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Home , Equivalence class, Ideal class group

LECTURE NOTES, WEEK 5 MATH 222A, ALGEBRAIC

MARTIN H. WEISSMAN

Abstract. We discuss the concepts from Chapter 4 of Milne’s Notes.

1. The Suppose that O is a . Then O is a PID if and only if O is a UFD. The “” H = H(O) measures the extent to which O is not a PID (equivalently a UFD). In particular, H is the trivial group if and only if O is a PID. As we will see soon, H is a finite . The ideal class group may be defined using fractional ideals; this is unnecessary, however, if we are just interested in the definition. Consider the following two sets: I(O) is the of nonzero ideals of O. P (O) is the set of nonzero principal ideals. Define the following among ideals: if a and b are in I(O), we say that they are equivalent if there exist nonzero principal ideals (x), (y) ∈ P (O), such that (x)a = (y)b. Let H(O) denote the resulting set of equivalence classes of nonzero ideals. Proposition 1.1. The relation above is an equivalence relation. The product op- eration on ideals yields a well-defined commutative law of composition on H(O). With this law of composition, H(O) is a group. Proof. First, we check that the relation is an equivalence relation. It is obviously reflexive and symmetric. If a ∼ b and b ∼ c, then there exist nonzero w, x, y, z ∈ O such that: (w)a = (x)b, (y)b = (z)c. It follows that: (w)(y)a = (y)(x)b = (x)(z)c. Hence (wy)a = (xz)c, so a ∼ c. Next, we prove that the product operation: I(O) × I(O) → I(O) descends to a well-defined operation H(O) × H(O) → H(O). So suppose that a ∼ b and a0 ∼ b0. Then, there exist x, y, x0, y0 such that: xa = yb, and x0a0 = y0b0. It follows that: xx0aa0 = yy0bb0. Hence the of the product depends only on the equivalence class of the factors. Finally, since (1) is an for multiplication of ideals in I(O), we see that the principal class is an identity element for multiplication in H(O). The law of composition in H(O) is commutative and associative, since these properties 1 2 MARTIN H. WEISSMAN descend from the corresponding properties of multiplication of ideals in I(O). Thus, we are left to check that every ideal has an inverse. Suppose that a is an ideal in I(O). Suppose that a ∈ a is any nonzero element. Then a ⊃ (a), so we have unique factorizations of ideals into primes (following Corollary 3.18 of Milne):

r1 rm s1 sm (a) = p1 ····· pm , and a = p1 ····· pm .

The containment a ⊃ (a) implies that si ≤ ri for all 1 ≤ i ≤ m. Define:

∗ r1−s1 rm−sm a = p1 ····· pm . Then we see that aa∗ = (a). Hence a and a∗ are inverses, principal equivalence. ¤ There is another way that “class groups” arise in number theory – they originally arose in Gauss’s work on binary quadratic forms. The class groups of Gauss are closely related (and sometimes equal) to the class groups of O ⊂ K when K is a quadratic extension of Q. Gauss was interested in binary quadratic forms over Z: these are functions f : Z2 → Z, of the form: f(x, y) = ax2 + bxy + cy2, where a, b, c ∈ Z are fixed. We assume that GCD(a, b, c) = 1 as well. In particular, one is interested in the following question: does the f(x, y) = m (for some fixed f and m ∈ Z) have a solution? In particular, one would like to find a “primitive solution”: Gauss says that f properly represents m, if there exist x, y ∈ Z, with GCD(x, y) = 1, such that f(x, y) = m. One may change the basis of Z2, without changing the Diophantine nature of the equation f(x, y) = m. Namely, if g ∈ GL2(Z), define: gf(x, y) = f((x, y) · g), with the above denoting matrix multiplication. We say that two quadratic forms g f1, f2 are equivalent if there exists g ∈ GL2(Z) such that f1 = f2. Gauss also considers a stronger equivalence relation: f1, f2 are properly equivalent if there g exists g ∈ SL2(Z) (determinant 1, not just ±1) such that f1 = f2. Note that if f1 and f2 are equivalent, then the Diophantine equations f1(x, y) = m and f2(x, y) = m have the same number of solutions. If f(x, y) = ax2 +bxy +cy2, we let ∆(f) = b2 −4ac – it is called the of f. It is elementary to show that if f1 and f2 are equivalent, then ∆(f1) = ∆(f2). The natural question is the following: if ∆(f1) = ∆(f2), then are f1 and f2 (prop- erly) equivalent? Or less naive is the following: How many (proper) equivalence classes of quadratic forms are there, with a given discriminant ∆? The connection between equivalence classes of (positive definite) quadratic forms (the problem considered by Gauss), and ideal class groups (in imaginary quadratic rings) is given by the following, which is adapted from Theorem 7.7 in Cox’s “Primes of the form x2 + ny2. 2 2 (1) If f(x, y) = ax + bxy + cy is a primitive (GCD√(a, b, c) = 1) of discriminant ∆ <√0, then af = (a, (−b + ∆)/2) is an ideal in O, the of in Q( ∆). (2) The map sending f to af induces a bijection from the set of proper equiva- lence classes of quadratic forms of discriminant ∆ to the set of ideal classes LECTURE NOTES, WEEK 5 MATH 222A, 3

H(O). In particular, when ∆ = −4n, the quadratic form x2 +ny2 yields the principal class (1). Also, if f = ax2 + bxy + cy2, and f ∗ = ax2 − bxy + cy2, then af · af ∗ is a . (3) The Diophantine equation f(x, y) = m has a solution, if and only if m = N(a) = [O : a], for some ideal a of O, in the class of af . From the second item, we see that the group structure on H(O) transfers to a group structure on proper equivalence classes of quadratic forms of a given (nega- tive) discriminant. Such a composition law was originally considered by Gauss; a new construction of this composition law has recently been given by Manjul Bhar- gava - Bhargava’s construction has led to many interesting generalizations (read the recent papers in Annals of Mathematics!)√ In particular, we see that since O = Z[ −5] is not a PID, we have H(O) 6= {1}. Hence, there should exist two primitive quadratic forms of discriminant −20, which 2 2 are not equivalent. Two such forms are: f1 = x + 5y (corresponding to the 2 2 principal class), and f2 = 2x + 2xy + 3y .

2. Finiteness In Chapter 4, Milne proves the following fundamental theorem: Theorem 2.1. For any number field K, with O, the ideal class group H(O) is finite. The proof is essentially geometric. The basic geometric connection is established by the following: Proposition 2.2. Suppose that a is an ideal in O. Define a real vector space by: ∼ V = K ⊗Q R = O ⊗Z R. Let σ denote the inclusion of K in V , sending k ∈ K to σ(k) = k ⊗ 1. Then a is a (full-rank) lattice in V . In other words a is a Z-module of rank n = dim(V ), and a spans V as a vector space. Proof. As an abelian group, any ideal a is isomorphic to Zn, with r = [K : Q]. Indeed, a is torsion-free as a Z-module, and finitely generated since O is Noetherian. Thus a is free of finite-rank. The rank may thus be computed locally, at any . Locally, ap is a principal ideal in the DVR Op, and hence the rank is equal to the rank of Op as a Zp∩Z-module, which equals n. Thus, the inclusion a in the vector space V , corresponds to an inclusion Zr ,→ Rn. To show that a is a full-rank lattice, it suffices to show that it spans Rn. Milne proves this using the discriminant; we illustrate the proof directly in the real and imaginary quadratic cases. The important point is to describe the embedding σ : K → V with respect to a carefully chosen basis of V . Specifically, we have K ⊗ R =∼ Rr × Cs, for some non-negative integers r and s. Thus, we have field embeddings σ1, . . . , σr+s, of K into R or C. √ 2 First, suppose√ that a is an ideal√ in O ⊂ Q( D), with D > 0. Then V = R . Let a1 = α1 + β1 D, a2 = α2 + β2 D be a basis of a. The inclusion of ai in V is given by: √ √ σ(ai) = (αi + βi D, αi − βi D). 4 MARTIN H. WEISSMAN

The determinant of the following matrix determines whether (a1, a2) spans V : µ √ √ ¶ α + β D α − β D det 1 1√ 1 1√ . α2 + β2 D α2 − β2 D This determinant can be computed: √ det = α1α2 − β1β2D − α1α2 + β1β2D + (β1α2 − α1β2 + β1α2 − α1β2) D. This equals: √ det = 2(β1α2 − α1β2) D. This equals zero if and only if: µ ¶ α β det 1 1 = 0. α2 β2

But this contradicts the assumption that a1, a2 are a basis of the rank 2 Z-module a. Secondly, if we suppose that D < 0, then V = C. With ai as before, the inclusion of ai in C looks like: √ zi = σ(ai) = αi + βi D.

A pair z1, z2 of complex numbers spans C as a real vector space, if and only if: µ ¶ z z det 1 2 6= 0. z¯1 z¯2 The proof in the real quadratic case now adapts without difficulty. ¤ Now, we may view any ideal a ∈ I(O) as a full-rank lattice in V =∼ Rr ⊕ Cs =∼ Rn. Given any full-rank lattice Λ ⊂ V , there exists a “fundamenatal domain” (parallelopiped) P for Λ. For example, if λ1, . . . , λn is a Z-basis of Λ, we may take:

P = {t1λ1 + ··· + tnλn such that 0 ≤ λi ≤ 1}. p Define V ol(V/Λ) = V ol(P ). Then V ol(V/O) = |∆K |, where ∆K is the discrimi- nant of the number field K/Q. If a is an ideal in O, and N(a) = [O : a] (its index as a subgroup), then we have: p V ol(V/a) = V ol(V/O)[O : a] = |∆K |N(a). In to prove the finiteness of the class group, we use the following algorithm: (1) Begin with an ideal a of O. (2) By a geometric result of Minkowski, there exists an element α ∈ a, and a constant M depending only on K, such that: |Nm(α)| ≤ MN(a). (3) Since (α) ⊂ a, we have (α) = ab for some ideal b (equivalent to a−1 in H(O)). Properties of the norm show that: N(a)N(b) = N((α)) = |Nm(α)| ≤ MN(a). Hence N(b) ≤ M. (4) Replacing a by its inverse, every ideal a is equivalent to one whose norm is bounded by M. (5) The set of ideals whose norm is bounded by M is finite, correspondingp to elements of the finite set of sublattices of O of index at most M/ |∆K | LECTURE NOTES, WEEK 5 MATH 222A, ALGEBRAIC NUMBER THEORY 5

361B Baskin, Department of Mathematics, University of California, Santa Cruz, CA 95064 E-mail address: [email protected]