A family of cyclic quartic fields with explicit fundamental units
aa r X i v : . [ m a t h . N T ] S e p A FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICITFUNDAMENTAL UNITS
STEVE BALADY AND LAWRENCE C. WASHINGTON
Abstract.
We construct a family of quartic polynomials with cyclic Galoisgroup and show that the roots of the polynomials are fundamental units orgenerate a subgroup of index 5.
The goal of this paper is to prove the following.
Theorem 1.
Let s be an integer such that s − s + 4 is a square. Let K s be thesplitting field of F s ( t ) = t + (4 s − s + 8 s − t + ( − s − t + 4 t + 1 , Then Gal ( K s / Q ) is cyclic of order 4. If s + 2 is squarefree and s = 0 , then ± andthe roots of F s ( t ) generate either the unit group of the ring of algebraic integers in K s or a subgroup of index 5. Computational evidence (see the end of Section 5) indicates that the index 5case does not occur, but we have not yet proved this.Families of cyclic quartic fields with explicit units have been studied in the past(for example, [4], [8], [9]), but it does not seem that this family has been studiedpreviously. In [1], families of cyclic cubic fields were constructed and the methodled naturally to studying integral points on a model of the elliptic modular surface X (3). In the present case, we are led to study integral points on a degree 4 coverof the surface X (2), but the Diophantine properties are not as transparent. Thefamily we study lives above a singular fiber of X (2). It would be interesting toknow if there are other families living as curves on the surface.1. The polynomials
As in [1], we start with the action of the Galois group given by a linear fractionaltransformation, but this time we take the matrix to have order 4 in PGL ( Z ). Let f, g be integers and let M = (cid:18) f − f + g − g (cid:19) . Then M = f − g (cid:18) f + g − f + g − g − f (cid:19) , M = ( f − g ) (cid:18) g − f + g − f (cid:19) . The action of the Galois group is to be θ M θ . We want θ to be a unit, so wetake it to have norm 1:(1) θ · f θ − f + g θ − g · ( f + g ) θ − f + g ) θ − g − f · gθ − f + g θ − f = 1 . If f g ( f + g ) = 0, this relation can be rearranged to say that θ is a root of t − ( f + g ) + 4( f + g ) + 16 f g f g ( f + g ) t (2) + 3(( f + g ) + 4)4 f g t − ( f + g ) − f g + 42 f g ( f + g ) t + 1 = 0 . (3)Since this is symmetric in f and g , we make the substitutions s = f + g and p = f g to obtain t − ( s − p ) + 4 s + 8 p sp t + 3(( s − p ) + 4)4 sp t − s − p + 42 sp + 1 . Let L = − s − p +44 sp . The polynomial becomes(4) t + (2 s + Ls − ps + 2 Lp ) t + ( − s − Ls + 6 p ) t + 2 Lt + 1 . Therefore, if L ∈ Z , or if 2 L ∈ Z and s is even, the roots of the polynomial areunits. Of course, since s = f + g and p = f g , there is the extra condition that s − p is a square. Remark.
The divisibility condition 4 sp | s − p + 4 that makes L integral is thesame type of condition that appears in the cubic case, where the paper [1] has thecondition f g | f + g + 1. These both seem to be analogues of the condition r | n for the Richaud-Degert real quadratic fields Q ( √ n + r ) (see [12]).Note that changing s to − s while holding p constant corresponds to changingthe signs of f and g . This has the effect of L
7→ −
L, t
7→ − t in (4). Therefore, we can for simplicity assume that L > L = 0 cannotoccur).The family considered in [4] (the “simplest quartic fields”) corresponds to thematrix M with f = 1 and g = −
1. However, the relation (1) becomes trivial inthis case and does not yield the polynomial defining the family.2.
The surface
When the analogous construction was done for the cubic case in [1], the expres-sion for one of the coefficients yielded an equation for X (3). In the present case,we are looking for integral points ( f, g, L ) on the surface X : ( f + g ) − f g + 4 + 2 Lf g ( f + g ) = 0 . This is a double cover of the surface s − p + 4 + 2 Lps = 0 , which can be transformed (see, for example, [3]) to y = ( x + 4)( x − x + L ) . Therefore, X is a degree 4 cover of the elliptic surface X (2), whose fibers areLegendre elliptic curves. The bad fibers L = ± f, g ) for which L is integral, andalmost all of them had L = ± FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICIT FUNDAMENTAL UNITS 3 f g s p L polynomial1 − − − − t − t − t − t + 15 − − −
85 2 t − t − t + 4 t + 15 − − − − / t − t − t − t + 117 − − − − t − t − t − t + 165 − − − t − t − t + 4 t + 1As mentioned above, a simple transformation changes the examples with L = − L = 2. 3. A family of fields
From now on, we make the assumption: L = 2 . Since s − p + 4 Lps + 4 = 0 is a quadratic in p , the quadratic formula yields p = Ls ±
12 ( s + 2) . If p = s + ( s + 2), the polynomial in Equation (4) becomes ( t + 1) , so we alwaystake(5) p = s −
12 ( s + 2) . We now have the polynomial(6) F s ( t ) = t + (4 s − s + 8 s − t + ( − s − t + 4 t + 1 , with the side condition that(7) 3 s − s + 4 is a square(this is a rewriting of s − p ). This condition immediately implies the following: s is even , s + 2 ≡ , p is odd , where the last follows from (5). Since p is odd, we must have f and g are odd . The original parameters f and g are given by(8) f, g = s ± √ s − s + 42 . The choice of which is f and which is g does not have much significance, but itaffects the choice of generator of the Galois group in the following.For future reference, we note the following consequence of L = 2: Lemma 2. If L = 2 , then s + 22 = (cid:18) f + 12 (cid:19) + (cid:18) g + 12 (cid:19) , ( f − g ) = 3 s − s + 4 . Proof.
The right side of the first equation is14 (cid:0) s − p + 2 s + 2 (cid:1) = 14 (cid:0) s + 4 (cid:1) = 12 (cid:0) s + 2 (cid:1) , where the first equality uses Equation (5). The second equation follows from Equa-tion (8). (cid:3) STEVE BALADY AND LAWRENCE C. WASHINGTON
Suppose now that Equation (7) holds. Then v = 3 s − s + 4 , for some v , which forces v = 2 w and s = 2 u for some u, w . This yields(3 u − − w = − . The solutions are given by(9) (3 u − ± w √ − n +1 (1 + √ √ n , with n ∈ Z (the first choice of signs is arbitrary; the second sign is chosen in orderto make the right side congruent to − √ s :4 , − , , − , , − , , − . Every third value (4, − s + 2 divisible by 3 . The other valueslisted yield squarefree values of s +2, although it is not known whether the sequenceyields infinitely many values of s such that s + 2 is squarefree. Questions of thistype seem similar to questions about squarefree Mersenne numbers, most of whichare unsolved. For the first 100 values of s (that is, for s arising from 1 ≤ n ≤ s + 2 or ( s + 2) / F s ( t ) in Equation (6) is(10) 256(3 s − s + 4) ( s + 2) . Since 3 s − s + 4 is a square, the discriminant is a square times s + 2. But s + 2is never a square, so k s = Q ( √ s + 2) ⊆ K s . Therefore, once we show that theGalois group is cyclic, we know that k s is the unique quadratic subfield.We first show that F s is irreducible, then identify the Galois action. Lemma 3.
Let | s | ≥ . The roots of F s ( t ) satisfy r = − s + 4 s − s + 4 − (3 / s − − (3 / s − + θ s − , with ≤ θ ≤ ,r = 1 + √ s − + 3 + √ s − − √ s − + θ s − , with − / ≤ θ ≤ − / ,r = (1 / s − + (1 / s − − θ s − , with ≤ θ ≤ ,r = 1 − √ s − + 3 − √ s − + 13 √ s − + θ s − , with − / ≤ θ ≤ / . Proof.
Let r = − s + 4 s − s + 4 − (3 / s − − (3 / s − . Substitute r + s − into F s ( t ). The result is a degree 21 polynomial P ( s ) divided by s . The real roots of P ( s ) all have absolute value less than 2.1, and P ( s ) has a positive top coefficient.Since P ( s ) is positive for large s and does not change sign in the interval (2 . , ∞ ),we see that P ( s ) > s ≥
3. Similarly, P ( s ) < s ≤ − r + 2 s − into F s ( t ). The result is a degree 21 polynomial Q ( s )divided by s . The real roots of Q ( s ) all have absolute value less than 1, and Q ( s )has a negative top coefficient. It follows that Q ( s ) < s ≥ Q ( s ) > s ≤ − s with s ≥
3. Then F s ( r + s − ) > > F s ( r + 2 s − ). Therefore, there isa zero r of F s ( t ) that satisfies the stated conditions. The case where s ≤ − r , r , r are similar. (cid:3) FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICIT FUNDAMENTAL UNITS 5
These expansions of r and r were found by letting s = 10 and finding theroots of F s ( t ) numerically. The coefficients of the expansions were then easy todeduce from the decimal expansions of the roots. For r and r , the expansionsof r + r and ( r − r ) / √ f = (cid:0) s + √ s − s + 4 (cid:1) / g = (cid:0) s − √ s − s + 4 (cid:1) /
2, then thelinear fractional transformation M maps r j to r j +1 . Given the approximation to r , we could obtain the other approximations from the action of M , but estimatingthe error terms would be harder. Lemma 4.
Let s ∈ Z . Then F s ( t ) is irreducible in Q [ t ] .Proof. The only possible rational roots of F s are ±
1. But F s ( ± = 0 when s ∈ Z . Therefore, F s does not have a linear factor, so, if it factors, it must havetwo quadratic factors, and they must have integer coefficients. This means that r + r j ∈ Z for some j .Let | s | ≥
10, say. The cases with | s | <
10 can be checked individually. FromLemma 3, we see that r + r is not an integer, so r , r cannot be the roots of aquadratic factor. Also, 0 < | r + r | < < | r + r | <
1, so these sums cannotbe integers. Therefore, F s cannot factor into quadratic factors. (cid:3) Lemma 5.
Let K s be the splitting field of F s and assume s − s + 4 is a square.Then Gal ( K s / Q ) is cyclic, and the linear fractional transformation M gives theGalois action on the roots of F s ( t ) .Proof. Since 3 s − s +4 is a square, the parameters f and g exist (see Equation (8)).Let θ be a root of F s . Then θ satisfies Equation (1). Let θ ′ = M θ , the result ofapplying the linear fractional transformation M to θ . Since M cyclically permutesthe factors in Equation (1), we see that θ ′ also satisfies this equation, and therefore F s ( θ ′ ) = 0. Therefore, Q ( r ) contains all the roots of F s . Since F s is irreducible, itfollows that the Galois group of F s is cyclic of order 4 and is generated by M . (cid:3) The discriminant
Proposition 6.
Suppose that s + 2 is squarefree and s − s + 4 is a square.Then the discriminant of K s is ( s + 2) .Proof. The discriminant of the polynomial F s ( t ) is2 ( s + 2) (3 s − s + 4) = 2 ( s + 2) ( f − g ) , where we have used Lemma 2 to obtain the second expression. We need to showthat the factor ( f − g ) can be removed. Let q be an odd prime dividing f − g .Since − spL = s − p + 4, we cannot have f ≡ g ≡ q ).If f g ( f + g ) q ) (equivalently, f q )), then Equation (2)becomes F s ( t ) ≡ ( t − /f ) ( t − f ) (mod q ) . Also, Lemma 2 implies that(11) 0 ≡ ( f − g ) = 3 s − s +4 = 3( f + g ) − f + g )+4 ≡ f − f +4 (mod q ) . We claim that the roots 1 /f and f of F s ( t ) (mod q ) are distinct. Suppose 1 /f ≡ f (mod q ). The resultant of f − f − f + 4 is 2 ·
3, so we must have q = 3. STEVE BALADY AND LAWRENCE C. WASHINGTON
Equation (11) now tells us that 0 ≡ f − f + 4 (mod 3), so f ≡ − f ≡ g , we also have g ≡ − s + 2 ≡ s + 3 is squarefree.Therefore, 1 /f f (mod q ).Let q be a prime of K s dividing q and let I be the inertia subgroup of Gal( K s / Q )for q . If q divides the discriminant of K , then σ ∈ I , where σ generates Gal( K s / Q ).Let r j ≡ f (mod q ). Then the other three roots are congruent to 1 /f mod q . But σ ∈ I means that f ≡ r j ≡ σ ( r j ) ≡ /f (mod q ), contradicting the fact that f /f (mod q ). Therefore, q does not divide the discriminant of K s , so f − g contributes no odd prime factors to the discriminant of K s .We have proved that the discriminant D of K s divides a power of 2 times ( s +2) .The subfield k s = Q ( √ s + 2) ⊂ K s has conductor 4( s + 2), since s + 2 issquarefree and congruent to 2 mod 4. Let χ be a Dirichlet character of order 4attached to K s . Then χ is the quadratic character attached to k s . Since χ hasconductor 4( s + 2), it follows that χ and χ − have conductor divisible by 4( s + 2).The conductor-discriminant formula implies that D is divisible by 4 ( s + 2) . Wehave therefore proved that D is a power of 2 times (( s + 2) / .Since 2 ramifies in k s / Q and K s / Q is cyclic, 2 is totally ramified in K s / Q . Thismeans that χ is the product of a character of conductor 16 and a character of oddconductor. The conductor-discriminant formula implies that 2 is the exact powerof 2 dividing D . Since ( s + 2) contributes 2 , this completes the proof. (cid:3) Remark.
In order to determine the ramification and discriminant of K s , we couldconsider the extension K s ( i ) / Q ( i ). Order the roots r , r , r , r , so that M r j = r j +1 . Let ρ = r + r i − r − r i. Computationally, it appears that(12) ρ = − i ( f + gi ) π π, where π = ( f + 1) / − i ( g + 1) /
2. This would suffice to remove the factor (3 s − s + 4) , since √ ρ generates the extension K ( i ) / Q ( i ), and since the odd parts ofthe discriminants of K s ( i ) / Q ( i ) and K s / Q are equal. However, the verificationof Equation (12) seems to be potentially quite involved, which is why we had theincentive to find the above proof.5. Fundamental units
The purpose of this section is to prove that ± F s ( t ) generatethe units of K s . Throughout this section, we assume that 3 s − s + 4 is a square. Lemma 7.
Let s = 0 and suppose s + 2 is squarefree. Then ǫ = − r r = s + 1 + | s | p s + 2 is the fundamental unit of the ring of integers of Q ( √ s + 2) .Proof. Since s + 2 Z [ √ s + 2] (that is,there is no 2 in the denominator). If ǫ = a + b √ s + 2 is the fundamental unit, so a, b >
0, then ǫ > s + 1 + s √ s + 2 = ǫ . Since ǫ is a power of ǫ , we must have ǫ = ǫ .From Lemma 3, we see that r r ≈ − s . In particular, 1 < − r r < ǫ , so r r = − ǫ . (cid:3) FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICIT FUNDAMENTAL UNITS 7
Note:
We did not need to know the ordering of the r j under the Galois group toobtain this result, since − r r is the only combination with the same approximatesize as ǫ . In fact, once we know this, if σ is a generator of Gal( K s / Q ) then σ maps r to r , and hence σ or σ − permutes the roots r , r , r , r cyclically (thatis, r j r j +1 ). Of course, we know that M = σ or σ − , depending on the choice ofsigns in Equation (8). Proposition 8.
Let K be a totally real number field with Gal ( K/ Q ) cyclic of order4. Let R K and D K be the regulator and discriminant of K . Let k be the quadraticsubfield of K and let ǫ and d k be the fundamental unit and discriminant of k . Then
14 log (cid:18) D K d k (cid:19) ≤ R K log ǫ . If d k > , then
14 log (cid:18) D K . d k (cid:19) < R K log ǫ . Proof.
This result is implicit in [2], [4], and [14], for example. However, since it doesnot seem to be explicit in the literature, we include the proof for the convenienceof the reader.The result is easily verified for Q ( ζ ) + , the maximal real subfield of the 16thcyclotomic field. In all other cases, an odd prime ramifies in K/ Q , so we mayassume that K/k ramifies at some prime of k that divides an odd prime of Z .Let E K and E k be the units of K and k , and let σ generate Gal( K/ Q ). Then E K /E k is a Z [ i ]-module, where i acts as σ . We claim that this module is torsion-free. If α ∈ Z [ i ] maps u ∈ E K into E k , then so does Norm( α ), so it suffices to showthat the Z -torsion is trivial.If u ∈ E K and u n ∈ E k , then ( σ u ) n = u n , so σ u = ± u , since k is real.Therefore, σ ( u ) = u , so u ∈ E k . Therefore, k ( u ) /k ramifies at most at theprimes above 2, hence is trivial since K/k is assumed to ramify at an odd prime.Therefore, u ∈ E k , so E K /E k is torsion-free.Since E K /E k has Z -rank 2, it has Z [ i ]-rank 1, so there is a unit η that generatesit as a Z [ i ]-module. Let η σ = δ ∈ E k , and let η ′ = σ ( η ). Then {± , ǫ, η, η ′ } generates E K as a Z -module.A calculation shows that the regulator of K is R K = 2 log( ǫ ) (cid:18) (log | η | −
12 log | δ | ) + (log | η ′ | + 12 log | δ | ) (cid:19) . The different of
K/k divides η − σ ( η ) = η − δ/η , so the discriminant of K/k dividesNorm
K/k ( η − δ/η ) = − ( η − δ/η ) . Since D K = d k Norm k/ Q ( D K/k ), we find that D K /d k divides ( η − δ/η ) ( η ′ ± / ( δη ′ )) ≤ ( x + 1 /x ) ( y + 1 /y ) , where x = Max( | η | / | δ | / , | δ | / / | η | ) and y = Max( | η ′ || δ | / , / | δ | / | η ′ | ).The conductor-discriminant formula implies that d k | D K . If 1 ≤ x, y ≤ √ d k ≤ ( x + 1 /x ) ( y + 1 /y ) < . STEVE BALADY AND LAWRENCE C. WASHINGTON
Therefore, if d k >
150 then at least one of x, y is larger than √
10. If x > √
10, then x + 1 /x < . x . If 1 ≤ x ≤ √
10, then x + 1 /x ≤ x . Therefore,( x + 1 /x ) ( y + 1 /y ) < . xy, so log( D K /d k ) ≤ log ( x + 1 /x ) ( y + 1 /y ) < . x + log y ) ≤ (cid:16) log 2 . √ (cid:0) log x + log y (cid:1) / (cid:17) (Cauchy-Schwarz)= 2 (cid:16) log 2 . R K / log ǫ ) / (cid:17) . This yields the last statement of the proposition. Note that by increasing the lowerbound for d k , we could replace 4 .
84 by any number larger than 4.If we do not require d k > x, y ≥
1, so x +1 /x ≤ x +1 ≤ x and similarly for y . Therefore,( x + 1 /x ) ( y + 1 /y ) ≤ xy. The above argument yields 14 log (cid:18) D K d k (cid:19) ≤ R K log ǫ . This inequality suffices for our purposes. (cid:3)
Lemma 9.
Assume s + 2 is squarefree. Let E be the units of K s and let U be thesubgroup generated by ± and r , r , r , r . Then [ E : U ] = 2 , , , , , .Proof. Let E k be the units of Q ( √ s + 2). By Lemma 7, E k ⊂ U . Let E = E/E k and U = U/E k . Then [ E : U ] = [ E : U ]. As in the proof of Proposition 8, E ≃ Z [ i ].The subgroup U is an ideal of Z [ i ] under this isomorphism.If [ E : U ] = 2, then there exists η that generates E as a Z [ i ]-module and such that(1 + i ) η ≡ r mod E k . This means that there exists δ ∈ E k such that η σ = r δ .Since σ fixes δ , we have ± η σ + σ + σ = r r δ = − ǫδ . But ǫ is the fundamental unit of Q ( √ s + 2), so this is impossible.Now suppose [ E : U ] = 4. The only ideal of index 4 in Z [ i ] is (2), so there exists η ∈ E such that η = r δ , with δ ∈ E k s . Taking norms to k s yields N ( η ) = − ǫδ .This is impossible because ǫ is the fundamental unit of k s .If [ E : U ] = 8, then there exists η such that η σ ) = r δ , with δ ∈ E k s . Takingnorms from K s to k s , we find1 = (cid:16) η σ + σ + σ (cid:17) = r σ δ σ = − ǫδ , which is impossible.Finally, every ideal in Z [ i ] has index that is a norm from Z [ i ] to Z . Since 3, 6, 7are not norms, [ E : U ] = 3 , , (cid:3) We can now show that ± r , r , r , r generate the units of K s . We assumethat | s | ≥ . The cases where | s | < can be checked individually. FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICIT FUNDAMENTAL UNITS 9
Let R be the regulator of K s and let R ′ be the regulator for the group U inLemma 9. Then R ′ /R = [ E : U ]. We have R ′ = ± det log | r | log | r | log | r | log | r | log | r | log | r | log | r | log | r | log | r | . This equals (see [15, Lemma 5.26(c)])14 (log | r | + i log | r | − log | r | − i log | r | ) (log | r | − log | r | + log | r | − log | r | ) × (log | r | − i log | r | − log | r | + i log | r | )= 14 (cid:0) log | r /r | + log | r /r | (cid:1) (2 log ǫ ) . Lemma 3 implies that(13) R ′ / log ǫ ≤ (cid:0) log (9 s ) + log (4) (cid:1) < | s | when | s | ≥ . Also,log D K s d k s ! = log ( s + 2) ≥ | s | when | s | ≥
1. Proposition 8 implies thatlog | s | ≤ R ′ / [ E : U ]log ǫ < E : U ] log | s | . Therefore, [ E : U ] < . By Lemma 9, [ E : U ] = 2 , , , , ,
8, so [ E : U ] = 1 or 5. This completes the proofof Theorem 1.When s ≡ , s + 2 ≡ s + 2 is not squarefree.However, if ( s + 2) / D K s is small enough that theabove proof shows that we still obtain either the full group of units or a subgroupof index 5, except when s = 4 (where the index [ E K s : U ] = 40). Computationalevidence suggests that the index 5 case does not occur.Since the units of K s are fairly small, we expect the class numbers to be large.Here is a table of the class groups for the first few values of s (for the class groups, a × b means Z /a Z × Z /b Z ). These calculations, and others in this paper, were doneusing GP/PARI [13].For all of the examples in the above table, − F s ( t ) generate the full group of units for the ring of algebraic integers of the cor-responding field, except for s = 4, where they generate a subgroup of index 40. Inthis case, s +2 = 18 is not squarefree, and the unit 17+4 √
18 = ( s +1)+ s √ s + 2is the fourth power of the fundamental unit 1 + √ K s /k s is ramifiedat 2, the class number of k s divides the class number of K s . Since the class numberof Q ( √ s + 2) goes to ∞ as s → ∞ (this can be made explicit with at most onepossible exception; see [12]), the class number of K s also goes to ∞ . However, thepotential effect of Siegel zeroes for quadratic fields can be overcome since we havea quartic field, so we obtain a result with no exceptions. s class group of K s −
12 448 4 × × −
176 60 × × × − × × × × − × × × × × − × × × × − × × × × × − × × × × × Proposition 10.
Assume s + 2 is squarefree and s − s + 4 is a square. Let h be the class number of K s and let h be the class number of k s . If | s | ≥ , then hh ≥ s + 2log ( | s | ) log (8( s + 2) /π ) . Proof.
Let χ be a quartic Dirichlet character associated with K s . Then8 hR p D K s = L (1 , χ ) L (1 , χ ) L (1 , χ ) . We have L (1 , χ ) = 2 h log ǫ p d k s . Louboutin [11] has shown that for a non-quadratic primitive Dirichlet character χ of conductor c ≥ | L (1 , χ ) | ≥
110 log( c/π ) . Therefore, h ≥ (cid:18) D K s d k s (cid:19) / log ǫR ( c/π ) , where c = 8( s + 2). By Equation (13), R/ log ǫ ≤ R ′ / log ǫ ≤ | s | . Puttingthese together, we obtain the result. (cid:3) It follows that h > | s | ≥ . Since we have computed the class numberof K s for | s | < , we obtain as a corollary that K s has class number 1 only for s = 4.St´ephane Louboutin pointed out to us the following: Let χ be an even Dirichletcharacter of conductor f >
1. Then there is the upper bound (see [10]) | L (1 , χ ) | ≤
12 ( c + log f ) , FAMILY OF CYCLIC QUARTIC FIELDS WITH EXPLICIT FUNDAMENTAL UNITS 11 where c = 2 + γ − log(4 π ) ≈ . · · · (and where γ = . · · · ). Let K be atotally real quartic field of conductor f K corresponding to the quartic character χ K , and let k be its quadratic subfield. Then h K /h k = f K R/ log ǫ | L (1 , χ K ) | ≤ (cid:18) c + log( f K )2 log( f K / (cid:19) f K , where we have used Proposition 8 to bound R/ log ǫ . Therefore, h K /h k < f K when f K > e c ≈ .
01, and a quick search using GP/PARI [13] shows that h K < f K also when f K ≤ f K = 8( s + 2) in the situation of Proposition 10, we seethat the lower bound estimate given there is, up to log factors, the correct order ofmagnitude. 6. Back to cubics
The construction of F s ( t ) was inspired by the cubic case. See [1], which startswith the element (cid:18) f − h ( f + g − f g ) /h − g (cid:19) ∈ PGL ( Q ) , where f, g, h can be taken to be distinct integers with h = 0. Assume that theassociated linear transformation gives the action of a Galois group on a number θ ,and assume that θ has norm 1 to Q . Then we obtain the equation θ · f θ − hθ ( f + g − f g ) /h − g · gθ − hθ ( f + g − f g ) /h − f = 1 , which can be rearranged to say that θ is a root of t + 3( f + g − f g ) − λh ( f + g ) h t + λt − , where λ = ( f + g + h ) / ( f gh ). If h = 1 and λ ∈ Z , we have a polynomialwith integral coefficients. Therefore, we want integral points ( x, y,
1) on the ellipticsurface x + y + 1 = λxy , which is the elliptic modular surface X (3).Following the lead of the quartic case, we look at the singular fiber λ = 3. If f is an integer, we need g to be a root of X − f X + f + 1 = ( X + f + 1)( X − ( f + 1) X + f − f + 1) . When f = 1, the second factor is irreducible over Q , so we must have g = − f − G f ( t ) = t + (9 f + 9 f + 6) t + 3 t − , which could be regarded as the analogue of our family of quartic polynomials. Thiscubic family appears in [7].Let r be a root of the polynomial x + (3 f + 3) x + 3 f x −
1. As pointed out in [7], − r − r is a root of G f ( t ). This can easily be verified by representing r by the matrix A = − f −
30 1 − f , then computing the characteristic polynomial of − A − A .The fields obtained from the polynomials G f ( t ) are some of Shanks’s “simplest cubicfields.” However, − G f ( t ) generate a subgroup of index 3 in thegroup of units generated by the roots of the polynomial x + (3 f + 3) x + 3 f x − In the cubic case, there are also many families of polynomials corresponding tocurves on the surface X (3) that do not lie in the singular fiber λ = 3 (see [1]). Itwould be interesting to find similar families in the quartic case. Acknowledgment.
The authors thank St´ephane Louboutin for several helpfulsuggestions.
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Department of Mathematics, University of Maryland, College Park, MD 20742
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