RINGS OF INTEGERS WITHOUT A POWER BASIS

RINGS OF INTEGERS WITHOUT A POWER BASIS KEITH CONRAD Let K be a number field, with degree n and ring of integers O K. When O K = Z[α] for some α O K, the set {1, α,..., α n 1 } is a Z-basis of O K. We call such a basis a ower basis. When K is a quadratic field or a cyclotomic field, O K admits a ower basis, and the use of these two fields as examles in algebraic number theory can lead to the imression that rings of integers should always have a ower basis. This is false. While it is always true that O K = Ze 1 Ze n for some algebraic integers e 1,..., e n, quite often we can not choose the e i s to be owers of a single number. The first examle of a ring of integers lacking a ower basis is due to Dedekind. It is the field Qθ) where θ is a root of T T 2 2T 8. The ring of integers of Qθ) has Z-basis {1, θ, θ+θ 2 )/2} but no ower basis. We will return to this historically distinguished examle later, but the main urose of the discussion here is to give infinitely many examles of number fields whose ring of integers does not have a ower basis. Our examles will be Galois cubic extensions of Q. Fix a rime 1 mod. The field Qζ )/Q has cyclic Galois grou Z/Z), and in articular there is a unique cubic subfield F, so F /Q is Galois with degree. The Galois grou GalF /Q) is the quotient of Z/Z) by its subgrou of cubes. In articular, for any rime q, q slits comletely in F if and only if its Frobenius in GalF /Q) is trivial, which is equivalent to q being a cube modulo. Theorem 1 Hensel). If 1 mod and 2 is a cube in Z/Z, then O F α O F. Z[α] for any Proof. Suose O F = Z[α] for some α. Let α have minimal olynomial ft ) over Q, so f is an irreducibe cubic in Z[T ]. Then O F = Z[α] = Z[T ]/ft ). Since 2 is a cube mod, 2 slits comletely in F, so f slits comletely in Z/2Z)[T ]. However, a cubic in Z/2Z)[T ] can t slit comletely: there are only two monic) linear olynomials mod 2. The set of rimes that fit the hyotheses of Theorem 1 are those 1 mod such that 2 1)/ 1 mod. These are the rimes that slit comletely in the slitting field of T 2 over Q, and by the Chebotarev density theorem there is a ositive roortion recisely, 1/6) of such rimes. The first few of them are 1, 4, 109, and 1. For each such, Theorem 1 tells us the ring of integers of F does not have a ower basis. Hensel actually roved a result that is stronger than Theorem 1: for 1 mod, 2 is a cube mod if and only if the index [O F : Z[α]] is even for all α O F Z. 1

2 KEITH CONRAD The roof that the integer ring of Dedekind s field Qθ) lacks a ower basis oerates on the same rincile as Theorem 1: show 2 slits comletely in the integers of Qθ), and that imlies there is no ower basis for the same reason as in Theorem 1. However, to show 2 slits comletely in Qθ) requires different techniques than the ones we used in the fields F, since Dedekind s field does not lie in a cyclotomic field. If K is a number field, a criterion for O K not to have a ower basis is that some rime number divides every index [O K : Z[α]] for α O K. When K is a cubic field, the only such rime could be 2 theorem of Engstrom), and this criterion alies that is, [O K : Z[α]] is even for all α O K ) if and only if 2 slits comletely in O K. See [1] for further details, including a roof that there are infinitely many cubic fields without a ower basis for which this even-index criterion does not aly. The integer ring of F for 1 mod ) has some Z-basis, which we now know need not be a ower basis. What is a Z-basis for this ring? Theorem 2. For 1 mod let η 0 = Tr Qζ)/F ζ ) = Fixing an element r Z/Z) with order, let η 1 = ζ, t η 2 = Then O F = Zη 0 + Zη 1 + Zη 2. t 1)/ r mod t 1)/ 1 mod ζ t. t 1)/ r 2 mod The numbers η i are examles of cyclotomic) eriods [5,. 16 17]. Proof. For c Z/Z), c 1)/ is a cube root of unity and therefore is in {1, r, r 2 }. For suitable c each of the three values is achieved. Let σ c GalQζ )/Q) send ζ to ζ. c Then σ c η i ) = σ c ζ t = ζ ct = ζ, t t 1)/ r i mod t 1)/ r i mod ζ t. t 1)/ c 1)/ r i mod so σ c ermutes {η 0, η 1, η 2 } the same way multilication by c 1)/ ermutes {1, r, r 2 }. Therefore η 0, η 1, and η 2 are Q-conjugates, and since F is the unique cubic subfield of Qζ ) any of η 0, η 1, η 2 generates F as a field extension of Q. The standard basis of Qζ ) over Q is {1, ζ,..., ζ 2 }, and this is also a basis for Z[ζ ] over Z. It is more convenient to multily through by ζ and use instead B = {ζ, ζ,..., ζ 1 } as a basis of Qζ ) over Q or Z[ζ ] over Z, since this is a basis of Galois conjugates a normal basis over Q). For examle, η 0, η 1, η 2 are sums of different numbers in B, so η 0, η 1, and η 2 are linearly indeendent over Q and thus form a Q-basis of F. Each x O F can be written as x = a 0 η 0 + a 1 η 1 + a 2 η 2 for rational a 0, a 1, and a 2. On the left side, since x is an algebraic integer in Qζ ) it is a Z-linear combination of the numbers in B. Exanding the right side in terms of B, the coefficients are a 0, a 1, and a 2, so comaring coefficients of the numbers in B on both sides shows a 0, a 1, and a 2 are all integers. To measure how far Z[η 0 ] is from the full ring of integers O F we d like to comute the index [O F : Z[η 0 ]]. This can be exressed in terms of discriminants: if ft ) Z[T ] is the

minimal olynomial of η 0 over Q then RINGS OF INTEGERS WITHOUT A POWER BASIS 1) discft )) = [O F : Z[η 0 ]] 2 disco F ). What are these discriminants? Theorem. For 1 mod, disco F ) = 2. Proof. The discriminant of O F is the determinant Trη0 2 ) Trη 0η 1 ) Trη 0 η 2 ) Trη 1 η 0 ) Trη1 2 ) Trη 1η 2 ) Trη 2 η 0 ) Trη 2 η 1 ) Trη2 2), where Tr = Tr F/Q. Since η 0, η 1, η 2 are Q-conjugates, as are η 0 η 1, η 0 η 2, and η 1 η 2, we have Trη0 2 disco F ) = ) Trη 0η 1 ) Trη 0 η 1 ) Trη 0 η 1 ) Trη0 2 ) Trη 0η 1 ) Trη 0 η 1 ) Trη 0 η 1 ) Trη0 2) 2) = a ab 2 + 2b, where a = Trη0 2) and b = Trη 0η 1 ). The trace of η 0 is Trη 0 ) = η 0 + η 1 + η 2 = t,)=1 ζt = 1. To comute Trη0 2 ), we comute η0 2 = ) where = = a 1)/ =1 b 1)/ =1 a 1)/ =1 b 1)/ =1 b 1)/ =1 a 1)/ =1 = 1 = 1 + ζ a+b ζ a1+b) ζ a1+b) b 1)/ =1,b 1 σ 1+b η 0 ) + c 0 η 0 + c 1 η 1 + c 2 η 2, c 0 = {b 0, 1 : b 1)/ = 1, 1 + b) 1)/ = 1}, c 1 = {b 0, 1 : b 1)/ = 1, 1 + b) 1)/ = r}, c 2 = {b 0, 1 : b 1)/ = 1, 1 + b) 1)/ = r 2 }. Recall r and r 2 are the elements of order in Z/Z).) Taking the trace of both sides of ) gives Trη 2 0) = 1) + c 0 + c 1 + c 2 ) Trη 0 ) = 1 c 0 + c 1 + c 2 ). The sum of the c i s is the number of solutions to b 1)/ = 1 in Z/Z excet for b = 1, so ) 1 4) Trη0) 2 = 1 1 = 2 1) + 1.

4 KEITH CONRAD 5) Writing Trη0) 2 = η0 2 + η1 2 + η2 2 = η 0 + η 1 + η 2 ) 2 2η 0 η 1 + η 1 η 2 + η 0 η 2 ) = Tr η 0 ) 2 2 Trη 0 η 1 ) = 1 2 Trη 0 η 1 ), we comare 4) and 5) to see that Trη 0 η 1 ) = 1)/. Feeding the formulas for Trη0 2) and Trη 0η 1 ) into the discriminant formula 2) gives ) 2 ) ) 2 1 2 ) 1 disco F ) = 1) + 1 1) + 1 2 = 2. Remark 4. The discriminant of F can be calculated using ramification rather than a basis: the extension Qζ )/Q is ramified only at, so if K is an intermediate field then it is ramified only at too. Since Qζ )/Q is totally ramified at, K is totally ramified at as well. If a number field is totally ramified at a rime and has degree n not divisible by then it can be shown that its discriminant is divisible by n 1 but not n. Therefore if [K : Q] = d, disck) = ± d 1. The sign of disck) is 1) r 2K) [5, Lemma 2.2], and when 1 mod the cubic field F has r 2 = 0 since a Galois cubic with a real embedding has r 2 = 0, so discf ) = 1 = 2. The first few rimes 1 mod are 7, 1, 19, 1, 7, 4, 61, 67, 7, 79, 97. For every rime 1 mod we can write 4 = A 2 +B 2 and such an equation determines A and B u to sign [,. 119]. The table below gives the ositive A and B for the above. 7 1 19 1 7 4 61 67 7 79 97 A, B) 1,1) 5,1) 7,1) 4,2) 11,1) 8,2) 1,) 5,) 7,) 17,1) 19,1) Numerically, for each of the above a calculation of ft ) as T η 0 )T η 1 )T η 2 ) shows the discriminant of ft ) is B) 2. By 1) and Theorem, the formula discft )) = B) 2 is equivalent to [O F : Z[η 0 ]] = B, so by the above table Z[η 0 ] has index 2 in O F when is 1 and 4, the index is when is 61, 67, and 7, and the index is 1 i.e., O F = Z[η 0 ]) for the other in the table. For a more rigorous discussion of the formula discft )) = B) 2, see [2]. Claude Quitte observed numerically for the above an index formula using A: [O F : Z[η 0 η 1 ]] = A. The formula discft )) = B) 2 leads to a formula for ft ). In terms of its roots η i, ft ) = T η 0 )T η 1 )T η 2 ) = T η 0 + η 1 + η 2 )T 2 + η 0 η 1 + η 0 η 2 + η 1 η 2 )T η 0 η 1 η 2 = T Trη 0 )T 2 + Trη 0 η 1 )T η 0 η 1 η 2 = T 1)T 2 1 T η 0η 1 η 2 = T + T 2 1 T η 0η 1 η 2. We want to write the constant term of ft ) in terms of. The general formula disc T + T 2 + at + b ) = 4a + a 2 + 18ab b 2 4b

with a = 1)/ and b = η 0 η 1 η 2 is 4 1 ) 2 2 + 9 6b b 2 + 2b 1 RINGS OF INTEGERS WITHOUT A POWER BASIS 5 Setting 4 = A 2 + B 2, this discriminant is 2 A 2 + B 2 + b 1 ) ) 2 = 2 A 2 ) = 2 6b 1) b 1)2 4 9 2 = 2 4 + b 1 ) ) 2. + b 1 ) ) 2 + B) 2. Therefore discft )) = B) 2 + b 1 = ±A. Since 1 mod we have + b 1)/ 1 mod, so if we choose the sign on A to make A 1 mod then + b 1)/ = A. Rewrite this as b = 1 A + ))/, so the minimal olynomial of η 0 over Q is 6) ft ) = T + T 2 1 1 A + ) T + with 4 = A 2 + B 2 and A 1 mod. Examle 5. The first fitting the hyotheses of Theorem 1 is = 1, for which 4 = 124 = 4) 2 + 2) 2. Therefore the cubic subfield of Qζ 1 ) is Qη 0 ) where, by 6), η 0 has minimal olynomial T + T 2 1 1 1 14 + ) T + = T + T 2 10T 8. The field Qη 0 ) is a cyclic cubic extension of Q in which 2 slits comletely and its ring of integers has no ower basis. Examle 6. The second fitting the hyotheses of Theorem 1 is = 4, for which 4 = 172 = 8) 2 + 2) 2 we use 8 so that A = 8 1 mod ). Thus the cubic subfield of Qζ 4 ) is Qη 0 ) where η 0 has minimal olynomial T + T 2 4 1 1 4 8 + ) T + = T + T 2 14T + 8 and this cubic field has the same roerties as at the end of the revious examle. The minimal olynomial of η 0 η 1 over Q is T η 0 η 1 ))T η 1 η 2 ))T η 2 η 0 )). Exanding this out, we get 7) T + η 0 η 1 + η 0 η 2 + η 1 η 2 η 2 0 η 2 1 η 2 2)T + η 0 η 1 )η 1 η 2 )η 2 η 0 ). The coefficient of T is Trη 0 η 1 ) Tr η 0 ) 2 = 1)/ 1) 2 = and the constant term is discft )) = ± B. The sign of the constant term in 7) is sensitive to the choice of ζ and nontrivial cube root of unity r mod in the definition of the η i, e.g., changing r can change η 1 into η 2 but η 0 η 1 and η 0 η 2 are not Q-conjugates. In fact η 0 η 2 has minimal olynomial T T B with constant term of oosite sign to the minimal olynomial of η 0 η 1. When the definition of the η i uses ζ = e 2πi/ and r is the numerically least nontrivial cube root of unity mod in {1,..., 1} then for all < 100 such that 1 mod the constant term of 7) turns out to be B excet when = 61.

6 KEITH CONRAD The only < 500 for which 1 mod and the class number of F is greater than 1 are 16, 7, 1, 49, and 97. For these the class grou of F is Z/2Z Z/2Z excet for = 1, when it is Z/7Z. References [1] D. S. Dummit and H. Kisilevsky, Indices in Cyclic Cubic Fields,. 29 42 of Number Theory and Algebra H. Zassenhaus, ed.), Academic Press, New York, 1977. [2] M-N. Gras, Méthodes et algorithmes our le calcul numérique du nombre de classes et des unités des extensions cubiques cycliques de Q, J. Reine Angew. Math. 7 1975), 89 116. [] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, 2nd ed., Sringer- Verlag, New York, 1990. [4] P. Samuel, Algebraic Number Theory, Houghton Mifflin, Boston, 1969. [5] L. Washington, An Introduction to Cyclotomic Fields, 2nd ed., Sringer-Verlag, New York, 1997.