Tools
3.6 Stickelberger’s Theorem
3.6.6 The Eisenstein Reciprocity Law
In this subsection I follow quite closely the exposition of [Ire-Ros]. We restate the theorem on the Stickelberger ideal, but now emphasizing the dependence onp. Thus, instead of lettingpbe a prime,q=pf, andma divisor ofq−1, we fix an integerm2 such thatm≡2 (mod 4), we let Θ be as in Proposition 3.6.13, and we setγ=mΘ, in other words,
γ=
1tm−1,gcd(t,m)=1
tσ−1t .
For all primespnot dividingmwe letf be the order ofpmodulom, we set q=pf ≡1 (modm), so thatm|(q−1), and as above we setd= (q−1)/m.
Important Warning. Since m is fixed, we denote simply by p instead of pm a prime ideal ofKm =Q(ζm) abovep. Note that previously,p denoted the unique prime ideal ofK=Q(ζp) abovep, but here the context is different and pis not given but will vary, onlym is fixed. Note also that sincef is the order ofpmodulom, by Proposition 3.5.18 we havef(p/p) =f, hence N(p) =pf =q.
For any prime idealPof L=Q(ζq−1, ζp) above p (hence abovep) recall that we have definedωPandτ(ω−Pd). In addition, in the sequel we will make the following common abuse of terminology: we will say that two (possibly fractional) idealsaandb(or elements) are coprime if for every prime idealq we have eithervq(a) = 0 orvq(b) = 0.
Definition 3.6.23. Forx∈Km coprime to p we set x
p
m
=ωdP(x) and G(p) =τ(ωP−d)m.
In addition, to simplify notation we will often writeχp(x)instead ofx
p
−1 m , so thatG(p) =τ(χp)m.
This notation is justified by Lemma 3.6.3 (3) and (4), which tells us that ωP(x)d andτ(ωP−d)m∈Km depend only onp=P∩Km. The above defini- tion generalizes the well-known quadratic reciprocity symbolx
p
studied in Section 2.2, hence is naturally called themth-power reciprocity symbol. The following series of definitions and formulas is completely analogous to what is done in the classical case of quadratic reciprocity.
Lemma 3.6.24. (1) x
p
m is characterized by the fact that it is a character of orderm such that
x p
m
≡x(q−1)/m (modp). (2) We have
τ(χp) =
t∈(ZK m/p)∗
χp(t)ψ1(t),
where as usualψ1(t) =ζpTr(ZK m /p)/(Z/ pZ)(t).
Proof. (1) is the translation of the corresponding properties of the char- acterωP, and (2) from the fact that the natural inclusion map from Km to Linduces a canonical isomorphism betweenZKm/p andZL/P.
Definition 3.6.25. Leta be an integral ideal ofKmcoprime tom, andxbe coprime toa. We definex
a
m andG(a)by the formulas x
a
m=
p|a
x p
vp(a) m
and G(a) =
p|a
G(p)vp(a).
If a = αZKm is a principal ideal, we will write x
α
m and G(α) instead of x
αZK m
m andG(αZKm).
Thus by definition we havex
ab
m=x
a
m
x
b
m andG(ab) =G(a)G(b).
Proposition 3.6.26. We have:
(1) |G(a)|2=N(a)m.
(2) G(a)ZKm =aγ, whereγ=mΘis as above.
(3) If α∈ZKm there exists a unitε(α)ofZKm such that G(α) =ε(α)αγ. Proof.Ifpis a prime, by Proposition 2.5.9 we have|G(p)|2=|τ(ωP−d)|2m= qm = N(p)m, so (1) follows by multiplicativity. Statement (2) for a prime ideal is exactly Proposition 3.6.13, and the general result follows by multi- plicativity. By (2) we have G(α)ZKm =αγZKm, hence G(α) = ε(α)αγ for
some unitε(α).
We now want to show thatε(α) is a root of unity. For this we need two lemmas.
Lemma 3.6.27. For any integral idealaprime tomand anyσ∈Gal(Km/Q) we haveG(a)σ=G(aσ).
Proof. By definition, if p is a prime ideal coprime tom we have G(p) = τ(ω−dP )m. Thus by Lemma 3.6.3 we haveG(p)σ=τ(ωσ(P)−d )m=G(σ(p)) since σ(p) is the prime ideal ofKm below σ(P). As usual, the lemma follows by
multiplicativity.
Lemma 3.6.28. Ifα∈ZKm then |αγ|2=N(α)m.
Proof. Sinceσ−1sendsζmtoζm−1 it is complex conjugation. Thus
|αγ|2=αγσ−1(αγ) =α(1+σ−1)γ . Denoting by∗
t a sum for 1tm−1 such that gcd(t, m) = 1 we have (1 +σ−1)γ=*
t
tσ−1t +*
t
tσ−1−t =*
t
tσt−1+*
t
(m−t)σ−1t
=m*
t
σt−1=m
σ∈Gal(Km/Q)
σ .
It follows that
α(1+σ−1)γ =
σ∈Gal(Km/Q)
σ(α)m=NKm/Q(α)m.
Proposition 3.6.29. The element ε(α) is a root of unity. In other words, there existi∈Zand a sign ±such thatG(α) =±ζmi αγ.
Proof. By Proposition 3.6.26 (1) and (3) we have
|G(α)|2=| N(α)|m=|ε(α)|2|αγ|2=|ε(α)|2N(α)m,
hence|ε(α)| = 1 (note that N(α) is automatically positive). On the other hand, applying Lemma 3.6.27 toa=αZKm and using Proposition 3.6.26 (3) we deduce that for allσ∈Gal(Km/Q),
ε(α)σαγσ=G(α)σ=G(ασ) =ε(ασ)αγσ,
so thatε(α)σ=ε(ασ). Since we have shown that|ε(α)|= 1 for allα∈ZKm, and in particular forασ, it follows that|ε(α)σ|= 1 for allσ. We conclude by Kronecker’s theorem (Corollary 3.3.10) thatε(α) is a root of unity, and the
proposition follows from Corollary 3.5.12.
We can now proceed to the statement and proof of Eisenstein’s reciprocity law. As the reader will notice, the proof of the following proposition is essen- tially identical to that of the classical proof of the quadratic reciprocity law (see the proof of Lemma 2.2.2).
Proposition 3.6.30. Letp1andp2be two distinct prime numbers not divid- ingmand letp1 andp2 be prime ideals of Km above p1 andp2 respectively.
Then
G(p1) p2
m
=
N(p2) p1
m
.
Proof. Let f1 and f2 respectively be the orders ofp1 and p2 modulom, so that by Proposition 3.5.18 we havef(pi/p) =fi. Setqi=N(pi) =pfii ≡1 (modm) and recall the notation of Definition 3.6.23. Sinceχp1(t) is anmth root of unity andm|(q2−1), by Lemma 3.6.24 we have
τ(χp1)q2 ≡
t∈(ZK m/p1)∗
χp1(t)q2ψ1(t)q2
≡
t∈(ZK m/p1)∗
χp1(t)ψ1(q2t)≡χp1(q2)−1τ(χp1) (modp2ZKm).
On the other hand, again by Lemma 3.6.24 we have τ(χp1)q2−1=G(p1)(q2−1)/m≡
G(p1) p2
m
≡χp2(G(p1))−1 (modp2). Since|τ(χp1)|2 =q1 is coprime top2, it follows from these two congruences that
χp2(G(p1))−1≡χp1(q2)−1≡χp1(N(p2))−1(mod p2).
Since both sides aremth roots of unity and sincep2is coprime tom, it follows
that they are equal, proving the proposition.
Corollary 3.6.31. For all ideals aandbcoprime to msuch that N(a)and N(b)are coprime we haveG(a)
b
m=N(b)
a
m. Proof. Clear since by definition·
a
m is a character and is multiplicative ina, andG(a) andN(a) are also multiplicative.
Corollary 3.6.32. In addition to the assumptions of the preceding corollary, assume thata=αZKm is a principal ideal. Then
N(b) α
m
= ε(α)
b
m
α N(b)
m
.
Proof. By multiplicativity we have G(α)
b
m
= ε(α)
b
m
αγ b
m
= ε(α)
b
m1tm−1,gcd(t,m)=1
A(t),
where
A(t) =
σt−1(α) b
t
=σt
σ−1t (α) b
m
= α
σt(b)
m
by Lemma 3.6.3. The result follows sinceN(b) =
1tm−1,gcd(t,m)=1σt(b).
From now on we will assume that m=is a prime number, we will let R=Z[ζ] be the ring of integers ofQ(ζ), and we denote byL= (1−ζ)R the unique prime ideal ofRabove .
Lemma 3.6.33. Ifa is an ideal coprime to thenG(a)≡ ±1 (modR).
Proof. Ifpis a prime ideal coprime to we have as above G(p) =τ(χp)≡τ(χp, ψ)≡
t∈(Z/Z)∗
ψ1(t)≡ −1 (modR)
since ψ1 is a nontrivial additive character, so the lemma follows by multi-
plicativity.
Definition 3.6.34. Let α∈R. We say thatαis primary if it is not a unit and if there existsx∈Zwith xsuch that α≡x(mod L2).
Lemma 3.6.35. If α∈R is coprime tothere exists i∈Z, unique modulo , such thatζiαis primary.
Proof. Since f(L/) = 1 we have R/L Z/Z, so there exists a ∈ Z, evidently unique modulo, such thatα≡a(mod L). Thusβ= (α−a)/(1− ζ) ∈ R, so in the same way there exists b ∈ Z, unique modulo , such that β ≡ b (mod L). It follows that α ≡ a+b(1−ζ) (modL2). By the binomial theorem we haveζi= (1−(1−ζ))i≡1−i(1−ζ) (modL2), hence ζiα≡a+ (b−ai)(1−ζ) (modL2), so we choosei=ba−1mod, which is possible sincea, otherwiseαwould not be coprime to . Lemma 3.6.36. Ifαis primary we haveε(α) =±1.
Proof.SinceLis the unique prime ideal above, for allσ∈Gal(Q(ζ)/Q) we haveσ(L) =L. By definition ofγ it follows thatLγ⊂ L. Thus
αγ ≡xγ≡x1t−1t≡x(−1)/2(mod L2).
By Fermat’s theorem we have x(−1)/2 ≡ ±1 (mod), hence αγ ≡ ±1 (modL2). On the other hand, by the preceding lemma we have G(α) = ε(α)αγ ≡ ±1 (modR), hence ε(α) ≡ ±1 (modL2). Now by Proposition 3.6.29 we have ε(α) = ±ζi for some integer i, hence ζi ≡ ±1 (mod L2). I claim that|i(this follows in fact from the uniqueness statement of Lemma 3.6.35). Indeed, as we have already seen, by the binomial theorem
ζi= (1−(1−ζ))i≡1−i(1−ζ) (modL2).
Thus the + sign must hold in the congruence (otherwiseL |2); hence i(1− ζ)≡0 (modL2), so that L | i, and hence | i as claimed. It follows that
ε(α) =±1.
Proposition 3.6.37. If α is primary and b is an ideal coprime to such thatN(b)is coprime to αRthen
α N(b)
=
N(b) α
.
Proof. Since α is primary the above lemma tells us that ε(α) = ±1.
On the other hand, for all p coprime to , χp is a character of order , we have χp(±1) = 1 and χp(±1)2 = χp(1) = 1, and since is odd it follows by multiplicativity that for all b prime to we have χb(±1) = 1, and in particularχb(ε(α)) = 1. The result thus follows from Corollary 3.6.32.
The Eisenstein reciprocity law immediately follows:
Theorem 3.6.38 (Eisenstein). Let be an odd prime, a ∈ Z an integer not divisible by,α∈Ra primary element, and assume thatαRandaRare
coprime. Then α
a
= a
α
.
Proof.As usual, by multiplicativity it is enough to prove this whena=p is a prime number different from and prime toαR. Letp be a prime ideal ofRabovep,f =f(p/p), so thatN(p) =pf. By the above proposition with b=p we haveχp(α)f =χα(p)f. Since f |(−1) = [Q(ζ)/Q],f is coprime to , and since bothχ-values are th roots of unity it follows that they are
equal.