MODULI SPACES AND DEFORMATION THEORY, CLASS 20

RAVI VAKIL

Contents

1. Special lecture by Ragni Piene 1 2. Embedded deformations (of X ↪ 𝔸^{n}) 1 3. Relations Criterion 2

1. Special lecture by Ragni Piene

Most of this class was a special lecture by Ragni Piene, visiting from Oslo.

2. Embedded deformations (of X ↪ 𝔸^{n})

(I mentioned an observation of Brian that a family of smooth affine varieties is formally locally trivial, even when the family is not trivial!)

We were proving:

Proposition. Suppose we have a fiber diagram

X ↪ X_{A}
↓  ↓
Spec k ↪ Spec A

where A ∈ C. Suppose X ↪ 𝔸^{n}. Then there is a closed immersion X_{A} ↪ 𝔸_{A}^{n} := 𝔸^{n} × Spec A. (Translation.)

This was reduced to:

Proposition. Suppose A′ → A is a square-zero extension in C, with kernel J. Suppose we have a fiber diagram

X_{A} ↪ X_{A′}
↓  ↓
Spec A ↪ Spec A′.

Suppose X ↪ 𝔸_{A}^{n}. Then there is a closed immersion X_{A′} ↪ 𝔸_{A′}^{n}. Brian loves this sort of question.

Date: Thursday, November 16, 2000.

Let’s use the notation O(X) to be the ring of global sections of the structure sheaf of a scheme X.

The argument went as follows. Let x_{1}, …, x_{n} ∈ O(X_{A}) be the functions on X_{A} mapping it to 𝔸_{A}^{n}.

The first step is to lift them to functions of O(X_{A′}); this will give a map X_{A′} → 𝔸_{A′}^{n}, restricting to the map over A.

Here’s how that happened. Let |X| be the underlying topological space of X_{A} and X_{A′}. Let I be the ideal sheaf of X_{A} in X_{A′}, so

0 → I → O_{X_{A′}} → O_{X_{A}} → 0.

Then J^{2} = 0 meant I^{2} = 0. Thus I is actually a coherent sheaf of O_{X_{A}} modules. X_{A} is affine, so H^{1}(X_{A}, I) = H^{1}(X, I) = 0. Thus the long exact sequence begins:

0 → H^{0}(X, I) → O(X_{A′}) → O(X_{A}) → 0.

Thus we can lift x_{1}, …, x_{n}.

So now we have a map X_{A′} → 𝔸_{A′}^{n}, and we wish to show that this is a closed immersion. On the level of topological spaces, it’s already an inclusion of X_{A′} as a closed subset (as on the level of topological spaces, this is the same as X_{A} ↪ 𝔸_{A}^{n}).

Hence we only need to check that this separates tangent vectors. This is left as an exercise.

The upshot of this is that the natural morphism of functors (Emb def. X) → (Def X) is formally smooth.

3. Relations Criterion

So our fundamental question is this.

Suppose we have X ↪ 𝔸^{n}. Suppose we have a fiber diagram

X ↪ X_{A}
↓  ↓
Spec k ↪ Spec A

where A ∈ C. Suppose X_{A} ↪ 𝔸_{A}^{n}, restricting to X ↪ 𝔸^{n}.

How can we tell when X_{A} is flat over A?

(Some description here.)

An A-module M is flat if N → M ⊗_{A} N is exact. It’s easy to show that this is equivalent to Tor_{1}^{A}(M, N) = 0 for all finitely-generated N; this doesn’t use any property of C.

However, an A-module that’s finitely generated can be filtered so that successive quotients are all k, so M is flat if Tor_{1}(M, k) = 0. (Warning: Here we are using A ∈ C!)

Now suppose we have a presentation for the ideal of definition (call it I_{A}) of X_{A} in 𝔸_{A}^{n}:

A[x_{1}, …, x_{n}]^{l} → A[x_{1}, …, x_{n}]^{m} → A[x_{1}, …, x_{n}] → O(X_{A}) → 0.

(Explain.)

A[x_{1}, …, x_{n}]^{l} → A[x_{1}, …, x_{n}]^{m} → I_{A} → 0,
0 → I_{A} → A[x_{1}, …, x_{n}] → O(X_{A}) → 0.

Lemma. O(X_{A}) is A-flat iff the above presentation for I_{A}, tensored by k, is a presentation for I.

Proof. Break it up into two pieces, using I_{A} (see above). (Short check.)

We also note that flatness is the same as I_{A} ⊗_{A} k = I.

The following is a fundamental useful fact. (I didn’t really state it during class

— although Jason effectively did. I’ll state it next time.)

Proposition (Relations criterion for flatness). Let O(X) = k[x_{1}, …, x_{n}]/(f_{1}, …, f_{m}), and O(X_{A}) = A[x_{1}, …, x_{n}]/(f′_{1}, ..., f′_{m}) where f′_{i} are liftings of f_{i}. Then X_{A} is flat over A iff every relation among (f_{1}, …, f_{m}) lifts to a relation among (f′_{1}, …, f′_{m}).

I’ll now prove the Relation criterion. Later, we’ll use it to prove stuff.

For the rest of this discussion, suppose we are given O(X) = k[x_{1}, …, x_{n}]/(f_{1}, …, f_{m}) and liftings f′_{i} of f_{i} to A[X].

This translates to: We have an exact sequence

A[x_{1}, …, x_{n}]^{m} → A[x_{1}, …, x_{n}] → O(X_{A}) → 0,

which, when tensored with k, gives

k[x_{1}, …, x_{n}]^{m} → k[x_{1}, …, x_{n}] → O(X) → 0.

Note that this need not imply that I_{A} ⊗ k = I (where I_{A}, and I, are as defined earlier). If it did, we’d already have flatness!

A bit more explanation: Suppose we have a complete set of relations for the f_{i}’s. In other words, an exact sequence

k[x_{1}, …, x_{n}]^{l} → k[x_{1}, …, x_{n}]^{m} → k[x_{1}, …, x_{n}] → O(X) → 0.

Then giving a lifting of these relations means giving

A[x_{1}, …, x_{n}]^{l} → A[x_{1}, …, x_{n}]^{m} → A[x_{1}, …, x_{n}] → O(X_{A}) → 0,

which is exact except at A[x_{1}, …, x_{n}]^{m}, where it is only assumed to be a complex. In other words, there may be more relations! We’re requiring that the relations lift, not that there aren’t any more relations upstairs.