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.
If a measurable region has area > 1, then there exist distinct such that .
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 . ♦
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.
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.
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.
Let be a finite extension. Then the Picard group of is finite; its cardinality is called the class number of K.
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 .
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 .
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.
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. ♦
If we replace by a general algebraically closed field k, would the proof still work? What additional conditions (if any) need to be imposed?
No maximal ideal of A is principal.
If is generated by f then , which is impossible by lemma 1. ♦
For any points and on E, is principal if and only if
When that happens, we write .
(⇐) 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. ♦
If satisfy , then .
Let as in corollary 2. Then . By the given condition so by corollary 2 again we have and hence . ♦
For any with , there is a unique such that
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 . ♦
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.