Commutative Algebra 47

Minkowski Theory: Introduction

Suppose K/\mathbb Q is a finite extension and \mathcal O_K is the integral closure of \mathbb Z in K.

In algebraic number theory, there is a classical method by Minkowski to compute the Picard group of \mathcal O_K (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 X\subseteq \mathbb R^2 has area > 1, then there exist distinct x, y\in X such that x-y\in \mathbb Z^2.

Proof

Use the following picture:

minkowski_lemma

Since the area > 1, there exist two points which overlap in the unit square on the right. The corresponding x,y\in X then give x-y\in \mathbb Z^2. ♦

Minkowski’s Theorem.

Let X\subseteq \mathbb R^2 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 \{\frac 1 2 x : x\in X\}, which has area > 1. By Minkowski’s lemma, there exist distinct x, y\in X such that \frac 1 2 x - \frac 1 2 y \in \mathbb Z^2. Since X is symmetric about the origin replace y by –y (so x\ne -y) to give \frac 1 2 x + \frac 1 2 y \in \mathbb Z^2. And since X is convex, x,y\in X \implies \frac 1 2 x + \frac 1 2 y \in X. Finally since x \ne -y, \frac 1 2 x + \frac 1 2 y is not the origin. ♦

By applying a linear transform to \mathbb R^2 we obtain the more useful version of Minkowski’s theorem.

Minkowski’s Theorem B.

Take a full lattice \mathbb Z^2 \cong L \subset \mathbb R^2 (i.e. discrete subgroup which spans \mathbb R^2). Taking a basis (v_1, v_2) of L, we obtain a fundamental domain

\{ \alpha_1 v_1 + \alpha_2 v_2 : 0 \le \alpha_1, \alpha_2 < 1\}

of area D. If X\subseteq \mathbb R^2 is convex, symmetric about the origin, and has area > 4D, then X has a non-zero point of L.

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Picard Group of Number Rings

Now we use this to compute \mathrm{Pic} A for A =\mathcal O_{\mathbb Q(\sqrt{-5})} = \mathbb Z[\sqrt{-5}].

First note that any non-zero ideal \mathfrak a \subseteq A has finite index since if x\in \mathfrak a - \{0\}, then N(x) \in \mathfrak a is a non-zero integer in \mathfrak a so N(x)A \subseteq \mathfrak a, where N is the norm function. We let

N(\mathfrak a) := [A : \mathfrak a].

Recall that if \mathfrak a = \mathfrak m_1^{k_1} \ldots \mathfrak m_n^{k_n}, by proposition 4 here, the composition factors of A/\mathfrak a comprise of exactly k_i copies of A/\mathfrak m_i for each i, so

N(\mathfrak a) = N(\mathfrak m_1)^{k_1} \ldots N(\mathfrak m_n)^{k_n}.

In particular, N(\mathfrak {ab}) = N(\mathfrak a)N(\mathfrak b) for any non-zero ideals \mathfrak a, \mathfrak b \subseteq A, and we can extend the norm function to the set of all fractional ideals of A.

Exercise

Prove that if \alpha = x + y\sqrt{-5} \in A - \{0\} then

N(\alpha A) = x^2 + 5y^2 = N(\alpha)

so we can consider N(\mathfrak a) as an extension of the norm function to the set of ideals.

To apply Minkowski’s theorem B, we identify x+y\sqrt{-5} \in A with (x,y) \in \mathbb R^2. Let \mathfrak a\subseteq A be any non-zero ideal with norm N, considered as a full lattice in \mathbb R^2. We take

S = \{(x,y) \in \mathbb R^2 : x^2 + 5y^2 \le t\}

where t will be decided later. Note that S is convex, symmetric about the origin and has area \frac{\pi t}{\sqrt 5}. If t = \frac {4\sqrt{5}}\pi N + \epsilon where \epsilon > 0 then Minkowski’s theorem B assures us there exists a\in \mathfrak a - \{0\} with N(a) \le t. Since this holds for all \epsilon > 0 we have:

N(a) \le \frac {4\sqrt 5}\pi N(\mathfrak a), \quad a\in \mathfrak a - \{0\}.

Now \mathfrak b := a\mathfrak a^{-1} \subseteq A has norm \le {4\sqrt 5}\pi, i.e. \le 2. Hence every element of Pic A can be represented by an ideal \mathfrak a of norm 1 or 2. Since A has norm 1 and \mathfrak m = (2, 1 + \sqrt{-5}) has norm 2, we have proven:

\mathrm{Pic} (\mathbb Z[\sqrt{-5}]) = \mathbb Z / 2\mathbb Z.

More generally, one can show the following.

Theorem.

Let K/\mathbb Q be a finite extension. Then the Picard group of \mathcal O_K is finite; its cardinality is called the class number of K.

Exercise

Prove that K = \mathbb Q(\sqrt{-23}) has class number 3. Note that \mathcal O_K = \mathbb Z[ \frac{1 + \sqrt{-23}}2].

Prove that K = \mathbb Q(\sqrt{-163}) has class number 1.

Prove that K = \mathbb Q(\sqrt{10}) has class number 2.

[ Hint: identify a + b\sqrt{10} \in \mathbb Z[\sqrt{10}] with (a + b\sqrt{10}, a - b\sqrt{10}) \in \mathbb R^2. Pick the square \{ (x, y) \in \mathbb R^2 : |x| + |y| \le t \} for a suitable t. You could also pick \{(x, y) \in \mathbb R^2 : |x|, |y| \le t\} but it is not as efficient. ]

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Geometric Example: Elliptic Curve Group

Take the elliptic curve E over \mathbb C given by \{(x,y) : y^2 = x^3 - x\} and let A = \mathbb C[E] be its coordinate ring. In Exercise B.1 here, we showed A is a normal domain. Clearly it is noetherian. By Noether normalization theorem, \dim A = 1. 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 \mathfrak m \subset A, write [\mathfrak m] for its image in Pic A. Recall that points P\in E correspond bijectively to maximal ideals \mathfrak m_P \subset A.

Lemma 1.

Suppose f\in A is not a unit. Then taking A/(f) as a complex vector space, \dim_{\mathbb C} A/(f) \le 2 if and only if f can be represented by a linear function in X, in which case

A/(f) \cong \mathbb C[Y]/(Y^2 - \beta)

for some \beta\in \mathbb C.

Note

The intuition is that the curve f(X, Y) = 0 and the elliptic curve cannot have less than 3 intersection points (with multiplicity) unless we take a vertical line.

Proof

Since Y^2 = X^3 - X in the ring A, without loss of generality we can write

f(X, Y) = Y\cdot g(X) + h(X)

for g(X), h(X) \in \mathbb C[X]. The condition \dim_{\mathbb C} A/(f) \le 2 implies Y^2 - X^3 + X and Y\cdot g(X) + h(X) have at most two intersection points. Solving gives us

g(X)^2 (X^3 - X) = h(X)^2.

If g(X) \ne 0, 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 g(X) = 0 and \deg h(X) = m, then \dim_{\mathbb C} A/(f) = 2m so h(X) is linear in X, and we are done. ♦

Exercise

If we replace \mathbb C 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 \mathfrak m \subset A is generated by f then A/(f) \cong \mathbb C, which is impossible by lemma 1. ♦

Corollary 2.

For any points P = (\alpha_1, \beta_1) and Q = (\alpha_2, \beta_2) on E, \mathfrak m_P \mathfrak m_Q is principal if and only if

\alpha_1 = \alpha_2, \beta_1 = -\beta_2.

When that happens, we write P = -Q.

Proof

(⇐) If \alpha_1 = \alpha_2, \beta_1 = -\beta_2 then setting f = X - \alpha_1 gives

A/(f) \cong \mathbb C[X, Y]/(Y^2 - X^3 + X, X - \alpha_1) \cong \mathbb C[Y]/(Y^2 - (\overbrace{\alpha_1^3 - \alpha_1}^{\beta_1^2})).

If \beta_1 \ne 0, this ring has exactly two maximal ideals, corresponding to maximal ideals \mathfrak m_P and \mathfrak m_Q of A. If \beta_1 = 0, it has exactly one maximal ideal so we still have (f) = \mathfrak m_P^2 = \mathfrak m_P \mathfrak m_Q.

(⇒) If \mathfrak m_P \mathfrak m_Q = (f) is principal, then \dim_{\mathbb C} A/(f) = 2 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 P,Q\in E satisfy [\mathfrak m_P] = [\mathfrak m_Q], then P=Q.

Proof

Let R = -P as in corollary 2. Then [\mathfrak m_P][\mathfrak m_{R}] = 1. By the given condition [\mathfrak m_Q][\mathfrak m_R] = 1 so by corollary 2 again we have R = -Q and hence P = Q. ♦

Lemma 2.

For any P, Q\in E with P\ne -Q, there is a unique R\in E such that

[\mathfrak m_P]\cdot [\mathfrak m_Q]\cdot [\mathfrak m_R] = 1.

Proof

First suppose P\ne \pm Q so P and Q have different x-coordinates. Let f = Y - cX - d be the equation of PQ. We get:

A/(f) \cong \mathbb C[X,Y]/(Y^2 - X^3 + X, Y - cX - d) \cong \mathbb C[X]/((cX+d)^2 - X^3 + X),

which has complex dimension 3. Since (f) is divisible by \mathfrak m_P \mathfrak m_Q we have (f) = \mathfrak m_P \mathfrak m_Q \mathfrak m_R for some R\in E. Geometrically R is the third point of intersection of PQ with E, which can be equal to P or Q.

If P = Q = (\alpha, \beta) with \beta \ne 0, we can similarly pick a line through P of gradient c = \frac{3\alpha^2 - 1}{2\beta}. Then as above A/(f) \cong \mathbb C[X]/((cX+d)^2 - X^3 + X) where (cX + d)^2 - X^3 + X has a double root for X = \alpha (this requires some algebraic computation). Hence (f) = \mathfrak m_P^2 \mathfrak m_R for some R\in E. ♦

Summary.

The Picard group of A is given by

\{ [\mathfrak m] : \mathfrak m \subset A \text{ maximal} \} \cup \{1\}.

In particular it is infinite.

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2 Responses to Commutative Algebra 47

  1. Vanya says:

    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.

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