MATH 248A. RINGS OF WITHOUT A POWER

KEITH CONRAD

Let K be a number field, with degree n and of integers OK . When OK = Z[a] for n−1 some a ∈ OK , the set {1, a, . . . , a } is a Z-basis of OK . We call such a basis a power basis. When K is a quadratic field or a cyclotomic field, OK admits a power basis, and the use of these two fields as examples in theory can lead to the impression that rings of integers should always have a power basis. This is false. While it is always true that

OK = Ze1 ⊕ · · · ⊕ Zen for some algebraic integers e1, . . . , en, quite often we can not choose the ei’s to be powers of a single number. The first example of a ring of integers lacking a power basis is due to Dedekind. It is the field Q(θ) where θ is a root of X3 + X2 − 2X + 8. The ring of integers of Q(θ) has Z-basis {1, θ, 4/θ} but no power basis. We will return to this historically distinguished example later, but the main purpose of the discussion here is to give infinitely many examples of number fields whose ring of integers does not have a power basis. Our examples will be Galois cubic extensions of Q. × Fix a prime p ≡ 1 mod 3. The field Q(ζp)/Q has cyclic Galois group (Z/pZ) , and in particular there is a unique cubic subfield Kp, so Kp/Q is Galois with degree 3. The Galois × group Gal(Kp/Q) is the quotient of (Z/pZ) by its subgroup of cubes. In particular, for any prime q 6= p, q splits completely in Kp if and only if its Frobenius in Gal(Kp/Q) is trivial, which is equivalent to q being a cube modulo p.

Theorem 1 (Hensel). If p ≡ 1 mod 3 and 2 is a cube in Z/pZ, then OKp 6= Z[a] for any a ∈ OKp .

Proof. Suppose OKp = Z[a] for some a. Let a have minimal polynomial f(X) over Q, so f is an irreducibe cubic in Z[X]. Then ∼ OKp = Z[a] = Z[X]/f(X).

Since 2 is a cube mod p, 2 splits completely in Kp, so f splits completely in (Z/2Z)[X]. However, a cubic in (Z/2Z)[X] can’t split completely: there are only two (monic) linear polynomials mod 2.  The set of primes which fit the hypotheses of Theorem 1 are those p ≡ 1 mod 3 such that 2(p−1)/3 ≡ 1 mod p. These are the primes which split completely in the splitting field of X3 − 2 over Q, and there is a positive proportion of such primes. The first few of them are 31, 43, 109, and 127. These primes give different cubic Galois extensions of Q whose ring of integers lacks a power basis. Hensel actually proved a result which is stronger than Theorem 1: for p ≡ 1 mod 3, 2 is a cube mod p if and only if the index [OKp : Z[a]] is even for all a ∈ OKp . The proof that Dedekind’s example lacks a power basis operates on the same principle as Theorem 1: show 2 splits completely in the integers of Q(θ), and that implies there is no power basis for the same reason as in Theorem 1. However, to show 2 splits completely 1 2 KEITH CONRAD requires different techniques than the ones we used, since Dedekind’s field does not lie in a cyclotomic field. For details, see [2, p. 101].

The reader may naturally want to see an example of a Z-basis of OKp . From now on we forget about whether or not OKp has a power basis and simply take for p any prime that is 1 mod 3. Theorem 2. Let X a η0 = ζp . a(p−1)/3≡1 mod p Fixing an element r ∈ (Z/pZ)× with 3, let X a X a η1 = ζp , η2 = ζp . a(p−1)/3≡r mod p a(p−1)/3≡r2 mod p

Then OKp = Zη0 + Zη1 + Zη2.

The numbers ηi are examples of (cyclotomic) periods [3, pp. 16–17]. × c Proof. For c ∈ (Z/pZ) , let σc ∈ Gal(Q(ζp)/Q) send ζp to ζp. It is left to the reader to check that η0, η1, η2 are Q-conjugates, so any of them generates Kp/Q (there is a unique cubic subfield of Q(ζp)).

Write L = Zη0 + Zη1 + Zη2, so L is a rank 3 sublattice of OKp . Therefore 2 (1) disc(L) = [OKp : L] disc(OKp ). We now compute the discriminant of L, which is the 3 × 3 determinant 2 Tr(η0) Tr(η0η1) Tr(η0η2) 2 disc(L) = Tr(η1η0) Tr(η1) Tr(η1η2) , 2 Tr(η2η0) Tr(η2η1) Tr(η2) where Tr = TrKp/Q. Since η0, η1, η2 are Q-conjugates, as are η0η1, η0η2, and η1η2, we have 2 Tr(η0) Tr(η0η1) Tr(η0η1) 2 disc(L) = Tr(η0η1) Tr(η0) Tr(η0η1) 2 Tr(η0η1) Tr(η0η1) Tr(η0) = a3 − 3ab2 + 2b3, 2 where a = Tr(η0) and b = Tr(η0η1). P a 2 The trace of η0 is η0 + η1 + η2 = (a,p)=1 ζp = −1. To compute the trace of η0, we compute 2 X X a+b η0 = ζp a(p−1)/3=1 b(p−1)/3=1 X X a(1+b) = ζp a(p−1)/3=1 b(p−1)/3=1 X X a(1+b) = ζp b(p−1)/3=1 a(p−1)/3=1 p − 1 X = + σ (η ) 3 1+b 0 b(p−1)/3=1,b6=−1 p − 1 = + c η + c η + c η , 3 0 0 1 1 2 2 where (p−1)/3 (p−1)/3 c0 = #{b 6= 0, −1 : b = 1, (1 + b) = 1}, MATH 248A. RINGS OF INTEGERS WITHOUT A POWER BASIS 3

(p−1)/3 (p−1)/3 c1 = #{b 6= 0, −1 : b = 1, (1 + b) = r}, (p−1)/3 (p−1)/3 2 c2 = #{b 6= 0, −1 : b = 1, (1 + b) = r }. (Recall r and r2 are the elements of order 3 in (Z/pZ)×.) 2 Taking the trace of η0 gives 2 Tr(η0) = (p − 1) + (c0 + c1 + c2) Tr(η0) = p − 1 − (c0 + c1 + c2). (p−1)/3 The sum of the ci’s is the number of solutions to b = 1 in Z/pZ except for b = −1, so p − 1  2 (2) Tr(η2) = p − 1 − − 1 = (p − 1) + 1. 0 3 3 Writing 2 2 2 2 Tr(η0) = η0 + η1 + η2 2 = (η0 + η1 + η2) − 2(η0η1 + η1η2 + η0η2) 2 = (Tr η0) − 2 Tr(η0η1)

(3) = 1 − 2 Tr(η0η1), we compare (2) and (3) to see that Tr(η0η1) = −(p − 1)/3. 2 Feeding in the formulas for the trace of Tr(η0) and Tr(η0η1) into the discriminant for L gives (after some careful algebra) disc(L) = p2.

Substitute this into (1). Since disc(OKp ) must have a prime factor (in fact, since Kp ⊂ Q(ζp) 2 we know p divides disc(OKp )), the only way for (1) to hold is for disc(OKp ) = p and [OKp : L] = 1. Thus OKp = L. 

We know now that η0 has minimal polynomial p − 1 f(X) = X3 + X2 − X − η η η , 3 0 1 2 and 2 2 disc(Z[η0]) = [OKp : Z[η0]] p . The first few p ≡ 1 mod 3 are 7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97. For each of these primes, using PARI it appears that f(X) has discriminant (pB)2, where 4p = A2 + 27B2 (such an equation, valid for p ≡ 1 mod 3, determines A and B up to sign). For a more rigorous discussion of this, see the paper of Gras [1].

This discriminant formula translates into an index formula: [OKp : Z[η0]] = |B|. Let η0η1η2 = cp. The general formula 4 1  2 1 disc(X3 + X2 − ((p − 1)/3)X + c) = p3 − p2 + 6c + p − 27c2 + 2c − , 27 3 9 27 2 with c = cp and set equal to (pB) leads, after some algebra, to

27cp = 1 − 3p ± Ap. Up to here A is only determined up to sign, so this serves to fix a choice of sign, namely if 27cp = 1 − 3p − Ap then A ≡ 1 mod 3. Therefore if p ≡ 1 mod 3 and we write 4p = A2 + 27B2 where A ≡ 1 mod 3, then the period η0 has minimal polynomial X3 + X2 − ((p − 1)/3)X + (1 − (A + 3)p)/27 4 KEITH CONRAD

2 and disc(Z[η0]) = (pB) . In particular, this cubic polynomial defines the cubic extension of Q inside Q(ζp) when p ≡ 1 mod 3. The first such p for which 2 mod p is a cube and the class number of Kp is greater than 3 2 1 is p = 277. The cubic subfield of Q(ζ277) is generated by a root of T + T − 92T + 236 and the class number of this cubic field is h = 4.

References [1] M-N. Gras, M´ethodes et algorithmes pour le calcul num´eriquedu nombre de classes et des ´esdes extensions cubiques cycliques de Q, J. Reine Angew. Math. 277 (1975), 89–116. [2] P. Samuel, “,” Houghton Mifflin, Boston, 1969. [3] L. Washington, “An Introduction to Cyclotomic Fields,” 2nd ed., Springer-Verlag, New York, 1997.