# Minkowski Theory: Introduction

Suppose is a finite extension and is the integral closure of in *K*.

In algebraic number theory, there is a classical method by Minkowski to compute the Picard group of (note: in texts on algebraic number theory, this is often called the *divisor class group*; there is a slight difference between the two but for Dedekind domains they are identical).

We will only consider the simplest cases here to give readers a sample of the theory.

Minkowski’s Lemma.If a measurable region has area > 1, then there exist distinct such that .

**Proof**

Use the following picture:

Since the area > 1, there exist two points which overlap in the unit square on the right. The corresponding then give . ♦

Minkowski’s Theorem.Let be a measurable region which is convex, symmetric about the origin and has area > 4. Then X has a lattice point other than the origin.

**Proof**

Take , which has area > 1. By Minkowski’s lemma, there exist distinct such that . Since *X* is symmetric about the origin replace *y* by –*y* (so ) to give . And since *X* is convex, . Finally since , is not the origin. ♦

By applying a linear transform to we obtain the more useful version of Minkowski’s theorem.

Minkowski’s Theorem B.Take a full lattice (i.e. discrete subgroup which spans ). Taking a basis of L, we obtain a fundamental domain

of area D. If is convex, symmetric about the origin, and has area > 4D, then X has a non-zero point of L.

# Picard Group of Number Rings

Now we use this to compute for .

First note that any non-zero ideal has finite index since if , then is a non-zero integer in so , where *N* is the norm function. We let

.

Recall that if , by proposition 4 here, the composition factors of comprise of exactly copies of for each *i*, so

In particular, for any non-zero ideals , and we can extend the norm function to the set of all fractional ideals of *A*.

**Exercise**

Prove that if then

so we can consider as an extension of the norm function to the set of ideals.

To apply Minkowski’s theorem B, we identify with . Let be any non-zero ideal with norm *N*, considered as a full lattice in . We take

where *t* will be decided later. Note that *S* is convex, symmetric about the origin and has area . If where then Minkowski’s theorem B assures us there exists with . Since this holds for all we have:

Now has norm , i.e. . Hence every element of Pic *A* can be represented by an ideal of norm 1 or 2. Since *A* has norm 1 and has norm 2, we have proven:

.

More generally, one can show the following.

Theorem.Let be a finite extension. Then the Picard group of is finite; its cardinality is called the

class number of.K

**Exercise**

Prove that has class number 3. Note that .

Prove that has class number 1.

Prove that has class number 2.

[ Hint: identify with . Pick the square for a suitable *t*. You could also pick but it is not as efficient. ]

# Geometric Example: Elliptic Curve Group

Take the elliptic curve *E* over given by and let be its coordinate ring. In Exercise B.1 here, we showed *A* is a normal domain. Clearly it is noetherian. By Noether normalization theorem, . Hence, *A* is a Dedekind domain.

We will show how computation of Pic *A* leads to point addition on the elliptic curve. For each maximal ideal , write for its image in Pic *A*. Recall that points correspond bijectively to maximal ideals .

Lemma 1.Suppose is not a unit. Then taking as a complex vector space, if and only if can be represented by a linear function in X, in which case

for some .

**Note**

The intuition is that the curve and the elliptic curve cannot have less than 3 intersection points (with multiplicity) unless we take a vertical line.

**Proof**

Since in the ring *A*, without loss of generality we can write

for . The condition implies and have at most two intersection points. Solving gives us

.

If , the LHS has odd degree while the RHS has even degree; hence the equation has at least 3 roots and each corresponds to at least one point on the elliptic curve. If and , then so is linear in *X*, and we are done. ♦

**Exercise**

If we replace by a general algebraically closed field *k*, would the proof still work? What additional conditions (if any) need to be imposed?

Corollary 1.No maximal ideal of A is principal.

**Proof**

If is generated by *f* then , which is impossible by lemma 1. ♦

Corollary 2.For any points and on E, is principal if and only if

When that happens, we write .

**Proof**

(⇐) If then setting gives

If , this ring has exactly two maximal ideals, corresponding to maximal ideals and of *A*. If , it has exactly one maximal ideal so we still have .

(⇒) If is principal, then so by lemma 1, *f* can be represented by a linear function in *X* so we must have *P* and *Q* as described. ♦

Corollary 3.If satisfy , then .

**Proof**

Let as in corollary 2. Then . By the given condition so by corollary 2 again we have and hence . ♦

Lemma 2.For any with , there is a unique such that

**Proof**

First suppose so *P* and *Q* have different *x*-coordinates. Let be the equation of *PQ*. We get:

which has complex dimension 3. Since is divisible by we have for some . Geometrically *R* is the third point of intersection of *PQ* with *E*, which can be equal to *P* or *Q.*

If with , we can similarly pick a line through *P* of gradient . Then as above where has a double root for (this requires some algebraic computation). Hence for some . ♦

Summary.The Picard group of A is given by

.

In particular it is infinite.

Could you please clarify the statement of Lemma 1 in the section Geometric Example. We knew that the dimension of A is 1, so I don’t know how make sense of dim A/(f) \leq 2.

Hmm, the notation is potentially confusing. Here refers to dimension as a complex vector space. Maybe I’ll add a note to that.