ARITHMETIC IN PURE CUBIC FIELDS AFTER DEDEKIND. IAN KIMING We will study the rings of integers and the decomposition of primes in cubic number fields K of type K = Q( 3 d) where d Z. Cubic number fields of this type are also called pure cubic number fields. Apart from the interest of the results in themselves, our objective is also to illustrate the power of the theory of ideals and prime decomposition: We determined in an exercise the ring of integers of K in a special case. We now deal with the general case and will see that things go much more smoothly when we use the theory of prime decomposition as our main tool. Obviously we may assume that d is cube free and > 1. One sees that one can then (in a unique fashion) write d in shape: d = ab 2 where a, b N, gcd(a, b) = 1, ab > 1, and ab square free. It will then be convenient to define: α := 3 ab 2 β := 3 a 2 b, so that α 2 = bβ. So, both α and β are both non-rational numbers in K; each of them is root of a certain monic polynomial of degree 3 with integer coefficients so they are also both algebraic integers in K. The above equation shows that 1, α, β is a Q-basis of K. It will be convenient to consider the number θ := α a O K. One sees that θ satisfies the equation: ( ) θ 3 + 3aθ 2 + 3a 2 θ + a(a 2 b 2 ) = 0 which must be the minimal equation for θ as clearly K = Q(θ). We denote by D K the discriminant of K, i.e., the discriminant of the ring O K of integers in K. Theorem 1. (i). Suppose that a 2 b 2 (9). Then O K = Z(1, α, β) and D K = 27a 2 b 2. (ii). Suppose a 2 b 2 (9). Put γ := 1 + aα + bβ 3. Then O K = Z(α, β, γ) and D K = 3a 2 b 2. This result was first proved by Dedekind, cf. [1]. We will break the proof up into a number of smaller steps. 1
2 IAN KIMING Lemma 1. We have (1, α, β) = 27a 2 b 2, and the index [O K : Z(1, α, β)] is a divisor of 3ab. Proof. Let ζ be a primitive 3 rd root of unity, i.e., a root of x 2 + x + 1. The 3 embeddings K C are then characterized by (α, β) (α, β), (α, β) (ζα, ζ 2 β), and (α, β) (ζ 2 α, ζβ), respectively. Now we compute: 1 α β 1 1 1 det 1 ζα ζ 2 β = αβ det 1 ζ ζ 2 = αβ (ζ 1)(ζ 2 1)(ζ 2 ζ) 1 ζ 2 α ζβ 1 ζ 2 ζ = αβ (ζ 3 ζ ζ 2 + 1)(ζ 2 ζ) = 3αβ (ζ 2 ζ), where we used the formula for a Vandermonde determinant. We know that the discriminant (1, α, β) is the square of this determinant so the first claim now follows from α 2 β 2 = a 2 b 2 and (ζ 2 ζ) 2 = ζ + ζ 2 2 = 3. By general theory we have then: 27a 2 b 2 = (1, α, β) = D K [O K : Z(1, α, β)] 2 which first implies (since D K is an integer) that 3D K is a square, and then that [O K : Z(1, α, β)] 2 is a divisor of (3ab) 2. Hence the second claim. So now the task at hand is to determine the index [O K : Z(1, α, β)]. The main idea is that studying the decomposition of prime divisors of 3ab will be helpful in this determination. Proposition 1. Suppose that p is a prime number dividing ab. The p decomposes in K as: with p a prime ideal of O K. po K = p 3 Proof. Suppose that p j a and let p be a prime divisor of p in K. Then p divides ab 2 = α 3, so that p must divide α. Then p 3 divides α 3 = ab 2. As gcd(a, b) = 1 and as ab is square free we have that ab 2 has form pm where p m. Then m is not divisible by p (since otherwise m Z p = Zp). Considering the prime decomposition of pm we then deduce that p 3 divides p. But then p necessarily has the stated decomposition as [K : Q] = 3 (remember that if po K = p e1 1 p es s where the prime ideals p i has inertia degree f i then 3 = [K : Q] = i e if i ). If p j b we can work similarly with β O K to obtain the desired conclusion. Proposition 2. The prime decomposition of 3 in K is as follows: { p 3, if a 2 b 2 (9) 3O K = p 2 1p 2, if a 2 b 2 (9).
ARITHMETIC IN PURE CUBIC FIELDS AFTER DEDEKIND. 3 Proof. If 3 j ab then a 2 b 2 (9) since ab is square free. The desired conclusion then follows from the previous proposition. So assume now that 3 ab. It is convenient to use some of our standard notation: For ω O K, ω 0, the integer ν p (ω) is defined as the power of p in the prime factorization of ω (that is, of the ideal ωo K ). Similarly we have ν p (c) for integers c 0 defined as the maximal power of p dividing c. Consider the number θ = α a satisfying the equation ( ) and let θ, θ, θ be the three conjugates of θ. Since now a 2 b 2 is divisible by 3 (as 3 ab) we see then that 3 divides θ 3 (as ideals in O K ). On the other hand we obtain from ( ) that ( ) 3a 2 = θθ + θθ + θ θ which implies 3 θ: For if 3 j θ then also 3 j θ and 3 j θ (consider an embedding g of K into C. Then g maps K to some algebraic number field K, and in fact any ideal of O K to an ideal of O K ; so g maps 3O K to 3O K because 3 Q. But it maps θo K to gθo K ). But then ( ) would imply 9 j 3a 2 and so 3 j a, contrary to assumption. Now we see that 3 can not split as a product of three distinct prime ideals: For if 3O K = p 1 p 2 p 3 with the p i prime ideals then p i j θ 3 for each i because 3 j θ 3 ; thus p i j θ for each i; were the p i distinct we would obtain 3O K = p 1 p 2 p 3 j θ, contradicting the above conclusion. It follows that 3 has a prime divisor p 1 such that p 2 1 j 3. We can then write: 3O K = p 2 1p 2 where a priori p 2 is just some ideal. However, upon taking norms we obtain 3 3 = N(3O K ) = N(p 1 ) 2 N(p 2 ); since N(p 1 ) is a non-trivial power of 3 we are forced to conclude N(p 2 ) = 3 which means that p 2 is a prime ideal (recall that N(p 2 ) = [O K : p 2 ]; since this index is 3 the ideal p 2 must be maximal). The decisive question is now whether the prime ideals p 1 and p 2 are equal. We see that we can finish the proof by showing p 1 = p 2 a 2 b 2 (9). To prove this it is convenient to write the equation ( ) as ( ) θ 3 + 3aαθ + a(a 2 b 2 ) = 0 (recall that θ := α a), and to define r := ν 3 (a 2 b 2 ) and s := ν p1 (θ). Suppose that p 1 = p 2 so that 3O K = p 3 1. Then 1 s 2: For we have 3 j θ 3 so that surely s 1; but we also know 3 θ, i.e., s < 3. Now, ν p1 (α) = 0 since otherwise 3O K = p 3 1 would divide α 3 = ab 2 contrary to our assumption 3 ab. Then we have: ν p1 (θ 3 ) = 3s, ν p1 (3aαθ) = 3 + s, ν p1 (a 2 b 2 ) = 3r. We must have 3s 3 + s (since s is not divisible by 3). Thus: ν p1 (θ 3 + 3aαθ) = minf3s, 3 + sg
4 IAN KIMING and by ( ) this minimum must coincide with 3r. But we can not have 3r = 3 + s since s is not divisible by 3. Hence, 3r = 3s < 3 + s 5 which gives first s = 1, and then r = 1. I.e., 9 a 2 b 2. Suppose then that p 1 p 2. Since 3 j θ 3 we also have p 2 j θ. Thus ν p2 (θ 3 + 3aαθ) 2. But by ( ) this valuation coincides with ν p2 (a(a 2 b 2 )) = ν p2 (a 2 b 2 ) = ν 3 (a 2 b 2 ). Thus 9 j a 2 b 2. Lemma 2. Suppose that p is a prime number dividing ab. Then p does not divide [O K : Z(1, α, β)]. Proof. Suppose that p divides both ab and the index [O K : Z(1, α, β)]. Since p divides this index there is a number ω O K such that ω Z(1, α, β) but pω Z(1, α, β). We will obtain a contradiction from this. ( ) As pω Z(1, α, β) we can write: pω = x + yα + zβ with x, y, z Z. Now, as p divides ab we have according to the previous proposition that p splits as po K = p 3. This decomposition implies: ν p (c) = 3 ν p (c), for c Z. Since p j ab we have that p divides either a or b. Suppose first that p j a. Then 3 ν p (α) = ν p (α 3 ) = ν p (ab 2 ) = 3 ν p (ab 2 ) = 3 as gcd(a, b) = 1 and ab square free. It follows that ν p (α) = 1. Working similarly with β 3 = a 2 b we find that ν p (β) = 2. Consider now the equation ( ). The equation implies x + yα + zβ 0 (p 3 ) whence in particular x 0 (p) as α β 0 (p). Since x Z we deduce p j x. Then x 0 (p 3 ) so that now yα + zβ 0 (p 3 ). Since β 0 (p 2 ) we have then yα 0 (p 2 ). As ν p (α) = 1 we must then have ν p (y) > 0, i.e., that p j y. But then y 0 (p 3 ) and so which implies p j z because ν p (β) = 2. zβ 0 (p 3 ) We now have that each of x, y, and z is divisible by p. But then ( ) shows that ω is in Z(1, α, β), contrary to assumption.
ARITHMETIC IN PURE CUBIC FIELDS AFTER DEDEKIND. 5 If p divides b we obtain a contradiction in an entirely similar fashion (we will then have ν p (α) = 2 and ν p (β) = 1). Proof of Theorem 1. By lemmas 1 and 2 the index k := [O K : Z(1, α, β)] is a divisor of 3ab but is not divisible by any prime divisor of ab. This implies that k is 1 or 3, and that k = 1 if 3 j ab. Also, by general theory and lemma 1 we have D K = (1, α, β)/k 2 = 3 (3/k) 2 a 2 b 2. We conclude that the proof of the theorem is finished if we can show k = 1 a 2 b 2 (9), and that O K = Z(α, β, γ) if k = 3 (where γ is defined as in the statement of the theorem). We may assume 3 ab since 3 j ab clearly implies a 2 b 2 (9), and also k = 1 as we just noted. Suppose first that a 2 b 2 (9). By Proposition 2 we have then 3O K = p 3 with p a prime ideal. From equation ( ) for θ = α a we first see that p j θ, and then that p 2 θ: For otherwise 3 ν 3 (a 2 b 2 ) = ν p (a 2 b 2 ) would be greater than 3 and so at least 6 whence ν 3 (a 2 b 2 ) 2, contradicting our assumption a 2 b 2 (9). Let now ω be an arbitrary number in O K. Since k is a divisor of 3 we certainly have 3ω Z(1, α, β), i.e., we have: 3ω = x + yα + zβ with x, y, z Z. It follows that 3bω = bx + by(θ + a) + z(θ + a) 2 since θ = α a and bβ = α 2. Thus: ( ) 3bω = x + y θ + z θ 2 where ( ) x := bx + aby + a 2 z, y := by + 2az, z := z. Now, ( ) certainly implies x + y θ + z θ 2 0 (p 3 ). Since p j θ but p 2 θ we can then deduce x y z 0 (3) in a completely similar way as in the proof of Lemma 2. Then ( ) implies x y z 0 (3) (as 3 ab). But then we have ω Z(1, α, β). We conclude that k = 1. Suppose then that a 2 b 2 (9). By Proposition 2 we have now 3O K = p 2 1p 2 with p 1, p 2 prime ideals. Equation ( ) for θ = α a certainly gives 3 j θ 3. We can then deduce p 1 p 2 j θ whence 3p 2 = p 2 1p 2 2 j θ 2, and so in particular: 3 j θ 2 = (α a) 2 = α 2 2aα + a 2 = bβ 2aα + a 2 = (1 + aα + bβ) + (a 2 1 3aα). As 3 a we have a 2 1 (3) so that certainly 3 j (a 2 1 3aα). We conclude that 3 j (1 + aα + bβ), i.e., that the number γ := 1 + aα + bβ 3
6 IAN KIMING is in O K. Since clearly γ Z(1, α, β) we cannot have k = 1, so k = 3. Since this index is a prime number we have O K = Z(1, α, β, η) if η is any number in O K that is not in Z(1, α, β). In particular we have O K = Z(1, α, β, γ) = Z(α, β, γ). Remark: Theorem 1 implies that the two distinct pure cubic fields Q( 3 6) and Q( 3 12) have the same discriminant 2 2 3 5. Thus, in contrast to the situation for quadratic number fields, the discriminant of a cubic field does not determine the field. It is further of interest to note the following theorem. Theorem 2. The prime numbers that ramify in K are precisely the prime divisors of 3ab. Proof. According to Propositions 1 and 2 every prime divisor of 3ab ramifies in K. On the other hand we have K = Q(θ) where θ satisfies the equation ( ). By the usual formula the discriminant of this equation is 3 3 a 2 b 4. I.e., (θ) = 3 3 a 2 b 4. We know then that if p is a prime not dividing 3ab then p does not ramify in K; cf. for instance my note http://www.math.ku.dk/~kiming/lecture_notes/2003-algebraic_number_theory_ koch/decomposition.pdf, or chapter 3.8 of [2]. The theorem follows. References [1] R. Dedekind: Ueber die Anzahl der Idealklassen in reinen kubischen Zahlkörpern. J. Reine Angew. Math. 121 (1900), 40 123. [2] H. Koch: Number Theory. Algebraic Numbers and Functions. Graduate Studies in Mathematics 24, AMS 2000. Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK- 2100 Copenhagen Ø, Denmark. E-mail address: kiming@math.ku.dk