? f = x^3 - 2;
? k = nfinit(f);
? i = idealfactor(k,3);
? j = idealfactor(k,5);
Next we form matrix whose rows correspond to a product of two primes, one dividing 3 and one dividing 5:
? m = matrix(2,2);
? m[1,] = i[1,];
? m[1,2] = 1;
? m[2,] = j[1,];
Note that we set m[1,2] = 1, so the exponent is 1 instead of 3. We apply the CRT to obtain a lift in terms of the basis for OK.
? ?idealchinese
idealchinese(nf,x,y): x being a prime ideal factorization and y a vector of elements, gives an element b such that
v_p(b-y_p)>=v_p(x) for all prime ideals p dividing x, and v_p(b)>=0 for all other p.
? idealchinese(k, m, [x,1]) [0, 0, -1]~
? nfbasis(f) [1, x, x^2]
Thus PARI finds the lift −(√3
2)2, and we finish by verifying that this lift is correct.
I couldn’t figure out how to test for ideal membership in PARI, so here we just check that the prime ideal plus the element is not the unit ideal, which since the ideal is prime, implies membership.
? idealadd(k, i[1,1], -x^2 - x) [3 1 2]
[0 1 0]
[0 0 1]
? idealadd(k, j[1,1], -x^2-1) [5 2 1]
[0 1 0]
[0 0 1]
6.3 Structural Applications of the CRT
The next lemma is an application of the Chinese Remainder Theorem. We will use it to prove that every ideal of OK can be generated by two elements. Suppose that I is a nonzero integral ideals of OK. If a ∈ I, then (a) ⊂ I, so I divides (a) and the quotient (a)I−1 is an integral ideal. The following lemma asserts that (a) can be chosen so the quotient (a)I−1 is coprime to any given ideal.
68 CHAPTER 6. THE CHINESE REMAINDER THEOREM Lemma 6.3.1. If I and J are nonzero integral ideals in OK, then there exists an a ∈ I such that the integral ideal (a)I−1 is coprime to J.
Before we give the proof in general, note that the lemma is trivial when I is principal, since if I = (b), just take a = b, and then (a)I−1 = (a)(a−1) = (1) is coprime to every ideal.
Proof. Let p1, . . . , pr be the prime divisors of J. For each n, let vn be the largest power of pn that divides I. Since pvnn 6= pvnn+1, we can choose an element an ∈ pvnn
that is not in pvnn+1. By Theorem 6.1.5 applied to the r + 1 coprime integral ideals pv11+1, . . . , pvrr+1, I ·³Y
pvnn´−1 , there exists a ∈ OK such that
a ≡ an (mod pvnn+1) for all n = 1, . . . , r and also
a ≡ 0 (mod I ·³Y
pvnn´−1 ).
To complete the proof we show that (a)I−1 is not divisible by any pn, or equiv-alently, that each pvnn exactly divides (a). Because a ≡ an (mod pvnn+1), there is b ∈ pvnn+1 such that a = an+ b. Since an ∈ pvnn, it follows that a ∈ pvnn, so pvnn divides (a). If a ∈ pvnn+1, then an= a − b ∈ pvnn+1, a contradiction, since an6∈ pvnn+1. We conclude that pvnn+1 does not divide (a), which completes the proof.
Suppose I is a nonzero ideal of OK. As an abelian group OKis free of rank equal to the degree [K : Q] of K, and I is of finite index in OK, so I can be generated as an abelian group, hence as an ideal, by [K : Q] generators. The following proposition asserts something much better, namely that I can be generated as an ideal in OK
by at most two elements.
Proposition 6.3.2. Suppose I is a fractional ideal in the ring OK of integers of a number field. Then there exist a, b ∈ K such that I = (a, b) = {αa+βb : α, β ∈ OK}.
Proof. If I = (0), then I is generated by 1 element and we are done. If I is not an integral ideal, then there is x ∈ K such that xI is an integral ideal, and the number of generators of xI is the same as the number of generators of I, so we may assume that I is an integral ideal.
Let a be any nonzero element of the integral ideal I. We will show that there is some b ∈ I such that I = (a, b). Let J = (a). By Lemma 6.3.1, there exists b ∈ I such that (b)I−1 is coprime to (a). Since a, b ∈ I, we have I | (a) and I | (b), so I | (a, b). Suppose pn| (a, b) with p prime. Then pn| (a) and pn| (b), so p - (b)I−1, since (b)I−1 is coprime to (a). We have pn| (b) = I · (b)I−1and p - (b)I−1, so pn| I.
Thus (a, b) | I, so I = (a, b), as claimed.
6.3. STRUCTURAL APPLICATIONS OF THE CRT 69 We can also use Theorem 6.1.5 to determine the OK-module structure of pn/pn+1. Proposition 6.3.3. Let p be a nonzero prime ideal of OK, and let n ≥ 0 be an integer. Then pn/pn+1∼=OK/p as OK-modules.
Proof. 1 Since pn6= pn+1, by unique factorization, there is an element b ∈ pn such that b 6∈ pn+1. Let ϕ : OK → pn/pn+1 be the OK-module morphism defined by ϕ(a) = ab. The kernel of ϕ is p since clearly ϕ(p) = 0 and if ϕ(a) = 0 then ab ∈ pn+1, so pn+1 | (a)(b), so p | (a), since pn+1 does not divide (b). Thus ϕ induces an injective OK-module homomorphism OK/p ,→ pn/pn+1.
It remains to show that ϕ is surjective, and this is where we will use Theo-rem 6.1.5. Suppose c ∈ pn. By Theorem 6.1.5 there exists d ∈ OK such that
d ≡ c (mod pn+1) and d ≡ 0 (mod (b)/pn).
We have pn| (d) since d ∈ pnand (b)/pn| (d) by the second displayed condition, so since p - (b)/pn, we have (b) = pn· (b)/pn| (d), hence d/b ∈ OK. Finally
ϕ µd
b
¶
= d
b · d (mod pn+1) = b (mod pn+1) = c (mod pn+1), so ϕ is surjective.
1Proof from [SD01, pg. 13].
70 CHAPTER 6. THE CHINESE REMAINDER THEOREM
Chapter 7
Discrimants and Norms
In this chapter we give a geometric interpretation of the discriminant of an order in a number field. We also define norms of ideals and prove that the norm function is multiplicative. Discriminants of orders and norms of ideals will play a crucial roll in our proof of finiteness of the class group in the next chapter.
7.1 Field Embeddings
Let K be a number field of degree n. By the primitive element theorem, K = Q(α) for some α, so we can write K ∼= Q[x]/(f ), where f ∈ Q[x] is the minimal polynomial of α. Because C is algebraically closed and f is irreducible, it has exactly n = [K : Q] complex roots. Each of these roots z ∈ C induces a homomorphism Q[x] → C given by x 7→ z, whose kernel is (f). Thus we obtain n embeddings of K ∼= Q[x]/(f ) into C:
σ1, . . . , σn: K ,→ C.
Let σ : K ,→ Cn be the map a 7→ (σ1(a), . . . , σn(a)), and let V = Rσ(K) be the R-span of the image σ(K) of K inside Cn.
Lemma 7.1.1. Suppose L ⊂ Rn is a subgroup of the vector space Rn. Then the induced topology on L is discrete if and only if for every H > 0 the set
XH = {v ∈ L : max{|v1|, . . . , |vn|} ≤ H}
is finite.
Proof. If L is not discrete, then there is a point x ∈ L such that for every ε > 0 there is y ∈ L such that 0 < |x − y| < ε. By choosing smaller and smaller ε, we find infinitely many elements x − y ∈ L all of whose coordinates are smaller than 1. The set X(1) is thus not finite. Thus if the sets XH are all finite, L must be discrete.
Next assume that L is discrete and let H > 0 be any positive number. Then for every x ∈ XH there is an open ball Bx that contains x but no other element of L.
Since XH is closed and bounded, it is compact, so the open covering ∪Bx of XH
has a finite subcover, which implies that XH is finite, as claimed.
71
72 CHAPTER 7. DISCRIMANTS AND NORMS Lemma 7.1.2. If L if a free abelian group that is discrete in a finite-dimensional real vector space V and RL = V , then the rank of L equals the dimension of V . Proof. If x1, . . . , xm ∈ L are a basis for RL, then M = Zx1 + · · · + Zxm has finite index in L, since otherwise the quotient L/M would be infinite, so there would be infinitely many elements of L in a fundamental domain for M , which by Lemma 7.1.1 would contradict discreteness of L. Thus the rank of L is m = dim(RL), as claimed.
Proposition 7.1.3. The R-vector space V = Rσ(K) spanned by the image σ(K) has dimension n.
Proof. We prove this by showing that the image σ(OK) is discrete. If σ(OK) were not discrete it would contain elements all of whose coordinates are simultaneously arbitrarily small. The norm of an element a ∈ OK is the product of the entries of σ(a), so the norms of nonzero elements of OK would go to 0. This is a contradiction, since the norms of elements of OK are integers.
Since σ(OK) is discrete in Cn, Lemma 7.1.2 implies that dim(V ) equals the rank of σ(OK). Since σ is injective, dim(V ) is the rank of OK, which equals n by Proposition 2.4.4.
Since σ(OK) is a lattice in V , the volume of V /σ(OK) is finite. Suppose w1, . . . , wn is a basis for OK. Then if A is the matrix whose ith row is σ(wi), then | det(A)| is the volume of V/σ(OK) (take this as the definition of volume).
Example 7.1.4. Let OK = Z[i] be the ring of integers of K = Q(i). Then w1 = 1,
As the above example illustrates, the volume V /σ(OK) need not be an integer, and it loses sign information. If we consider det(A)2 instead, we obtain a well-defined integer.