreducible curve C ⊂ X, the image i(C) ⊂ ℙ is not contained in any linear subspace of dimension n − 1.

0.7. Proof of claim. since L is very ample, the image of Γ(X, L) → Γ(C, L_{|C}) containes a base point free pencil V ⊂ Γ(C, L_{|C}). The map Sym^{m}V → Γ(C, L_{|C}^{⊗n}) has rank ≥ n + 1, therefore Γ(X, L^{⊗n}) → Γ(C, L_{|C}^{⊗n}) has rank ≥ n + 1, which is what we claimed.

0.8. Now we consider the collection of “bad” divisors and show that there are “good” ones left. So consider

T = {(H, y) ∈ ℙ^{∨}×P | dim f^{−1}y∩H = 1} ⊂ ℙ^{∨}×P.

It is clear that T is a Zariski closed subset. We can describe the fibers of pr_{2} : T → P using irreducible components of the fibers:

pr_{2}^{−1}(y) = ∪_{C⊂f^{−1}y}{H|i(C) ⊂ H}.

But by the fact that i(C) is not contained in any linear subspace of dimension n − 1, we have

codim(pr_{2}^{−1}(y), ℙ^{∨}) ≥ n.

Therefore dim T ≤ dim P + dim ℙ^{∨} − n. Thus if n is large enough, pr_{1}(T) ⊂ ℙ^{∨} is of large codimension (at least n − dim P). In particular pr_{1}(T) ≠ ℙ^{∨}. We now fix such large n. Thus there are plenty of H which map finitely to P.

For a fixed closed point y ∈ P (k) consider the set

U(y) = \{H ∈ ℙ^{∨}(k) | \begin{matrix}H∉pr_{1}(T) \\ H ∩ f^{−1}y ⊂ sm(X/P) \\ H ∩ f^{−1}y is reduced \end{matrix} \} .

This is clearly a nonempty open set of ℙ^{∨}. Moreover, if H ∈ U(y) then H ∈ U(y′) for all y′ in a neighborhood of y. If moreover n ≥ 3, then we have that #H ∩ f^{−1}y ≥ 3. so we are done for all points in a neighborhood V of y. We deal with points in P ^{\} V in the same way. Using Noetherian induction we are done.