As has been hinted in many previous posts, many facts about algebraic number theory tell us about geometric objects like elliptic curves. For instance, if you are working on a problem which primarily uses the affine geometry of a curve like the semistable reduction theorem for elliptic curves, the scheme you’re working on is opposite to what’s called a Dedekind Domain. We begin a series of posts on Dedekind Domains, beginning today with the very abstract and progressing to the concrete(which would of course be terrible for teaching this material but I mean these posts as more of a reference work).

Definition: A Dedekind Domain is an integral domain which is:

• Krull Dimension 1
• Noetherian
• Integrally Closed(Normal)

Many equivalent criteria exist for characterizing Dedekind domains $R$ among one-dimensional noetherian domains, two of which are:

• For each prime $\mathfrak{p}$, the localization $R_\mathfrak{p}$ is a DVR
• Every nonzero ideal admits a unique  factorization into a product of prime ideals

Then an affine ring of an elliptic curve (or any reasonable nonsingular curve) corresponds to a Dedekind domain as they are irreducible(integral) nonsingular(normal)  curves(1-dimensional and noetherian). This is already enough to be of interest to any number theorist, but rings of integers of number fields are also Dedekind domains.

Definition: A number field  $K$ is a finite algebraic extension of $\mathbf{Q}$. The ring of integers $\mathcal{O}_K$ is the integral closure of $\mathbf{Z}$ in $K$. This is sometimes also called the maximal order of $K$, in reference to the fact that we call subrings of $K$ that are also free $\mathbf{Z}$-modules of rank $[K:\mathbf{Q}]$ contained in $\mathcal{O}_K$ orders of $K$.

These are 1-dimensional since any prime ideal $\mathfrak{p}$ of an order $\mathcal{O}$ intersects with $\mathbf{Z}$ in an integral  prime ideal $(p)$  so $\mathcal{O}/\mathfrak{p}$ is a finite extension of $\mathbf{Z}/(p)$ so is a finite integral domain and thus a field. They are clearly noetherian since they are finitely generated $\mathbf{Z}$-modules.

From this we can already see that rings of integers are Dedekind domains, but much more is true. The following theorem shows that starting from any order (or affine ring of  a geometrically integral algebraic curve which could even be singular) the integral closure is a Dedekind domain. It also holds independent interest regarding the valuative criteria for properness and seperatedness (Hartshorne exercise II.4.11).

Theorem[Krull-Akizuki]:If $R$ is a one dimensional noetherian domain with field of fractions $K$ and $L$ is a finite extension of $K$ then the integral closure of $R$ in $L$ is a Dedekind domain.

For a proof consider first that we can simplify by taking $K=L$ by letting $l_1, \dots, l_n$ be a basis of $L/K$ contained in the integral closure $\mathcal{O}_L$ and considering $S= R[l_1, \dots, l_n]$. This is noetherian because it is a finitely generated $R$ module and $R$ is noetherian. It is one dimensional because $R$ is one-dimensional and if $\mathfrak{P} \supsetneq \mathfrak{P}' \supsetneq (0)$ is a chain of primes in $S$ then their intersections with $R$ give a chain of primes $\mathfrak{p} \supset \mathfrak{p}' \supset (0)$ in $R$. Since $\mathfrak{P},\mathfrak{P}'$ are nonzero ideals we could say about either of them that they contain a nonzero element $\alpha$ and thus the principal ideal generated by $\alpha$. Since every element of $S$ is integral over $R$(since $l_1, \dots,l_n\in \mathcal{O}_L$), $\alpha$ satisfies an irreducible monic polynomial over $R$ so $\alpha^n + a_{n-1}\alpha^{n-1} + \dots + a_1 \alpha = a_0 \in R$. Thus $\mathfrak{P}' \supset (\alpha) \ni a_0$ (respectively for $\mathfrak{P}$) and so $\mathfrak{p},\mathfrak{p}'$ are nonzero primes of $R$ so they must be equal by dimensionality. On the other hand consider that since $S/R$ is a finite integral extension so $(S/\mathfrak{P}')/(R/\mathfrak{p}')$ is also a finite integral extension, or more to the point since $\mathfrak{p}'$ is a maximal ideal,  a finite algebraic field extension. We thus arrive at a contradiction under the supposition that there is a chain of primes of length greater than 1 since $S/\mathfrak{P}'$ has the nonzero prime ideal $\mathfrak{P}/\mathfrak{P}'$. On the other hand, $S$ cannot be dimension zero because $R$ contains nonzero primes and is contained in $S$. Since a dimension zero domain is a field, if $p \in \mathfrak{p} \subset R$ then $1/p\in S$. This is in contradiction to the fact that $S$ is integral over $R$ whereas the minimal polynomial for $1/p$ over $R$ is $pX -1$.

Thus we have reduced ourselves to the case that $L=K$ and to the following statement, which is sometimes itself called the Krull-Akizuki Theorem. Let $R$ be a one dimensional noetherian domain with field of fractions $K$. Then any ring $S$ such that $R \subset S\subset K$ is also a one dimensional noetherian domain.

The idea of the proof is that if $I$ is a nonzero ideal of $S$ and $a$ is a nonzero element of $I\cap R$ then we can find an integer $k$ such that $a^kS \subset R$ and so as an $R$ submodule $S/aS \cong a^k S/ a^{k+1} S \subset R/$(something), and thus is finitely generated over $R$ so $I \bmod aS$ is finitely generated over $R$ and so over $S$. Thus by adding $a$ to the list of generators, $I$ is finitely generated over $S$. For the gritty details of the proof of this statement, I refer the reader to Theorem 4.9.2 in http://books.google.com/books?id=APPtnn84FMIC&lpg=PA83&ots=2L9MiWbIYZ&dq=krull%20akizuki&pg=PA85

Note that I make no claims as the the finite generation of $S$ over $R$. If $L/K$ is a separable field extension this holds true, but it’s quite possible to cook up an example of an integral closure $S/R$ which is not finitely generated if the field extension is inseparable(see Theorem 100 of Kaplansky’s book on Commutative Rings, where the example is actually a pair of DVRs). If this happens you’re typically stuck and can’t really do any algebra or geometry, but if we have finite generation( or a finite type morphism if that’s the terminology you prefer)  of $S/R$ then we can have much of what we could possibly want to be true.

For instance, take the case that we have a degree $m$ morphism of curves $f: X \to Y$ over an algebraically closed field $k$. Then if $x_1, \dots x_r$ are the preimages of $y$ with multiplicities $e_1 ,\dots , e_r$  in $x$ it is well-known that $m = \sum e_i$. We have a corresponding result once we allow for non-closed points.

Theorem: Let $R$ be a Dedekind Domain with fraction field $K$. Let $L$ be a finite field extension and $S$ the integral closure of $R$ in $L$. Then if $S$ is finitely generated over $R$ if a prime ideal $\mathfrak{p}$ of $R$ factors in $S$ as

$\mathfrak{p}S = \prod_{i=1}^g \mathfrak{P}_i^{e_i}$

then if $\dim_{R/\mathfrak{p}} S/\mathfrak{P}_i = f_i$

$[L:K] = \sum_{i=1}^g e_if_i$.

For a detailed proof we refer the reader to either Dino Lorenzini’s Invitation to Arithmetic Geometry Theorem III.3.5. The idea is that if $R$ is a P.I.D. then we can use classical techniques, like the structure theorem on finitely generated modules over a PID to get $\dim_{R/\mathfrak{p}} S/\mathfrak{p}S = [L:K]$ and the chinese remainder theorem to decompose $S/\mathfrak{p}S$ into a product of finite field extensions of $R/\mathfrak{p}$. Then we can reduce to the case of a PID by localization.

Having considered ideals and how they factor, we come to fractional ideals: a key point in algebraic number theory. An $R$-module $M$ contained in the fraction field $K$ is called a fractional ideal if it is finitely generated over $R$ or equivalently if there is a denominator $d \in R$ such that $dM \subset R$.

A fractional ideal $I$ is called invertible if there exists another fractional ideal $J$ such that the module product $IJ = qR$ for some $q \in K^\times$. We call ideals of the form $qR$ for $q\in K^\times$ principal fractional ideals. Note that invertible fractional ideals are projective $R$-modules of rank 1 and any projective $R$-module of rank 1 is isomorphic to a fractional ideal. Also note that two fractional ideals $I,J$ are isomorphic as $R$-modules iff there is $q \in K^\times$ such that $qI = J$. By definition the invertible fractional ideals of a ring form a group, and likewise after we mod out by the normal subgroup of principal fractional ideals. We call this quotient group the Picard Group $Pic(R)$ and if the reader proves the above claims about projective modules of rank 1 this name makes sense in a wider context.

The Picard Group of a Dedekind Domain(which we sometimes call the ideal class group to distinguish how nice it is) is particularly nice because of the following characterization:

• A one dimensional noetherian domain $R$ is Dedekind if and only if every fractional ideal is invertible.

This is easy enough to prove with the following lemma.

Lemma: A fractional ideal $I$ of a one-dimensional noetherian domain $R$ is invertible if and only if for all $\mathfrak{p} \in Spec(R)$, $IR_\mathfrak{p}$ is a principal ideal of $R_\mathfrak{p}$.

For a proof, see page 17 of http://math.uga.edu/~pete/8430notes2.pdf or else Neukirch’s book, section I.12.

We bring this up because if $R$ is a one-dimensional noetherian domain with field of fractions $K$ and $S$ is the integral closure of $R$ in $K$ and $S$ is finitely generated over $R$ then we can relate the Picard group of $R$(which might not be so easy to understand) with the Picard group of $S$ (which by the above is easier to understand).

Theorem: If we let $S_\mathfrak{p} =(R -\mathfrak{p})^{-1}S$, there is an exact sequence

$\displaystyle 0 \to S^\times/R^\times \to \bigoplus _{p \in Spec(R)} S_\mathfrak{p}/R_\mathfrak{p} \to Pic(R)/Pic(S) \to 0$.

To prove this first note that $K^\times$ surjects onto the group of principal fractional ideals in the obvious way and that if $qR = q'R$ then $q = rq'$ for some $r \in R^\times$. Thus the principal fractional ideals are isomorphic to $K^\times /R^\times$.

Moreover by the above Lemma, the invertible fractional ideals are isomorphic to the direct sum over all primes $\mathfrak{p}$ of the principal ideals of $R_\mathfrak{p}$. Thus the group of invertible fractional ideals is isomorphic to $\bigoplus_{\mathfrak{p} \in Spec(R)} K^\times/R_\mathfrak{p}^\times$.

Thus we have the exact sequence

$0 \to K^\times/R^\times \to \bigoplus_{\mathfrak{p} \in Spec(R)} K^\times/R_\mathfrak{p}^\times \to Pic(R) \to 0$

Likewise if we do the same for $S$ we get the sequence

$0 \to K^\times/S^\times \to \bigoplus_{\mathfrak{P} \in Spec(S)} K^\times/S_\mathfrak{P}^\times \to Pic(S) \to 0$

Now consider that $Spec(S_\mathfrak{p})$ consists only of (the finitely many) primes lying above $\mathfrak{p}$. It is not hard to prove that the localization of a Dedekind domain is Dedekind, and to use the Chinese Remainder Theorem to prove that a Dedekind domain with finitely many primes is a PID. If we let $P(A)$ denote the group of principal fractional ideals of a ring $A$, we have:

$P(S_\mathfrak{p}) = \bigoplus_{\mathfrak{P} \supset \mathfrak{p}} P(S_\mathfrak{P})$ and thus since every prime of $S$ lies over a prime of $R$, $\bigoplus_{\mathfrak{P} \in Spec(S)} P(S_\mathfrak{P}) = \bigoplus_{\mathfrak{p} \in Spec(R)}\bigoplus_{\mathfrak{P}\supset \mathfrak{p}} = \bigoplus_{\mathfrak{p} \in Spec(R)} P(S_\mathfrak{p})$. Therefore our sequence for $S$ becomes the following:

$0 \to K^\times/S^\times \to \bigoplus_{\mathfrak{p} \in Spec(R)} K^\times/S_\mathfrak{p}^\times \to Pic(S) \to 0$

Then we also have natural (surjective) quotient maps $\alpha: K^\times/R^\times \to K^\times/S^\times$ and $\beta: \bigoplus_{\mathfrak{p} \in Spec(R)} K^\times/R_\mathfrak{p}^\times \to \bigoplus_{\mathfrak{p} \in Spec(R)} K^\times/S_\mathfrak{p}^\times$ which commute and thus give a natural surjective quotient map $\gamma: Pic(R) \to Pic(S)$. Then we just apply the snake lemma for our result.

Next time we see what we can do with Dedekind Domains in separable extensions.

About these ads