2024-10-11

Yesterday

The diagonal of is a closed immersion, so is an immersion, and in fact a closed immersion because the map is finite. But this is impossible as is not surjective.

Discrete valuation rings

Let be a local domain. Then the following conditions are equivalent:

  1. is a PID.
  2. is Noetherian, Krull dimension , and normal.
  3. is Noetherian, Krull dimension , and .
  4. is a Noetherian valuation ring.

We call satisfying any of the above conditions a discrete valuation ring, or DVR.

If is a domain, we can drop the condition of Krull dimension in 3.: the dimension is bounded by , and Noetherian Krull dimension is equivalent to Artinian, but Artinian domains are fields, contradicting that is local.

Proof

Easy cases

1. 2. and 3. are easy checks:

  • 1. 2.) A PID is Noetherian, has Krull dimension because every non-zero proper ideal is maximal, and PID UFD normal (integer root theorem).
  • 2. 3.) because is principally generated.

So is 4. 1.: in a valuation ring, every finitely-generated ideal is principal. If is Noetherian, then every ideal is finitely-generated.

3. 4.

Because , then we can find such that the image of generates , i.e., . Nakayama gives that .

To see this claim, apply Nakayama’s Lemma to

and therefore since , we must have that .

Let , where and . If or , then or . Otherwise, neither , so they must belong to the maximal ideal . Write and , and therefore . And so on.

If this stops, then or as before. If it doesn’t stop, then

So .

Example

satisfies 3.. What is ? Let , . If , it must be “infinitely divisible by ”. So must be .

Krull’s Intersection Theorem

Exercise

The ideal . Use and Nakayama. (In general, this argument doesn’t work, i.e., if we looked at for an arbitrary ideal .)

The upshot

From , we get a discrete valuation

which is a (surjective) homomorphism. Under this valuation:

  • .
  • if and only if , where is called the uniformiser.
  • If then for some ( is certainly principal a priori).

Returning to

If then . We can rewrite this as

where we choose and . Then the valuation (the valuation of is as it is a unit).

Remark

If , a DVR, is a -algebra, and , then

Cohen structure theorem.

Definition (Non-singular curve)

Let be a curve. We say that is non-singular if for all closed points , is a DVR. This is equivalent to by the above remark.

#TODO diagram

Examples

, , are non-singular (since being non-singular is a local property, and we know what these curves look like locally).

However, is singular. To see this, we regard . Consider the monic polynomial

Now,

But , and : so is not normal. Alternatively, we can use the ring and consider the fact that in .

Looking at the ring more, the maximal ideal . We want to calculate . We can write

So .

Sections

Let be non-singular. Let be open. Then

The valuation tells us that if there is no pole of at , then the function is regular.

Proof

It suffices to assume that . Then means maximal (the closed points correspond to maximal ideals). But now if and only if . It thus suffices to prove that

The containment is clear. For , let and let . If , then so there is nothing to do. If , then for some maximal ideal. But , and so for and . That is, and , a contradiction to .

Hartog’s theorem. Poles occur in codimension . In our case, they only exist at closed points: if a function has no poles there, then the function is regular.

Remark: Normalisation

There exists a normal “curve”, non-singular (may be disconnected)

“Normal section”. C.f. integral closure of a ring in an extension.

Pullback on divisors

Let be a map of non-singular curves. Then we define the pullback

where is the uniformiser of the DVR , is the induced map on local rings:

Note that implies that .

Example

Consider the curves

and the map

⟻

Let be a closed point, i.e., for some . Then

Since and are two distinct primes in , each factor gets valuation . If instead , we get , and the valuation is : we get ramification.

Lemma

Let be a finite morphism (here) of non-singular curves. Then

Example

Consider , the usual inclusion given by the complement of . We know that we have

since they have the same function fields. However . But is not finite.

A finite, non-constant map of non-singular curves is flat.