ARITHMETIC IN PURE CUBIC FIELDS AFTER DEDEKIND. IAN KIMING

We will study the rings of integers and the decomposition of primes in cubic √ 3 number fields K of type K = Q( 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 = ab2 where a, b ∈ N, gcd(a, b) = 1, ab > 1, and ab square free. It will then be convenient to define: √ √ 3 3 α := ab2 β := a2 b , so that α2 = 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 ∈ OK . One sees that θ satisfies the equation: (∗)

θ3 + 3aθ2 + 3a2 θ + a(a2 − b2 ) = 0

which must be the minimal equation for θ as clearly K = Q(θ). We denote by DK the discriminant of K, i.e., the discriminant of the ring OK of integers in K. Theorem 1. (i). Suppose that a2 6≡ b2 (9). Then OK = Z(1, α, β) and DK = −27a2 b2 . (ii). Suppose a2 ≡ b2 (9). Put γ :=

1 + aα + bβ . 3

Then OK = Z(α, β, γ) and DK = −3a2 b2 . This result was first proved by Dedekind, cf. [1]. We will break the proof up into a number of smaller steps. 1

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Lemma 1. We have ∆(1, α, β) = −27a2 b2 , and the index [OK : 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 x2 + x + 1. The 3 embeddings K ,→ C are then characterized by (α, β) 7→ (α, β) ,

(α, β) 7→ (ζα, ζ 2 β) ,

respectively. Now we compute:    1 α β 1 1 2    det 1 ζα ζ β = αβ · det 1 ζ 1 ζ 2 α ζβ 1 ζ2

and (α, β) 7→ (ζ 2 α, ζβ) ,

 1 ζ 2  = αβ · (ζ − 1)(ζ 2 − 1)(ζ 2 − ζ) 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 α2 β 2 = a2 b2 and (ζ 2 − ζ)2 = ζ + ζ 2 − 2 = −3. By general theory we have then: −27a2 b2 = ∆(1, α, β) = DK · [OK : Z(1, α, β)]2 which first implies (since DK is an integer) that −3DK is a square, and then that [OK : 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 [OK : 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: pOK = p3 with p a prime ideal of OK . Proof. Suppose that p j a and let p be a prime divisor of p in K. Then p divides ab2 = α3 , so that p must divide α. Then p3 divides α3 = ab2 . As gcd(a, b) = 1 and as ab is square free we have that ab2 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 p3 divides p. But then p necessarily has the stated decomposition as [K : Q] = 3 (remember that if pOK = pe11 · · · pess where P the prime ideals pi has inertia degree fi then 3 = [K : Q] = i ei fi ). If p j b we can work similarly with β ∈ OK to obtain the desired conclusion. Proposition 2. The prime decomposition of 3 in K is as follows:  3 p , if a2 6≡ b2 (9) 3OK = 2 p1 p2 , if a2 ≡ b2 (9) .



ARITHMETIC IN PURE CUBIC FIELDS AFTER DEDEKIND.

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Proof. If 3 j ab then a2 6≡ b2 (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 ∈ OK , ω 6= 0, the integer νp (ω) is defined as the power of p in the prime factorization of ω (that is, of the ideal ωOK ). Similarly we have νp (c) for integers c 6= 0 defined as the maximal power of p dividing c. Consider the number θ = α − a satisfying the equation (∗) and let θ, θ0 , θ00 be the three conjugates of θ. Since now a2 − b2 is divisible by 3 (as 3 - ab) we see then that 3 divides θ3 (as ideals in OK ). On the other hand we obtain from (∗) that (∗∗)

3a2 = θθ0 + θθ00 + θ0 θ00

which implies 3 - θ: For if 3 j θ then also 3 j θ0 and 3 j θ00 (consider an embedding g of K into C. Then g maps K to some algebraic number field K 0 , and in fact any ideal of OK to an ideal of OK 0 ; so g maps 3OK to 3OK 0 because 3 ∈ Q. But it maps θOK to gθOK 0 ). But then (∗∗) would imply 9 j 3a2 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 3OK = p1 p2 p3 with the pi prime ideals then pi j θ3 for each i because 3 j θ3 ; thus pi j θ for each i; were the pi distinct we would obtain 3OK = p1 p2 p3 j θ, contradicting the above conclusion. It follows that 3 has a prime divisor p1 such that p21 j 3. We can then write: 3OK = p21 p2 where a priori p2 is just some ideal. However, upon taking norms we obtain 33 = N(3OK ) = N(p1 )2 N(p2 ); since N(p1 ) is a non-trivial power of 3 we are forced to conclude N(p2 ) = 3 which means that p2 is a prime ideal (recall that N(p2 ) = [OK : p2 ]; since this index is 3 the ideal p2 must be maximal). The decisive question is now whether the prime ideals p1 and p2 are equal. We see that we can finish the proof by showing p1 = p2 ⇔ a2 6≡ b2

(9) .

To prove this it is convenient to write the equation (∗) as ([)

θ3 + 3aαθ + a(a2 − b2 ) = 0

(recall that θ := α − a), and to define r := ν3 (a2 − b2 ) and s := νp1 (θ). Suppose that p1 = p2 so that 3OK = p31 . Then 1 ≤ s ≤ 2: For we have 3 j θ3 so that surely s ≥ 1; but we also know 3 - θ, i.e., s < 3. Now, νp1 (α) = 0 since otherwise 3OK = p31 would divide α3 = ab2 contrary to our assumption 3 - ab. Then we have: νp1 (θ3 ) = 3s , νp1 (3aαθ) = 3 + s , νp1 (a2 − b2 ) = 3r . We must have 3s 6= 3 + s (since s is not divisible by 3). Thus: νp1 (θ3 + 3aαθ) = minf3s, 3 + sg

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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 - a2 − b2 . Suppose then that p1 6= p2 . Since 3 j θ3 we also have p2 j θ. Thus νp2 (θ3 + 3aαθ) ≥ 2 . But by ([) this valuation coincides with νp2 (a(a2 −b2 )) = νp2 (a2 −b2 ) = ν3 (a2 −b2 ). Thus 9 j a2 − b2 .  Lemma 2. Suppose that p is a prime number dividing ab. Then p does not divide [OK : Z(1, α, β)]. Proof. Suppose that p divides both ab and the index [OK : Z(1, α, β)]. Since p divides this index there is a number ω ∈ OK such that ω 6∈ 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 pOK = p3 . This decomposition implies: νp (c) = 3 · νp (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 · νp (α) = νp (α3 ) = νp (ab2 ) = 3 · νp (ab2 ) = 3 as gcd(a, b) = 1 and ab square free. It follows that νp (α) = 1 . 3

2

Working similarly with β = a b we find that νp (β) = 2 . Consider now the equation (]). The equation implies x + yα + zβ ≡ 0

(p3 )

whence in particular x ≡ 0 (p) as α ≡ β ≡ 0 (p). Since x ∈ Z we deduce p j x. Then x ≡ 0 (p3 ) so that now yα + zβ ≡ 0

(p3 ) .

Since β ≡ 0 (p2 ) we have then yα ≡ 0 (p2 ). As νp (α) = 1 we must then have νp (y) > 0, i.e., that p j y. But then y ≡ 0 (p3 ) and so zβ ≡ 0

(p3 )

which implies p j z because νp (β) = 2. 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.

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If p divides b we obtain a contradiction in an entirely similar fashion (we will then have νp (α) = 2 and νp (β) = 1).  Proof of Theorem 1. By lemmas 1 and 2 the index k := [OK : 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 DK = ∆(1, α, β)/k2 = −3 · (3/k)2 · a2 b2 . We conclude that the proof of the theorem is finished if we can show k = 1 ⇔ a2 6≡ b2

(9) ,

and that OK = 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 a2 6≡ b2 (9), and also k = 1 as we just noted. Suppose first that a2 6≡ b2 (9). By Proposition 2 we have then 3OK = p3 with p a prime ideal. From equation (∗) for θ = α − a we first see that p j θ, and then that p2 - θ: For otherwise 3 · ν3 (a2 − b2 ) = νp (a2 − b2 ) would be greater than 3 and so at least 6 whence ν3 (a2 − b2 ) ≥ 2, contradicting our assumption a2 6≡ b2 (9). Let now ω be an arbitrary number in OK . 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 = α2 . Thus: (†)

3bω = x0 + y 0 θ + z 0 θ2

where (‡)

x0 := bx + aby + a2 z , y 0 := by + 2az , z 0 := z .

Now, (†) certainly implies x0 + y 0 θ + z 0 θ2 ≡ 0 (p3 ). Since p j θ but p2 - θ we can then deduce x0 ≡ y 0 ≡ z 0 ≡ 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 a2 ≡ b2 (9). By Proposition 2 we have now 3OK = p21 p2 with p1 , p2 prime ideals. Equation (∗) for θ = α − a certainly gives 3 j θ3 . We can then deduce p1 p2 j θ whence 3p2 = p21 p22 j θ2 , and so in particular: 3 j θ2 = (α − a)2 = α2 − 2aα + a2 = bβ − 2aα + a2 = (1 + aα + bβ) + (a2 − 1 − 3aα) . As 3 - a we have a2 ≡ 1 (3) so that certainly 3 j (a2 − 1 − 3aα). We conclude that 3 j (1 + aα + bβ), i.e., that the number γ :=

1 + aα + bβ 3

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IAN KIMING

is in OK . Since clearly γ 6∈ Z(1, α, β) we cannot have k = 1, so k = 3. Since this index is a prime number we have OK = Z(1, α, β, η) if η is any number in OK that is not in Z(1, α, β). In particular we have OK = Z(1, α, β, γ) = Z(α, β, γ) . √ 3



Remark: Theorem 1 implies that the two distinct pure cubic fields Q( 6) and √ Q( 3 12) have the same discriminant −22 · 35 . 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 −33 a2 b4 . I.e., ∆(θ) = −33 a2 b4 . 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 Zahlk¨ orpern.’ 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, DK2100 Copenhagen Ø, Denmark. E-mail address: [email protected]