THE SPLITTING FIELD OF X3 − 2 OVER Q KEITH CONRAD In this note, we calculate all the basic invariants of the number field p K = Q( 3 2;!); p where ! = (−1 + −3)=2 is a primitive cube root of unity. For example, the unit group has rank 2 by the Dirichlet Unit Theorem, and we'll find a basis. Here is the notation for the fields and Galois groups to be used. Let p k = Q( 3 2); p K = Q( 3 2;!); p F = Q(!) = Q( −3); ∼ G = Gal(K=Q) = S3; ∼ N = Gal(K=F ) = A3; H = Gal(K=k): First we work out the basic invariants for the fields F and k. Theorem 1. The field F = Q(!) has ring of integers Z[!], class number 1, discriminantp −3, and unit group {±1; ±!; ±!2g. The ramified prime 3 factors as 3 = −( −3)2. For p 6= 3, the way p factors in Z[!] = Z[X]=(X2 + X + 1) is identical to the way X2 + X + 1 factors mod p, so p splits if p ≡ 1 mod 3 and p stays prime if p ≡ 2 mod 3. We now turn to the field k, which will bep discussed in more detail.p 3 3 Though we will show below that Ok = Z[ 2], the inclusion Z[ 2]p⊂ Opk already is useful for computing the class number. It tells us that j disc(k)j ≤ j disc(1; 3 2; 3 4)j = 108 = 2233, so the Minkowski bound for k is at most p 3! 4 16 3 pj disc(k)j = ≈ 2:94: 33 π 3π p 3 3 In Ok, the rational prime 2 has the principal prime factorization (2) = ( 2) , so k has class number 1: the ring Ok hasp unique factorization. 3 Next we showpOk =pZ[ 2]. Let α = a+b 3 2+c 3 4 be an algebraic integer, with a; b; c all rational. Computing Tr p p k=Q of α, α 3 2, and α 3 4 we see 3a; 6b; 6c 2 Z. So the denominators of a, b, and c involve at most 2 and 3. To show 2 and 3 do not appear in the denominator, we consider the situation p-adically for p = 2 and p = 3. Lemma 1. Let L=L0 be an extension of local fields. If L=L0 is unramified and a is any 0 integer of L which generates the residue field extension, then OL = OL0 [a]. If L=L is totally ramified and λ is any prime of L, then OL = OL0 [λ]. Proof. By [2, Prop. 3, p. 59], which can be applied since the residue field extension is separable, if λ is any prime of L and a is any integer of L which generates the residue field 1 2 KEITH CONRAD 0 of L over that of L , then powers of a or powers of a + λ generate OL as an OL0 -algebra. When L=L0 is unramified, let λ be a prime of L0. When L=L0 is totally ramified, let a = 1 (or 0). p 0 0 3 3 We apply this with L = Q2 or Q3 and L = L ( 2). Since X − 2 is Eisenstein in 3 3 2 Q2[X] and (X − 1) −p2 = X −p3X + 3X − 3 is Eisenstein in Q3[pX], by Lemmap1 the 3 3 3 3 ring ofp integers ofpQ2( 2) is Z2[ 2] and the ring of integers of Q3( 2) = Q3(1 + 2) is 3 3 Z3[1 + 2] = Z3[ 2]. p 3 Using this to calculate the ring of integers of Q(p2), we see that the coefficients of α 3 have denominators prime to 2 and 3, hence Ok = Z[ 2] and p p disc(k) = disc(1; 3 2; 3 4) = −108 For a different approach (pun intended) to calculating the integers of k, see [2, Example, pp. 67-68]. We now turn to factorization of rational primes in Ok. First let's treat the ramified primes, 2p and 3. p In Q ( 3 2), 3 is totally ramified and 1 + 3 2 is a prime element. This number also 3 p def 3 generates the global prime ideal factor of 3 in Ok. Indeed, let π = 1 + 2 be the root of (X − 1)3 − 2 in k. Expanding this polynomial and substituting X = π, we see p p (1) π3 = 3π2 − 3π + 3 = 3(π2 − π + 1) = 3(1 + 3 2 + 3 4): p p p def 3 3 def 3 × The number u = 1 + 2 + 4 has inverse v = 2 − 1, so u 2 Ok and (2) 3 = π3v: Two calculations which we'll be using later are p p p (3) πv = 3 4 − 1; π2v = 1 − 3 2 + 3 4: p The way p factors in Z[ 3 2] can be described by how X3 − 2 factors mod p [2, Prop. 25, p. 27]. We need to know when 2 is a cube mod p and, if so (and p 6= 2), when there is a primitive cube root of unity mod p. The latter condition (for p 6= 3) is the same as determining when −3 is a squarep mod p, which by quadratic reciprocity occurs when p ≡ 1 mod 3. (Alternatively, that −3 2 Z=pZ , (Z=pZ)× contains an element of order 3, −3 p i.e. that 3jp − 1, proves a particular case of quadratic reciprocity: ( p ) = ( 3 ).) Now we factor primes p ≥ 5 (so p is unramified). If p ≡ 2 mod 3, then 2 is a cube mod p and there is no primitive cube root of unity in Z=pZ, so (since h(k) = 1) 2 (p) = (x1)(x2); N(x1) = p; N(x2) = p : Here, N denotes the norm of an ideal. Passing to norms of elements, since −1 is a norm from 2 k we can choose suitable unit multiples so p = x1x2 where p = Nk=Q(x1), p = Nk=Q(x2). For example, p p p 5 = (1 + 3 4)(1 + 2 3 4 − 3 4): If p ≡ 1 mod 3 and 2(p−1)=3 ≡ 1 mod p (the smallest such p is 31) then X3 − 2 splits completely mod p and p = x1x2x3; Nk=Q(xi) = p: If p ≡ 1 mod 3 and 2(p−1)=3 6≡ 1 mod p then X3 − 2 is irreducible mod p and p is prime in Ok. p What is the unit group of Z[ 3 2]? The only roots of unity are ±1, and by Dirichlet the unit group has rank 1, so there is a single fundamental unit > 1. We already have found a unit greater than 1, namely u, and happily this turns out to be the fundamental unit. To THE SPLITTING FIELD OF X3 − 2 OVER Q 3 prove this, we'll show u is less than the square of the fundamental unit, so it must equal the fundamental unit. (The unit u is not so hard to discover independently of any considerations of how 3 factors, which was the way we came across it. The norm form for k is p p 3 3 3 3 3 (4) Nk=Q(a + b 2 + c 4) = a + 2b + 4c − 6abc: Glancingp atp the coefficientsp 1; 2; 4; −6 on the right hand side, we see they add up to 1, so 1 + 3 2 + 3 4 is a unit of Z[ 3 2], with norm 1.) Lemma 2. Let k=Q be any cubic extension which is not Galois, with negative discriminant d, i.e. k has only one real embedding. View k as a subfield of R by its unique real embedding. 3 Then for the fundamental unit U > 1 of Ok, jdj=4 < U + 7. Proof. Let σ : k ! C be one of the complex embeddings of k, so Nk=Q(U) = Uσ(U)σ(U) = 2 Ujσ(U)j > 0, so Nk=Q(U) = 1. Let x = jσ(U)j = p1 2 (0; 1). Write U 1 σ(U) = xeiy; σ(U) = xe−iy;U = : x2 2 Since Z[U] ⊂ Ok, jdj ≤ j disc(1; U; U )j. We calculate the latter: 2 1 UU 2 2 disc(1; U; U ) = 1 σ(U) σ(U)2 1 σ(U) σ(U)2 = ((σ(U) − σ(U))(σ(U) − U)(σ(U) − U))2 = −4 sin2(y)(x3 + 1=x3 − 2 cos y)2 < 0: Taking absolute values, 1 jdj ≤ sin2(y)(x3 + 1=x3 − 2 cos y)2 4 = (1 − c2)(z − 2c)2 (c = cos y; z = x3 + 1=x3 > 2) = (z2 − 4cz + 4c2)(1 − c2) = z2 − c2z2 − 4cz(1 − c2) + 4c2(1 − c2) = z2 − (cz + 2(1 − c2))2 + 4(1 − c2)2 + 4c2(1 − c2) = z2 + 4 − (cz + 2(1 − c2))2 < z2 + 4 1 = x6 + 6 + x6 = U 3 + 6 + x6 < U 3 + 7: p 3 3 Applying this to k = Q( 2),p the fundamentalp unit U > 1 of Ok satisfies 108=4 < U + 7, 2 3 3 so U > 2, hence U > 4 > 1 + 2 + 4 = u. Thus U = u. p 2 3 By an explicit calculation, disc(1p ; u;p u ) = −108 so Z[u] = Z[ 2], which also follows from the transition matrix from f1; 3 2; 3 4g to f1; u; u2g having determinant −1; explicitly, 4 KEITH CONRAD p 3 2 = u2 − 3u + 2. The minimal polynomial for u over Q can be found as follows: 1 1 3 (5) u = p ) + 1 = 2 ) u3 − 3u2 − 3u − 1 = 0: 3 2 − 1 u p p 3 3 Theorem 2.