# 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:

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.

# 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. ]

# 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.

• limsup says:

Hmm, the notation is potentially confusing. Here $\dim_{\mathbb C}$ refers to dimension as a complex vector space. Maybe I’ll add a note to that.