Kummer Theory and Reciprocity Laws

Peter Stevenhagen

Abstract. Insert abstract here.

1. Introduction How can we find abelian extensions of a number field? For any such field K, we have the cyclotomic extension K ⊂ K(ζm); the Galois group will be abelian and a ∗ subgroup of (Z/m√ Z) . We might also adjoin a square root, but should we adjoin n a general root a, we no longer even have a Galois extension, unless ζn ∈ K already. But in fact, for K = Q, all quadratic extensions of Q are already cyclotomic: for p 6= 2, there is only one cyclotomic extension which is ramified only at p ∗ ∼ and only tamely ramified, namely, Q(ζp). This field has Galois group (Z/pZ√) = Z/(p−1)Z, so it has a unique quadratic subfield, which is easily seen to be Q( p∗), where p∗ = (−1)(p−1)/2p: it is the unique quadratic field ramified only at p. In √ √ √ addition, we have Q( −1, 2) = Q(ζ8), so we find Q( a) ⊂ Q(ζ4|a|). Already this discussion of quadratic subextensions of cyclotomic extensions gives us the classical quadratic reciprocity law. Given distinct odd primes p, q, then there is a Legendre symbol (p/q) which is 1 or −1, respectively, depending on whether or not p is a square modulo q or not, respectively. One has the quadratic reciprocity law where  q  , p or q ≡ 1 (mod 4), p  p = q q − , else.  p We can write this more compactly as p q = (−1)(p−1)(q−1)/4. q p To prove this, consider first the case q ≡ 1 (mod 4). We consider the cyclotomic extension (ζ )/ , with Galois group ( /q )∗. This extension contains the Q q √Q Z Z quadratic extension ( q). Let σ ∈ ( /q )∗ be the Frobenius at p, and note that Q √ √ Z Z (p/q) = 1 if and only if σp( q)/ q = 1. By the property of the Frobenius, we 2 Peter Stevenhagen

have √ √ √ p−1 σp( q)/ q ≡ ( q) (mod p), so this is equal to 1 if and only if q(p−1)/2 ≡ 1 (mod p), which holds if and only if (q/p) = 1 by Euler’s criterion. This proves quadratic reciprocity. The case when √ q ≡ 3 (mod 4) is similar, except now we work in Q( −q), and we end up with an extra factor (−1)(p−1)/2. Seen one way, then, the quadratic reciprocity law is none other than the statement that the quadratic extensions are all contained in cyclotomic extensions, over which we have control.

2. Kummer Theory We would like to generalize the quadratic reciprocity law; to do so, we need to construct abelian extensions of number ﬁelds K. √ n Throughout we assume that ζn ∈ K. Then L = K( a) (the splitting ﬁeld of Xn − a) is abelian over K, for we have a map

Gal(L/K) ,→ hζni √ σ( n a) σ 7→ √ . n a √ Note that this is independent of the choice of n a (they diﬀer by a root of unity, which since they are in K, drops out), and by a little work we see that in fact this is a group homomorphism. Kummer theory gives a certain converse to this statement: If L/K is cyclic of n order n, and ζn ∈ K, then there exists an α ∈ L such that L = K(α) and α ∈ K. This was basically discovered by Lagrange: If G = hσi, then for any x write down the Lagrange resultant

−1 −2 2 1−n n−1 α = x + ζn σ(x) + ζn σ (x) + ··· + ζ σ (x).

We see that σ(α) = ζnα, and if α 6= 0, then such an α will generate the extension. We can always ﬁnd such an α 6= 0; this is a result from Galois theory (known as ‘Artin-Dedekind’ or linear independence of characters). More generally, Kummer theory tells us that if L/K is Galois with group n G which is abelian of√ exponent n (meaning that for all σ ∈ G, σ = idL), and n ∗ n ∗ ζn ∈ K, then L = K( W ) for some subgroup (K ) ⊂ W ⊂ K . Then Kummer theory tells us that (within a ﬁxed algebraic closure) there is a bijection

{W :(K∗)n ⊂ W ⊂ K∗} o / {L ⊃ K abelian, exponent n (inside K)} √ W / K( n W ) (L∗)n ∩ K∗ o L. Kummer Theory and Reciprocity Laws 3

In this case, if W ↔ L, then we have a perfect pairing ∗ n Gal(L/K) × W/(K ) → hζni √ σ( n a) (σ, a) 7→ √ ; n a that is to say, ∼ ∗ n Gal(L/K) = Hom(W/(K ) , hζni).

If L/K is a cyclic extension of degree n but ζn 6∈ K, it can be very hard to describe. However, when one adjoins ζn to K, by√ Kummer theory, the extension n L(ζn) is now Kummer over K(ζn), so L = K(ζn, a) for some a ∈ K(ζn). Note that this extension is not abelian, but it is abelian in two steps, which is usually good enough. √ n L(ζn) = K(ζn, a) ppp ppp ppp ppp L p K(ζn) oo ooo ooo ooo K o

3. Norm Residue Symbol We now look at the local situation. Recall the statement of local class field theory: if F = Kp is a local field, and E ⊃ F is a finite abelian extension, then ∗ ∗ ∼ F /NE/F E −→ Gal(E/F ) and we have a bijection between open subgroups of F ∗ and finite abelian extensions of F . √By Kummer theory, if ζn ∈ F , then E ⊃ F of exponent n arises as E = F ( n W ) where (F ∗)n ⊂ W ⊂ F ∗. But by class field theory, E ⊃ F of exponent n ∗ n ∗ ∗ arises as (F ) ⊂ NE/F E ⊂ F . These two ways of viewing extensions are dual to each other. √ The maximal exponent n extension by Kummer theory is E = F ( n F ∗). This is a finite extension, for in the local case (F ∗)n has finite index in F ∗. Since the bijection in class field theory is inclusion-reversing, NE∗ = (F ∗)n. By the perfect pairing, we have ∗ ∗ n ∗ ∗ n F /(F ) × F /(F ) → hζni √ σ ( n β) (α, β) 7→ α√ n β

This is called the nth power norm residue symbol, and we denote it F ,(−, −)n,F . It has two useful properties: 4 Peter Stevenhagen

√ n ∗ • From local class field theory, (α, β)n,F = 1 if and only if α ∈ N(F ( β) ). ∗ ∗ ∼ • Since AF /NE/F AE −→ I, the inertia group, if the extension E/F is un- ∗ ramified and α ∈ AF , then (α, β)n,F = 1. ∗ ∗ Example 3.1. Take the case F = Qp, n = 2, p 6= 2. Then Qp = hpi × Zp, so ∗ 2 2 ∗ 2 ∗ 2 ∗ (Qp) = hp i × (Zp) , and so (Qp) has index 4 in Qp, where ∗ ∗ 2 ∼ ∼ Qp/(Qp) = hpi × hai = Z/2Z × Z/2Z, for a ∈ with (a/p) = −1. Z √ √ We have (x, y)2,Qp ) = σx( y)/ y. In the unramified case, y = a, we see that (a, a) = 1, and since πF 7→ FrobE/F , we see (p, a) = −1. In the ramified case, y = p the norms must generate a subgroup of index 2, so (a, p) = −1; and finally we √ ∗ √ wish to know if p ∈ N(Qp( p) ): but (p, p)(−1, p) = (−p, p) = 1 (N( p) = −p), (p−1)/2 so (p, p) = (−1, p) = (−1) . We summarize the values of (x, y)2,Qp in the following table: x \ y a −p a 1 −1 p −1 (−1)(p−1)/2 The symbol has other important properties: • (−α, α) = 1 for all α ∈ F ∗; • (1 − α, α) = 1 for α ∈ F ∗ \{1}; • (α, β) = (β, α)−1. The latter property, for example, follows from (−αβ, αβ) = (−α, α)(β, α)(α, β)(−β, β) = (β, α)(α, β) = 1. Note that we also have this pairing for archimedean F . For F = C this is trivial, for F = R we have the Artin isomorphism

∗ ∗ Gal(C/R) R /NC −−−−−−→ and a corresponding pairing ∗ ∗2 ∗ ∗2 R /R × R /R → h−1i

with (−1, −1)∞ = −1. Now, for a number ﬁeld K such that ζn ∈ K, for any prime p of K, there is a norm residue symbol ∗ ∗ n ∗ ∗ n (−, −)n,p : Kp /(Kp ) × Kp /(Kp ) → hζni. We then have: Proposition 3.2 (Product formula). For α, β ∈ K∗, Y (α, β)n,p = 1. p≤∞ Kummer Theory and Reciprocity Laws 5

Example 3.3. To complete the picture, we compute the quadratic symbols for Q∞ = ∗ ∗ 2 R and Q2. In the ﬁrst case, R /(R ) = h−1i, and (−1, −1)2,∞ = −1. For p = 2, ∗ ∗ ∗ ∗ 2 2 ∗ 2 ∗ 2 Q2 = h2i × Z2, and Q2/(Q2) = h2 i × (Z2) , now (Z2) = 1 + 8Z2, so we see ∗ ∗ 2 ∼ Q2/(Q2) = h2i × h−1i × h5i.

We summarize the values of (x, y)2,2 in the following table:

x \ y 2 −1 5 2 1 1 −1 −1 1 −1 1 5 −1 1 1

Proof of the product formula. Consider the following diagram:

0 Y ∗ M pn Kp / Gal(Kp( β)/Kp) p≤∞ p≤∞

√ J / Gal(K( n β)/K)

The left vertical map is the definition of the idèles. The top map is α 7→ ((α, β)n,p)p, which one sees is trivial for almost all p. The horizontal bottom arrow is the Artin Q map. The right vertical map is (σp)p 7→ p σp. ∗ The statement is then that for α ∈ K ⊂ J,√ the global Artin map (the ∗ n bottom arrow) factors as J → J/K = CK → Gal(K( β)/K); this is a nontrivial statement from class field theory.

Example 3.4. For the case n = 2, K = Q, p, q distinct odd primes, writing (p, q)2,` = (p, q)`, we have Y (p, q)` = (p, q)2(p, q)p(p, q)q(p, q)∞ = 1. `≤∞

Since p, q > 0, (p, q)∞ = 1. We see from the computation in the above table that (p−1)(q−1)/4 (p, q)2 = 1 unless p, q ≡ 3 (mod 4), i.e. (p, q)2 = (−1) . Also, we see √ qp q (p, q) = √ = q(p−1)/2 ≡ (mod p), p q p

p and similarly (q, p)q ≡ q (mod q). Altogether, q p (−1)(p−1)(q−1)/4 = 1 p q which is the statement of quadratic reciprocity. 6 Peter Stevenhagen

4. Higher Reciprocity Laws Are there higher reciprocity laws? In the nineteenth century, such a question was asked. But what should this mean? We begin with the cubic reciprocity law: for which p is 2 a cube modulo p? ∗ ∗ 3 ∗ ∗ 3 For Fp ⊃ (Fp) , if p ≡ 2 (mod 3) we see that already Fp = (Fp) , and this question is not interesting, for 2 is always√ a cube. We assume therefore that p ≡ 1 (mod 3). Then p splits as p = ππ in Q( −3). In this case, 2 is a cube modulo 3 p if and only if X − 2 splits completely√ modulo p, which happens if and only if 3 p splits completely in L = Q(ζ3, 2). This is a Galois extension of degree 6 with Galois group S3. √ 3 L = Q(ζ3, 2) N ooo NNN ooo NN ooo NNN oo N √ o 3 K = Q(ζ3) Q( 2) PPP p PPP ppp PPP ppp PPP pp PP ppp Q √ 3 But now the extension L/K = Q(ζn, 2)/Q(ζ3) is Kummer, so we can get a handle on it. In fact, one can show that this is the ray class ﬁeld of Q(ζ3) of conductor 6, and it follows from this that p splits completely in L if and only if p = x2 + 27y2. For an nth power reciprocity law, we assume that ζn ∈ K. What is the analogue of the Legendre symbol for nth powers? Given a prime p - n∞, we have ∗ ∗ the unit group (O/p) , which has order divisible by n since hζni ,→ (O/p) (the discriminant of the polynomial Xn − 1 is prime to p). Therefore we deﬁne: − ∗ :(O/p) → hζni p n α ≡ α(Np−1)/n (mod p). p n For the primes in

S(α) = {p : p | n∞ or ordp(α) 6= 0}

we do not deﬁne the symbol. We extend the symbol to any fractional OK -ideal b of K by multiplicativity, ord (b) α Y α p = . b n p p6∈S(α) This symbol is then multiplicative in b on the group I(α) generated by p 6∈ S(α). For any element β ∈ K∗, we also deﬁne α α = . β n (β) n Kummer Theory and Reciprocity Laws 7

∗ Lemma 4.1. For α, β ∈ K , with ζn ∈ K, we have −1 α β Y = (α, β) . β α p n n p∈S(α)∩S(β) In particular, if α and β are coprime, then

−1 α β Y = (α, β) . β α p n n p|n∞

To prove this, we use the following fact:

Lemma 4.2. If b ∈ I(α), then √ α σ ( n α) = b√ , b n n α √ n where σb ∈ Gal(K( α)/K) is the Artin symbol of b. √ n Proof. Suppose b = p ∈ I(α). Now α is a unit at p, so Kp( α) is an unramiﬁed extension of Kp. The Artin symbol of p in this case is √ n Np ( α) (Np−1)/n σpα = √ = α (mod p) n α which is exactly the deﬁnition of the power residue symbol.

Proof of power reciprocity. By the lemma, α Y = (β, α) . β n,p n p6∈S(α) Similarly, β Y Y = (α, β) = (α, β)−1 α n,p n,frakp n p6∈S(β) p∈S(β) by the product formula. For p 6∈ S(α) ∪ S(β), we clearly have (α, β)n,p = 1. For the quotient α β −1

β n α n we therefore obtain the quantity Y Y (β, α)n,p (α, β)n,p. p∈S(β)\S(α) p∈S(β)

Using (β, α)(α, β)−1 = 1 at the primes p ∈ S(β) \ S(α), the result follows. 8 Peter Stevenhagen

The expression for power reciprocity may look complicated, but F ∗/(F ∗)n is a finite abelian group, so the norm residue symbol relies on only a finite amount of information. For example, we have the supplementary law for a odd and positive:  2 2 1, a ≡ ±1 (mod 8) = (−1)(a −1)/8 = a −1, a ≡ ±3 (mod 8). We may now prove cubic reciprocity, a result originally due to Eisenstein. Suppose that α, β ∈ Z[ζ3] are coprime and coprime to 3. Since the ideal (3) ramifies 2 in Q(ζ3) as (3) = (1 − ζ3) , we have a map 2 Z[ζ3] → Z[ζ3]/3Z[ζ3] = Z[ζ3]/(1 − ζ3) Z.

This latter group is of order 6, so the map takes hζ6i = h−ζ3i injectively into the k quotient. Replacing α by ζ6 α we may assume α ≡ 1 (mod 3), and similarly with β. In this case, we actually have α β = . β 3 α 3 This follows from the power reciprocity law: α β −1 = (α, β)3,(1−ζ3); β 3 α 3 as this ﬁeld is totally complex, there are no inﬁnite primes. Evaluating this entry, we see that it is equal to 1. Example 4.1. As another example, we can take quintic reciprocity. Take α, β ∈ 4 Z[ζ5] coprime to each other and coprime to 5. We have (5) = (1 − ζ5) , and α β −1 = (α, β) . β α 5,(1−ζ5) Multiplying α, β by units, for which we have √ ∗ Z[ζ5] = h−ζ5i × h(1 + 5)/2i, 3 we can obtain α, β ≡ 1 (mod (1 − ζ5) ). Then (α, β)5,1−ζ5 = 1, and α β = . β 5 α 5

Exercises

Exercise 3.1. For odd primes p and q, we showed how to deduce the quadratic reciprocity law p p− 1 q− 1 q = (−1) 2 2 q p Kummer Theory and Reciprocity Laws 9

r −1 from the inclusion Q( q q) ⊂ Q(ζq) and properties of the Artin symbol. Give a similar proof for the supplementary law  2 1 p ≡ ±1 mod 8, = p −1 p ≡ ±3 mod 8,

where p denotes an odd prime. √ In the next two exercises, let K = Q( −3) be the number√ field generated by √ 3 the cube root of unity√ ζ3 = (−1 + −3)/2, and L = K( 2) the extension of K generated by a root 3 2 of X3 − 2. The unique primes of K lying over 2 and 3 are denoted by 2 and t, respectively. Exercise 3.2. (a) Prove that K ⊂ L is√ cyclic√ of degree 3, and that : Gal(L/K) → hζ3i defined by (σ) = σ( 3 2)/ 3 2 is a group isomorphism. (b) Show that the conductor F(L/K) divides 2t4. (c) Let p be a finite prime of K not dividing 2t, and let Np be the cardinality of its residue class field. Prove that ((p, L/K)) is the unique element of (Np−1)/3 hζ3i that is congruent to 2 modulo p. (d) Show that L is the ray class field of K with modulus 6 (= 2 · t2).

Exercise 3.3. Let p be a prime number, and let m be the number of distinct roots 3 of X − 2 in Fp. Prove the following: (a) m = 0 if and only if p ≡ 1 mod 3 and p is not represented by X2 + 27Y 2. (b) m = 1 if and only if p 6≡ 1 mod 3. (c) m 6= 2. (d) m = 3 if and only if p is represented by X2 + 27Y 2.

Exercise 3.4. Let K be a number field, and r the number of real primes of K. Denote by Cl the class group of K, and by Cl∗ the strict (or narrow) class group Q of K; i.e., the ray class group modulo the cycle f = p real p. The 2-rank rk2 A of a multiplicatively written abelian group A is defined to be the dimension of the 2 F2-vector space A/A . ∗ (a) Prove that rk2 Cl ≤ rk2√Cl ≤ rk2 Cl +r − 1 if r > 0. (b) Let H = {x ∈ K∗ : K( x) is unramified over K at all finite primes}, and let H+ be the set of elements of H that are positive at all real primes of ∗ ∗2 K. Prove that rk2 Cl equals the dimension of the F2-vector space H/K , + ∗2 and that rk2 Cl equals the dimension of the F2-vector space H /K . Exercise 3.5. Let the notation be as in the previous exercise.

(a) Let a, b ∈ H. Prove that the norm residue symbol (a, b)2, equals 1 for Q Q every ﬁnite prime Q of K, and that p real(a, b)2,p = 1. ∗ (b) Prove that rk2 Cl ≤ rk2 Cl +br/2c. 10 Peter Stevenhagen

Mathematisch Instituut, Universiteit Leiden, Postbus 9512, 2300 RA Leiden, The Netherlands E-mail address: [email protected]