SMALL WILD SETS OF SELF-EQUIVALENCES OF NUMBER FIELDS
MARIUS SOMODI
We prove that ifP is a non-dyadic prime ideal in an arbitrary number fieldK such that−1 is a local square at P, then either{P}or {P, P∗}, whereP∗ is a some prime ideal inK, is the wild set of a self-equivalence ofK.
AMS 2010 Subject Classification: 11E81, 11R99.
Key words: Hilbert-symbol equivalence, wild prime.
1. INTRODUCTION
The concept ofHilbert-symbol equivalencewas introduced in [5] to charac- terize the situation when two number fields have isomorphic Witt rings of qua- dratic forms. Recall that by a Hilbert-symbol equivalence between two number fields Kand Lwe understand a pair of mappings (T, t), whereT is a bijection between the sets of primes of the two number fields andt:K∗/K∗2→L∗/L∗2 is an isomorphism between the square-class groups of the two number fields that preserves the Hilbert symbols:
(t(a), t(b))T(P)= (a, b)P, ∀a, b∈K∗/K∗2.
By a self-equivalence of a number field K we understand a Hilbert-symbol equivalence between K and K. Any number field has at least one self-equiva- lence (with T and tthe identity mappings).
In [5], Perlis et al. prove that number fields K and L have isomorphic Witt rings of quadratic forms if and only if there is a Hilbert-symbol equiva- lence between K and L. Moreover, for any Hilbert-symbol equivalence (T, t) betweenK andL,T maps complex primes inK to complex primes inL, real primes in K to real primes inL, and dyadic primes in K to dyadic primes in L. In particular, the two number fields must have the same degree overQand the same number of dyadic primes. In addition,t(−1) =−1.
Carpenter shows in [1] that there is a Hilbert-symbol equivalence between two number fields K and L if and only ifK and L have the same number of real embeddings, −1 is a global square in both K and L or in neither, and there is a way of pairing the dyadic primes P of K with the dyadic primes
REV. ROUMAINE MATH. PURES APPL., 56(2011),1, 57–67
P of Lsuch that the corresponding dyadic completions have the same degree over the field Q2 and −1 is a local square in both KP and LP or in neither.
If (T, t) is a Hilbert-symbol equivalence between K and L, then with respect to this equivalence a non-archimedian prime P of K is called tameif
ordT(P)(t(a))≡ordP(a) (mod 2), ∀a∈K∗/K∗2.
Note that a non-archimedian non-dyadic prime in K is tame with respect to (T, t) if and only if t maps square-classes of units at P to square-classes of units at T(P) and square-classes of primes atP to square-classes of primes at T(P). If a non-archimedian prime P of K is not tame with respect to (T, t), then we call itwild. The set of all wild primes of a Hilbert-symbol equivalence is called the wild set of the equivalence. Carpenter proves in [1] that if there is a Hilbert-symbol equivalence between two number fields, then there is a Hilbert-symbol equivalence between the two fields that has a finite wild set.
A lower bound on the size of any wild set between two number fields can be found in [3] and a formula of the minimum size of any wild set between two number fields can be found in [6].
A necessary and sufficient condition for a non-archimedian primeP inK to be wild with respect to a Hilbert-symbol equivalence between K and some number fieldLis that−1 is a local square atP. It is unknown whether for any non-archimedian prime P in a number field K at which −1 is a local square, there is a Hilbert-symbol equivalence between K and some number field L with wild set W ={P}. We consider the following related question: given a number fieldK and a finite non-dyadic primeP inK such that −1 is a local square at P, is there a self-equivalence ofK with wild set equal to{P}? The question has an affirmative answer for two number fields: the field of rational numbersQ([7]) and the field of Gaussian numbersQ(√
−1) ([8]). The reader can easily check using Carpenter’s conditions that these fields are solitary in their Witt equivalence class, in the sense that the only number fields that are Witt equivalent with them are the fields themselves.
In this paper we show that ifP is a non-archimedian non-dyadic prime in a number field K, such that −1 is a local square at P, then either {P} is the wild set of a self-equivalence of K, or {P, P∗} (where P∗ is some other non-dyadic prime ideal inK) is the wild set of a self-equivalence ofK. We also show that there are number fieldsK in which this result cannot be improved.
Specifically, we show that there are non-archimedian non-dyadic primes P in K =Q(√
−14), at which −1 is a local square, such that {P} is not the wild set of any self-equivalence of K. We conclude this section by recalling that Czogala and Carpenter have classified independently all quadratic number fields up to Hilbert-symbol equivalence. They have shown that any quadratic number field is Hilbert-symbol equivalent to one of the number fields Q(√
d), where d=−1, ±2, ±7, ±17 (see [1, 2] for details).
2. MAIN RESULTS
We begin by introducing a few notations and recalling some results that will be used in this paper.
If (T, t) is a Hilbert-symbol equivalence between K and L, then for any prime P of K, the global square-class isomorphism t induces a local square- class group isomorphism tP : KP∗/KP∗2 → L∗T(P)/L∗2T(P) that preserves the Hilbert-symbols atP. By acorrespondencebetweenK andLwe mean an ob- jectC = (S,S, T,(tP)P∈S), whereS (S, respectively) is a finite set of primes inK (inL, respectively) that contains all the archimedian and dyadic primes, T :S → S is a bijection, and for eachP ∈ S,tP :KP∗/KP∗2→L∗T(P)/L∗2T(P) is a group isomorphism such that (tP(a), tP(b))T(P) = (a, b)P for anya, b.
The local square-class group KP∗/KP∗2 can be regarded as an F2-vector space. If P is non-archimedian non-dyadic then
KP∗/KP∗2 ={1, uP, πP, uPπP},
whereuP is the local square-class of a non-square unit atP andπP is the local square-class of a prime element atP(see Chapter 6, Theorem 2.22, in [4]). It is known that a non-dyadic prime idealP inK is tame with respect to a Hilbert- symbol equivalence between K and L if and only if tP(uP) =uT(P) (see, for instance, Remark 2 from [8], which can be generalized to any number field).
On the other hand, if P = D is dyadic, then dimF2(KD∗/KD∗2) = [KD : Q2] + 2, and an F2-base of KD∗/KD∗2 is {uD, u1, . . . , uj, πD}, whereu1, . . . , uj
are square-classes of non-square units at D,πD is the square-class of a prime atD, anduD is the square-class of a non-square unit atDsuch thatKD(√
uD) is the unique unramified quadratic extension ofKD. The local square-classuD
has the property that (uD, x)D = (−1)ordD(x)for anyx∈KD∗/KD∗2. Moreover, D is tame with respect to (T, t) if and only if tD(uD) =uT(D).
If K and L are Hilbert-symbol equivalent and S (S, respectively) is a finite set of primes in K (L, respectively) that contains all the archimedian and dyadic primes, consider the sets
EK(S) ={x∈K∗/K∗2: ordP(x)≡0 (mod 2), ∀P /∈ S}, EL(S) ={x∈L∗/L∗2 : ordQ(x)≡0 (mod 2), ∀Q /∈ S}.
Corollary 5 from [6] shows that
(1) rk2(EK(S)) =|S|+rk2(CK(S)),
whereCK(S) is theS-ideal class group ofK. IfP is a non-archimedian prime in K, we will denote by clS(P) the square-class of P inCK(S). Let
νS :EK(S)→
Q∈S
KQ∗/KQ∗2
be the diagonal mapping νS(x) = (x, x, . . . , x), and νQ, where Q∈ S, be the Q-component of νS. In [1] it is shown that
rk2 Q∈S
KQ∗/KQ∗2
= 2|S|.
If ωK(S) := Im(νS), then
(2) rk2(ωK(S)) =|S|
(see Lemma 13 from [6]).
For a correspondenceC = (S,S, T,(tP)P∈S) between two number fields K and L, let
HS ={(x)S ∈ωK(S) : (tP)S((x)S)∈ωL(S)} and consider the defectof the correspondence defined in [6] as
δC=|S| −rk2(HS).
Since rk2(ωK(S)) = rk2(ωL(S)) = |S|, it follows that δC = 0 if and only if (tP)P∈S(ωK(S)) =ωL(S).
Theorem 29 in [6] shows that any correspondence C can be extended to a Hilbert-symbol equivalence between K andL that has exactly
|W(C)|+δC+|rk2(CK(S))−rk2(CL(S))|
wild primes and no Hilbert-symbol equivalence that extends C exists, that has fewer wild primes. Here W(C) denotes the set of wild primes in the correspondence.
Having introduced these notations, we are ready to prove the main re- sults.
Theorem2.1. For any non-archimedian, non-dyadic prime P of an ar- bitrary number field K such that −1 is a local square at P there is a non- archimedian, non-dyadic prime P∗ of K, not necessarily distinct from P, and a self-equivalence of K that has the wild set equal to {P, P∗}.
Proof. LetSbe a set of primes inKthat does not containP but contains all the archimedian and dyadic primes of K and such that theS-class number hK(S) ofKis odd. Such a set of primes is calledsufficiently largein [1] and [5].
Lemma 5 from [6] implies EK(S) is anF2-vector space of dimensionm=|S|. Letx1, . . . , xm be anF2-base ofEK(S). Regarding the primeP, there are two possibilities:
Case1. Any base element xi is a local square atP.
Since S is sufficiently large, the S-class number hK(S) of K is odd.
Then the ideal class [P] is a square in CK(S), so clS(P) = 1. Therefore, by Lemma 6 from [6], [EK(S ∪ {P}) :EK(S)] = 2, so there is p∈K∗ such that
p is a local unit at each prime outside S ∪ {P} and a local prime element at P. Let πP =νP(p). Note that {x1, . . . , xm, p} is anF2-base of EK(S ∪ {P}).
We define a self-correspondence ofK as follows: S1=S1 =S ∪ {P},T : S1 → S1 is the identity mapping, and each local square-class isomorphismtQ : KQ∗/KQ∗2 → KQ∗/KQ∗2 is the identity mapping, when Q=P. Define the local square-class isomorphism tP :KP∗/KP∗2 → KP∗/KP∗2 as follows: 1 →1, uP → uPπP, πP → πP, uPπP → uP (therefore, P is wild by construction). Note that this correspondence has a defect equal 0. Indeed, for any base element xi
we find
(tQ)Q∈S1(νS1(xi)) = (tQ)Q∈S1(xi, . . . , xi,1) =νS1(xi) and
(tQ)Q∈S1(νS1(p)) = (tQ)Q∈S1(p, . . . , p, πP) =νS1(p), so rk2(HS1) =m+ 1 =|S1|, hence the defect is 0.
Theorem 29 from [6] shows that this correspondence can be extended to a self-equivalence of K which is tame outside S1. Of course, the only wild prime in S1 is P, soW ={P}.
Case2. There are base elements xi that are not local squares at P. After a change of base, we can assume that exactly one of the base elements (let’s say xm) is a local non-square unit at P and all the other base elements are local squares at P. Like in Case 1, let p ∈ K∗ be such that ordP(p) ≡ 1 (mod 2) and ordQ(p) ≡ 0 (mod 2) for any Q /∈ S ∪ {P}. Let πP =νP(p).
Using a result established in the proof of Theorem 2 from [5] and summa- rized as Lemma 27 in [6], there are infinitely many primesP∗ outsideS ∪ {P} and elements p∗ ∈K∗ such that
p∗ ∈KQ∗2, ∀Q∈ S, p∗=p inKP∗/KP∗2, ordP∗(p∗)≡1 (mod 2),
ordQ(p∗)≡0 (mod 2), ∀Q /∈ S ∪ {P, P∗}.
Let πP∗ =νP∗(p∗) and S∗ =S ∪ {P, P∗}. Note that {x1, . . . , xm, p, p∗} is an F2-base ofEK(S∗).
Let 1≤i≤m−1 and denote by Ω the set of all primes inK. By Hilbert reciprocity we find that
Q∈Ω
(xi, p∗)Q= 1.
In this product, all the symbols but (xi, p∗)P∗ vanish, as both xi and p∗ are local units outside S∗,xi is a local square atP, andp∗ is a local square at all the primes in S. Therefore, (xi, p∗)P∗ = 1, and sincep∗ is a local prime atP∗, xi∈KP∗2∗,∀1≤i≤m−1. Using a similar argument, we find−1∈KP∗2∗.
Also
Q∈Ω
(p, p∗)Q = 1 implies (p, p∗)P(p, p∗)P∗ = 1, so (p, p)P(p, p∗)P∗ = 1. But since −1 is a local square atP, we get (p, p)P = (−1, p)P = 1, which forces (p, p∗)P∗= 1. But p∗ is a local prime atP∗ and pis a local unit at P∗. Therefore, p∈KP∗2∗.
Finally,
Q∈Ω
(xm, p∗)Q = 1 implies (xm, p∗)P(xm, p∗)P∗ = 1. But xm is a non-square unit at P and p∗ is a local prime at P, so (xm, p∗)P = −1, which implies (xm, p∗)P∗=−1. Since p∗ is a prime atP∗,xm must be a local non-square unit at P∗.
We define a correspondence as follows: let T : S∗ → S∗ be the iden- tity mapping. For any Q ∈ S, let tQ : KQ∗/KQ∗2 → KQ∗/KQ∗2 be the identity mapping. Define tP :KP∗/KP∗2 → KP∗/KP∗2 as follows: 1 → 1, uP → uPπP, πP → πP, uPπP → uP and define tP∗ : KP∗∗/KP∗2∗ → KP∗∗/KP∗2∗ as follows:
1→1,uP∗ →πP∗uP∗,πP∗ →πP∗,uP∗πP∗→uP∗. Lett∗= (tQ)Q∈S∗. Note that
t∗(νS∗(xi)) =t∗(xi, . . . , xi,1,1) = (xi, . . . , xi,1,1) =νS∗(xi), ∀1≤i≤m−1, t∗(νS∗(xm)) =t∗(xm, . . . , xm, uP, uP∗) = (xm, . . . , xm, uPπP, uP∗πP∗) =
=νS∗(xmp∗),
t∗(νS∗(p)) =t∗(p, . . . , p, πP,1) = (p, . . . , p, πP,1) =νS∗(p), t∗(νS∗(p∗)) =t∗(1, . . . ,1, πP, πP∗) = (1, . . . ,1, πP, πP∗) =νS∗(p∗).
These identities show that the defect of this correspondence is 0, so we can extend it tamely to a self-equivalence (T, t) of K. By construction, the wild set of (T, t) isW ={P, P∗}.
Corollary 2.2. In any number field K there are infinitely many non- dyadic primesP such thatW ={P}is the wild set of a self-equivalence ofK. Proof. Consider a sufficiently large set of primes S in K and pick an F2-baseBofEK(S). There are infinitely many non-dyadic primesP such that
−1 and all the elements in B are local squares at P. For any such prime P, use the construction from Case 1 of the proof of Theorem 2.1.
Corollary 2.3. Let K be a number field. For any n ≥ 1 there is a self-equivalence of K that has exactly n wild non-dyadic primes.
Proof. We will use an argument by induction on n. For n = 1, we use Corollary 2.2. Suppose now that our statement holds for some n ≥ 1. Let (T1, t1) be a self-equivalence that has as its wild setW1={P1, . . . , Pn}, where thePi’s are distinct non-dyadic primes. By Corollary 2.2, there is a non-dyadic prime Pn+1 outside {T1(P1), . . . , T1(Pn)} such thatW2 = {Pn+1 } is the wild set for some self-equivalence (T2, t2). Let Pn+1 =T1−1(Pn+1 ).
We claim that the composition (T2T1, t2t1) is a self-equivalence of K whose wild set is {P1, . . . , Pn, Pn+1}. Indeed, for each 1 ≤ i ≤ n, (t1)Pi is a wild mapping (it maps the square-class of a non-square unit to the square-class of a prime element) and (t2)T1(Pi) is a tame mapping (it maps square-classes of units to square-classes of units and square-classes of primes elements to square-classes of primes elements). Therefore, (t2t1)Pi = (t2)T1(Pi)(t1)Piis wild.
Similarly, (t2t1)Pn+1 = (t2)P
n+1(t1)Pn+1 is wild as (t2)P
n+1 is wild and (t1)Pn+1 is tame. All the other non-archimedian primes are tame with respect to this composition (as the composition of two tame local mappings is tame).
In the remaining part of the paper we provide an answer to the question posed in the introduction: is any setW ={P}, whereP is a non-archimedian, non-dyadic prime in an arbitrary number fieldK, at which−1 is a local square, the wild set of self-equivalence of K? In other words, can we eliminate the prime P∗ from the previous theorem? We are showing that there are number fields in which this question has a negative answer. An example of such a field is Q(√
−14). The details are presented below. Recall thatQ(√
−14) and Q(√
−2) are Hilbert-symbol equivalent (see, for instance, [2]).
Let D = {A, D} and D = {A, D}, where A, A are the archimedian primes and D, D the dyadic primes of Q(√
−14) and Q(√
−2), respectively.
We begin with the following lemma (some of these properties are mentioned without a proof in [3]):
Lemma 2.4. Let K=Q(√
−14)and K =Q(√
−2).
a) Any Hilbert-symbol equivalence between K and K has at least one wild prime.
b) There is a Hilbert-symbol equivalence between K and K that has ex- actly one non-dyadic wild prime.
c) If P is a non-dyadic prime such that W = {P} is the wild set of a Hilbert symbol equivalence between K and K, then P generates CK(D).
d)IfP is a non-dyadic prime that generatesCK(D), thenEK(D∪{P}) = EK(D) and uP ∈νP(EK(D)).
Proof. a) The ideal class group CK of K is cyclic of order 4 and 2- rank equal to 1. Since D = (2,√
−14) is not principal but D2 = 2Z[√
−14]
is, [D] has order 2 in CK. It follows that CK(D) is cyclic of order 2 and rk2(CK(D)) = 1. On the other hand, rk2(CK(D)) = 0, as CK(D) is trivial.
Using Corollary 31 from [6], it follows that any Hilbert symbol equivalence between K and K has at least |rk2(CK(D))−rk2(CK(D))|= 1 non-dyadic wild prime.
b) Since KA KA C, KA∗/KA∗2 KA∗/KA∗2 {1}. Therefore, by abuse of notation, we will identifyKA∗/KA∗2×KD∗/KD∗2withKD∗/KD∗2. When we do so, the mappingνD is identified with the mappingνD :EK(D)→KD∗/KD∗2
and ωK(D) = Im(νD) with Im(νD). Similar identifications are made for K. Note that by (2), we have rk2(Im(νD)) =|D|= 2.
We will construct a correspondence C = (D,D, T, tD) of defect 0, in whichDis tame. (SinceKA∗/KA∗2 KA∗/KA∗2 {1}, the mappingtAis trivial, so we ignore it.) According to Theorem 29 from [6], this correspondence can be extended to a Hilbert-symbol equivalence between K andK that has exactly
|rk2(CK(D))−rk2(CK(D))|= 1 non-dyadic wild prime.
By (1),
rk2(EK(D)) =|D|+rk2(CK(D)) = 3. Obviously, −1,2 ∈EK(D). We claim that also 2 +√
−14∈EK(D). Indeed, NK/Q(2 +√
−14) = 18. Since 2 ramifies, as 2OK = D2, and 3 splits, as 3OK =P3P3, with P3 = (3,1 +√
−14) andP3 = (3,1−√
−14), we have (2 +√
−14)OK =DP3αP3β, withα+β = 2. Note that 2 +√
−14∈/ P3, soα= 0. Hence, (2 +√
−14)OK= DP32, so 2 +√
−14∈EK(D) and 2 +√
−14 is a local prime at D, which will be denoted by πD.
We claim that{−1,2,2 +√
−14} is anF2-base for EK(D). Indeed, sup- pose that (−1)u2v(2 +√
−14)w = 1 in EK(D), where u, v, w ∈ {0,1}. Since NK/Q(2) = 4 and NK/Q(2 +√
−14) = 18, then 4v ·18w = 1 in Q∗/Q∗2, so w = 0. Since −1 and ±2 are not global squares, we have u = v = 0 so the claim is proved. Then Im(νD) is generated by {−1,2, πD} and, as previously noted, rk2(Im(νD)) = 2. But −14 ∈ KD∗2, so 2 = −7 in KD∗/KD∗2. Since −7 is a square in Q2, it is a square in KD. It follows that 2 = 1 in KD∗/KD∗2, so {−1, πD} is anF2-base for Im(νD). Since −1,2 +√
−14 ∈EK(D), we get by Hilbert reciprocity that
(−1,−1)D = (−1, πD)D = 1.
Since (uD, πD)D =−1, we cannot haveuD =−1, so−1, uD, πDareF2-linearly independent in KD∗/KD∗2. Since (·,·)D is a non-degenerate symmetric bilinear form, there is an element vD ∈KD∗/KD∗2 such that (vD,−1)D = (vD, πD)D =
−1 and (vD, uD)D = 1. We have
(−1,−1)D = (uD,−1)D = (πD,−1)D = 1
and (vD,−1)D =−1. Hence,vD is not in the span of−1, uD, πD, which implies that −1, uD, vD, πD are F2-linearly independent so they form an F2-base for KD∗/KD∗2.
The following table summarizes the Hilbert symbols involving the base elements of KD∗/KD∗2:
−1 vD uD πD
−1 1 −1 1 1 vD −1 −1 1 −1 uD 1 1 1 −1 πD 1 −1 −1 1
An analysis of the dyadic local square class group of K = Q(√
−2) can be found in [5]. We use some parts of that analysis here. We know that rk2(EK(D)) = 2 and we can easily check that {−1,√
−2} is an F2-base of EK(D). As F2-vector spaces, KD∗/KD∗2 is generated by {−1, vD, uD, πD} andKD∗/KD∗2 is generated by{−1,1 +√
−2, uD,√
−2}(uD denotes the local square-class for which KD (√
uD) is the unique unramified quadratic exten- sion of KD ).
The table of dyadic Hilbert symbols involving the base elements in KD∗/KD∗2 is
−1 1 +√
−2 uD √
−2
−1 1 −1 1 1 1 +√
−2 −1 −1 1 −1
uD 1 1 1 −1
√−2 1 −1 −1 1
Consider now the correspondenceC= (D,D, TD, tD) betweenKandK, with TD :D → DmappingDtoDandAtoA, andtD mapping the base elements ofKD∗/KD∗2 as follows: −1→ −1,vD →1 +√
−2,uD →uD, andπD →√
−2.
Note thatδC = 0 astDmaps Im(νD) =−1, πDto Im(νD) =−1,√
−2. Hence, we can extend C to a Hilbert symbol equivalence between K and K that has exactly one wild (non-dyadic) prime.
c) Let (T, t) be a Hilbert symbol equivalence between K and K, with wild set W = {P}. Note that (T, t) induces a correspondence C1 = (D ∪ {P},D∪ {T(P)}, T1, t1), whereT1 is the restriction ofT toD ∪ {P}andt1 is the collection of local square-class group isomorphisms induced by t relative to D ∪ {P}. If P does not generate CK(D), then rk2(CK(D ∪ {P})) = 1 and rk2(CK(D ∪ {T(P)})) = 0, so any Hilbert-symbol equivalence that ex- tends C1, (T, t) in particular, must contain |rk2(CK(D ∪ {P}))−rk2(CK(D∪ {T(P)}))|= 1 additional non-dyadic wild prime, contradiction.
d) Consider a Hilbert-symbol equivalence (T, t) that has one (non-dyadic) wild primeP. SincePgeneratesCK(D),Pcannot be a square inCK(D), which means that clD(P) = 1. Therefore, if S = D ∪ {P}, then rk2(CK(S)) = 0.
Using (1),rk2(EK(S)) =rk2(EK(D)) = 3, soEK(D) =EK(S), which proves the first claim.
In order to prove the second claim, suppose that νP(EK(D)) = {1}. Since νD and νS have the same domain (EK(D) = EK(S)) and νP is trivial,
their images ωK(S) and ωK(D) are isomorphic. But this is impossible, as by (2), rk2(ωK(D)) =|D|= 2 and rk2(ωK(S)) =|S|= 3.
Proposition 2.5. There are finite non-dyadic primes P in Q(√
−14) at which −1 is a local square and such that {P} is not the wild set of a self- equivalence of Q(√
−14).
Proof. Using the notations from Lemma 2.4, we will show that if P is a non-dyadic prime ofKwith−1∈KP∗2such thatPgeneratesCK(D) then there is no self-equivalence ofKfor whichP is the sole wild prime. In particular, by Lemma 2.4 c), if {P} is the wild set of a Hilbert-symbol equivalence between K and K then it cannot be the wild set of a self-equivalence of K.
By way of contradiction, suppose that (T, t) is a self-equivalence of K with wild setW ={P}. LetP0 =T(P). Restrict (T, t) to a self-correspondence C1 = (S,S0, T1, t1), where S = D ∪ {P}, S0 = D ∪ {P0}, T1 = T|S, and t1 = (tA, tD, tP). SinceP is wild,tP(uP) =πP0, whereπP0 is the square-class of a prime element inKP0. SincePis the only wild prime, Theorem 2 from [6] im- plies that rk2(CK(S)) =rk2(CK(S0)) andδC1 = 0, i.e.,t1(ωK(S)) =ωK(S0).
ButPgeneratesCK(D), sork2(CK(S)) = 0, which forcesrk2(CK(S0)) = 0, so P0 generates CK(D) as well. By Lemma 2.4 d), EK(S0) = EK(D), so the P0 coordinate of every element in ωK(S0) = Im(νS0) belongs to{1, uP0}. In addition, by Lemma 2.4 d), uP ∈νP(EK(S)). Letz∈EK(S) be such that νP(z) =uP. SinceνS(z) ∈ωK(S), this forces the P0 coordinate of t1(νS(z)) to be tP(uP) =πP0. SinceπP0 ∈ {/ 1, uP0}, it follows that t1(νS(z))∈/ ωK(S0).
This contradicts t1(ωK(S)) =ωK(S0).
Acknowledgements.The author wishes to thank the referee for many helpful sug- gestions, particularly for proposing the current formulation of Lemma 4 d).
This project was supported by a Summer Fellowship of the University of Nor- thern Iowa.
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Received 10 January 2010 University of Northern Iowa
Department of Mathematics Cedar Falls, IA 50614, USA
somodi@uni.edu
