Definition 1.6: A is integrally closed or normal if its integral closure in K = Frac(A) is itself.
Proposition 1.7: If L is algebraic over K then every element of L can be written as ab where b ∈ B and a ∈ A. Thus L = Frac(B). In particular, for any extension L/Q, Frac(OL) = L.
Number Theory, §13.1.
Proof. Given α ∈ L, suppose that it satisfies the equation
P (x) := anxn+ an−1xn−1+ · · · + a0 = 0
with a0, . . . , an ∈ K and an 6= 0. Since Frac(A) = K, by multiplying by an element of A as
necessary we may assume a0, . . . , an ∈ A. Then
an−1n P x d
:= xn+ an−1xn−1+ anan−2xn−2+ · · · + an−1n a0.
Hence anα is integral over A, i.e. anα ∈ B. This shows α is in the desired form.
For the last part, take K = Q and A = Z.
For short we call (13.1) the “AKLB” setup if we further assume A is integrally closed in K. In the usual case, A is the integral closure of Z in K. in this case, we write A = OK.1
When F = Q, the algebraic closure of Q, a ∈ Q is called an algebraic number and a ∈ OQ
is an algebraic integer.
Theorem 1.8 (Rational Roots Theorem): A UFD is integrally closed.
Proof. Suppose R is a UFD with field of fractions K. Let x ∈ K be integral over R; suppose x satisfies
xn+ an−1xn−1+ · · · + a0 = 0
where a0, . . . , an−1 ∈ R. Write x = pq where p, q ∈ R are relatively prime. Then multiplying
the above by qn gives
pn+ an−1pn−1q + · · · + a1pqn−1+ a0qn = 0
q(an−1pn−1+ · · · + a0qn−1) = −pn
Thus q | p, possible only if q = 1. This shows x ∈ R.
Note that in the definition of integrality, an element is integral if it is the zero of any monic polynomial in A[x]. However, it suffices to check that its minimal polynomial is in A[x].
Proposition 1.9: Let L be an algebraic extension of K and A be integrally closed. Then α ∈ L is integral over A iff its minimal polynomial f over K has coefficients in A.
Proof. The reverse direction is clear. For the forward direction, note all zeros of f are integral over K since they satisfy the same polynomial equation that α satisfies. The coefficients of f are polynomial expressions in the roots so are integral over A, and hence in A (since they are already in K).
Proposition 1.10 (Finite generation):
1
Later on, when we take K to be an extension of the p-adic field Qp, we will use OKto denote the integral
1. Let A ⊆ B ⊆ C be rings. If B is finitely generated as an A-module and C is finitely generated as a B-module, then C is finitely generated as an A-module.
2. If B is integral over A and finitely generated as an A-algebra, then it is finitely generated as an A-module.
Proof.
1. Take products of generators.
2. Let algebra generators be β1, . . . , βm. Then
A ⊆ A[β1] ⊆ · · · ⊆ A[β1, . . . , βm]
is a chain of integral extensions, so item 2 follows from 1. Combining this proposition with Lemma 1.4 we get the following:
Proposition 1.11 (Transitivity of integrality): Let A ⊆ B ⊆ C be integral domains and K, L, M be their fraction fields.
1. If B is integral over A and C is integral over B, then C is integral over A.
2. Let A0 is the integral closure of A over B and A00 be the integral closure of A0 over C. Let A000 be the integral closure of A in C.
3. The integral closure of A is integrally closed. Proof.
1. For γ ∈ C, let bi be the coefficients of the minimal polynomial of C over B. Then γ
is integral over A[b0, . . . , bm], so by Proposition 1.10, item 2, A[b0, . . . , bm, γ] is finitely
generated over A. Since γA[b0, . . . , bm, γ] ⊆ A[b1, . . . , bm, γ], by Lemma1.4, γ is integral
over A.
2. By item 1 applied to A ⊆ A0 ⊆ A00, A00 is integral over A so A00 ⊆ A000. Conversely, any
element a ∈ A000 is integral over A so a fortiori integral over A00; thus A000 ⊆ A00.
3. Follows from item 2 applied to A = B = C.
§2 Norms and Traces
Let B be a free A-module of rank n. Then any element β ∈ B defines an A-linear map mβ
(or [β]), multiplication by β. It is helpful to think of β as a linear map because then we can apply results from linear algebra.
Definition 2.1: The trace, determinant, and characteristic polynomial of mβ are called the
Number Theory, §13.2.
These are computed by choosing any basis of e1, . . . , en for B over A, and then computing
the action of β on this basis.
Proposition 2.2 (Elementary properties): The following hold (a ∈ A; β, β0 ∈ B): 1. Tr(β + β0) = Tr(β) + Tr(β0)
2. Tr(aβ) = aTr(β) 3. Tr(a) = na
4. Nm(ββ0) = Nm(β) · Nm(β0) 5. Nm(a) = an
Proposition 2.3 (Behavior with respect to field extensions): Suppose L/K is a degree n field extension, M is a finite extension of L, and β ∈ L.
1. (Relationship with roots of minimal polynomial) If f (X) is the minimal polynomial of β over K and β1, . . . , βm are the roots of f (X) = 0 in a Galois closure of K, then
letting r = [L : K(β)] = mn, (a) charL/K(β) = f (X)r
(b) TrL/K(β) = r(β1+ · · · + βm)
(c) NmL/K(β) = (β1· · · βm)r
2. (Relationship with embeddings) Suppose L is separable over K, M is a Galois extension of K, and σ1, . . . , σn are the set of distinct embeddings L → M fixing K. Then
(a) TrL/K(β) = σ1(β) + · · · + σn(β)
(b) NmL/K(β) = σ1(β) · · · σn(β)
In particular, this is true when L = M is a Galois extension of K, and we can think of the σk as simply the elements of G(L/K).
3. (Transitivity of trace and norm) Suppose β ∈ M and M/K is separable.2 Then
(a) TrM/K(β) = TrL/K(TrM/L(β))
(b) NmM/K(β) = NmL/K(NmM/L(β))
4. (Integrality) Assume AKLB. If β ∈ B, then the coefficients of charL/K(β), and hence
TrL/K(β) and NmL/K(β), are integral over A. In particular, if A is integrally closed in
L then they are in A. Proof.
1. If r = 1, i.e. K[β] = L, then by the Cayley-Hamilton Theorem, f (mβ) = 0. Since
f (X) is irreducible, f (X) | charL/K(β). However, these are monic polynomials of the
same degree so they are equal.
In the general case, take a basis xi of K[β] over K and a basis yj of L over K[β]. Then
xiyj form a basis of L over K, and the matrix of mβ with respect to this basis is n
copies of A. This proves (a), which implies the rest of the statements.
2. Let β1, . . . , βm be the conjugates of β. There are m distinct imbeddings K(β) → M ;
they each take β to a different βk. Each of these imbeddings extend to r := [L :
K(β)] = mn imbeddings L → M . Now use item 1.
3. Note that for any finite extensions K ⊆ L ⊆ N with N Galois, an imbedding L ,→ N fixing K can be extended to a K-automorphism on N , and so be considered an element of the set G(N/K)/G(N/L).3
Let N be a Galois extension containing M . By item 2, TrM/K(β) = X σ∈G(N/K)/G(N/M ) σ(β) TrL/K(TrM/L(β)) = TrL/K X σ∈G(N/L)/G(N/M ) σ(β) = X τ ∈G(N/K)/G(N/L) X σ∈G(N/L)/G(N/M ) τ (σ(β))
where in the second sum we take arbitrary representatives τ ∈ G(N/K) and σ ∈ G(N/L). These are equal because for any choice of these representatives,
{σ ∈ G(N/K)/G(N/M )} = {τ σ | τ ∈ G(N/K)/G(N/L), σ ∈ G(N/L)/G(N/M )} when considered in G(N/K)/G(N/M ) (i.e. as imbeddings M ,→ N fixing K). The same is true of the norm.
4. The minimal polynomial of α has coefficients in A, by Proposition 1.9. Hence the result follows from item 1.
§3 Discriminant
Definition 3.1: If B is a ring and free A-module of rank m, and β1, . . . , βm ∈ B, then their
discriminant is
D(β1, . . . , βm) = det[TrB/A(βiβj)]1≤i,j≤m.
3Using the primitive element theorem, write L = K(β). The imbeddings L → N are those taking β to a
conjugate; there are [L : K] imbeddings. But we know G(N/K)/G(N/L) = [L : K], so all of the imbeddings must be extendable. We also use this fact (in addition to a counting argument) in the proof of 2.
Number Theory, §13.3.
Proposition 3.2: If the change of basis matrix from γi to βi is T , then
D(γ1, . . . , γm) = det(T )2· D(β1, . . . , βm).
Proof. Let M1 and M2 be the matrices of the bilinear form
(α, α0) = TrB/A(αα0)
with respect to the bases (β1, . . . , βm) and (γ1, . . . , γm), respectively. Then, using the change
of basis formula for bilinear forms,
D(β1, . . . , βm) = det(M1)
D(γ1, . . . , γm) = det(M2)
M2 = TtM1T
det(M2) = det(T )2· det(M1)
from which the result follows.
Consider the discriminant of an arbitrary basis of B over A. By the above fact, this is well-defined up to multiplication by the square of a unit. The residue in A/(A×)2 is called
the discriminant disc(B/A). The discriminant also refers to the ideal of A this element generates.
Note disc(B/A) can be thought of as the determinant of the matrix of the bilinear form (β, β0) = TrB/A(ββ0).
Proposition 3.3 (Criterion for integral basis): Let A ⊆ B be integral domains and B be a free A-module of rank m with disc(B/A) 6= 0. Then γ1, . . . , γm ∈ B form a basis for B as
an A-module iff
(D(γ1, . . . , γm)) = (disc(B/A))
as ideals.
Proof. Let βi be a basis. If the change of basis matrix from γi to βi is T , then by Proposi-
tion 3.2,
D(γ1, . . . , γm) = det(T )2· D(β1, . . . , βm) = det(T )2disc(B/A)
Now γi is basis iff T is invertible, iff det(T ) is a unit, iff (D(γ1, . . . , γm)) = (disc(B/A)).
Proposition 3.4 (Discriminants and Field Extensions):
1. (Relationship with embeddings) Let L be separable finite over K of degree m, and σ1, . . . , σm be the embeddings of L into a Galois extension M fixing K. Then for any
basis β1, . . . , βm of L over K,
D(β1, . . . , βm) = det(σiβj)2 6= 0.
2. (Nondegeneracy of trace pairing) If B is free of rank m over A (with fraction fields K, L as above), then the pairing
(β, β0) 7→ Tr(ββ0)
Here perfect means that the map a 7→ (b 7→ (a, b)) is an isomorphism L → L∗, and similarly for b 7→ (a 7→ (a, b)). This is equivalent to saying that the bilinear form is nondegenerate. Proof. Use Proposition 2.3(1b), and that σk, det are both multiplicative. Inequality follows
from independence of characters:
Let G be a group, F a field. Then the homomorphisms G → F×are linearly independent.
Thus for K of degree m over Q, we can talk of disc(OK/Z).
A closely related quantity to the discriminant is the different.
Definition 3.5: Assume AKLB, and suppose L/K is a finite separable extension. The codifferent of B with respect to A is
B∗ = {y ∈ L | Tr(xy) ∈ A for all x ∈ B}. The different of B with respect to A is
DB/A = (B∗)−1.
In other words, it is the largest B-submodule satisfying Tr(E) ⊆ A. Note that DB/A = (B∗)−1.
Remark 3.6: We will define the discriminant in general, when B is not necessarily a free A- module, in Chapter21. The relationship between the two definitions is the following: Let p be an ideal in A. Then Apis a principal ideal domain (in fact, a DVR). Let S = A−p; then S−1B
is free over S−1A by the structure theorem for modules. We have (disc(S−1B/S−1A)) = (pAp)m(p) for some m(p). Then
disc(B/A) =Y
p
pm(p).
§4 Integral bases
Proposition 4.1 (Finite generation of integral extensions): Let A be integrally closed and L separable of degree m over K. There are free finite A-submodules M and M0 of L such that M ⊆ B ⊆ M0. B is a finitely generated A-module if A is Noetherian, and free of rank m if A is a PID.4
Proof. Let {β1, . . . , βm} ⊆ B be a basis for L over K. Take a basis βi0 so that Tr(βiβj0) = δij.
Then
Aβ1+ · · · + Aβm ⊆ B ⊆ Aβ10 + · · · + Aβ 0 m.
The second inclusion follows because if β ∈ B, then writing β = P
jbjβ 0
j, we have that
bi = Tr(ββi) ∈ A. (In other words, the βi0 form a basis for the codifferent B
∗, which contains
B.)
Number Theory, §13.4.
If A is Noetherian, then M0 is finitely generated, so its submodule B is finitely generated over A. If A is a PID, then by the Structure Theorem for Modules (over PIDs), M is a direct sum of cyclic modules and a free module. Since it is contained in a free module of rank m and contains a free module of rank m, it must be free of rank m.
The following is immediate:
Theorem 4.2: If K is finite over Q (i.e. a number field), then OK is a finitely generated
Z-module. It is the largest subring that is finitely generated over Z.
Definition 4.3: A basis for OK as a Z-module is called an integral basis.
Proposition 4.4: Suppose K has characteristic 0 (so L separable over K), L = K[β], and f is the minimal polynomial of β over K. Let f (X) =Q(X − βi) in the Galois closure of L.
Then
D(1, β, . . . , βm−1) = Y
1≤i<j≤m
(βi− βj)2 = (−1)m(m−1)/2· NmL/K(f0(β)).
This is called the discriminant of f .5
Proof. Note the βi are conjugates of β; assume β = β1.
By Proposition 3.4, we have D(1, β, . . . , βm−1) = 1 β1 · · · β1m−1 1 β2 · · · β2m−1 .. . ... . .. ... 1 βm · · · βmm−1 2 = Y 1≤i<j≤m (βi− βj)2,
where the last statement follows by evaluating the Vandermonde determinant. For the second equality, note by Proposition 2.3(1c) that
NmL/K(f0(β)) = NmL/K((β1− β2) · · · (β1− βm)) = Y 1≤i≤m Y 1≤j≤m, j6=i (βi− βj) = (−1)m(m−1)2 Y 1≤i<j≤m (βi − βj)2.
Proposition 4.5: If K = Q[α], α ∈ OK, and D(1, α, . . . , αm−1) = disc(O/Z) then {1, α, . . . , αm−1}
is an integral basis.
Proof. Using change-of-basis and the correspondence between index and determinant, D(1, α, . . . , αm−1) = disc(OK/Z) · [OK : Z[α]]2.
Now disc(OK/Z) ∈ Z so [OK : Z[α]] = 1.
Theorem 4.6 (Stickelberger’s Theorem):
1. Let s is the number of complex (nonreal) embeddings K → C. Then sign[disc(K/Q)] = (−1)s/2.
2. disc(OK/Z) ≡ 0 or 1 (mod 4).
Proof. 1. Write K = Q[α] by the Primitive Element Theorem and α1, . . . , αr be the real
conjugates and β1, β1, . . . , βs, βs be the complex conjugates. By Proposition 4.4,
sign(D(1, α, . . . , αm−1)) = sign Y 1≤j≤s (βj − βj)2 ! = Y 1≤j≤s i2 = (−1)s/2.
2. Let α1, . . . , αm be an integral basis. Let P and −N be the sum of the terms in the
expansion of det(σiαj) corresponding to even and odd permutations, respectively:
P = X even π∈Sm m Y i=1 σiαπ(i) N = X odd π∈Sm m Y i=1 σiαπ(i). Then disc(OK/Z) = det(σiαj)2 = (P − N )2 = (P + N )2− 4P N. Take σ ∈ G(Kgal/Q). Note composition by σ permutes the σ
i, say by ν. Then P = X even π∈Sm m Y i=1 σiαν−1π(i) N = X odd π∈Sm m Y i=1 σiαν−1π(i)
and hence σ permutes {P, N }. Hence σ fixes P + N, P N and they are rational. Since they are integral over Z they are integers. Thus the above is congruent to 0 or 1 modulo 4.
Example 4.7 (Quadratic extensions): Any quadratic extension of Q is in the form Q(√m) for some squarefree integer m. We find the ring of integers of Q(√m). Consider two cases.
Number Theory, §13.4.
1. m ≡ 2, 3 (mod 4): The minimal polynomial of √m is X2− m, so
disc(1,√m) = (√m − (−√m))2 = 4m. Note disc(1,
√ m)
disc(Q(√m)/Q) must be a square by Proposition 3.2 so disc(Q(
√
m)/Q) equals m or 4m. However, by Stickelberger’s Theorem, disc(Q(√m)/Q) ≡ 0, 1 (mod 4). Hence disc(Q(√m)/Q) 6= m and disc(Q(√m)/Q) = 4m. By Proposition 3.3, 1,√m is an integral basis.
2. m ≡ 1 (mod 4): Note 1+
√ m
2 is integral with minimal polynomial X
2− X − m−1 4 , so disc 1,1 + √ m 2 = 1 + √ m 2 − 1 −√m 2 2 = m.
Since m is squarefree, disc(Q(√m)/Q) = m and Proposition 3.3 says 1,1+
√ m 2 is an
integral basis.
The following tells us about integral bases for products of fields. Proposition 4.8: Suppose that K, L are field extensions of Q such that
[KL : Q] = [K : Q][L : Q]. Let d = gcd(disc(K/Q), disc(L/Q)). Then
1. OK ⊆ d−1OKOL.
2. If OKL = OKOL, then disc(KL/Q) = disc(K/Q)[L:Q]disc(L/Q)[K:Q].
Proof. Let {α1, . . . , αm} be an integral basis for K and {β1, . . . , βn} be an integral basis for
L. By the degree assumption, we know that {αiβj} is a basis for KL over Q. Any element
of KL integral over Q can be written as
γ = X 1 ≤ i ≤ m 1 ≤ j ≤ m aij r αiβj (13.2) where gcd(r, gcd(aij)) = 1.
We need to show that r | d. Let xi =
Pn
j=1 aij
r βj. We will turn (13.2) into a system of
equations by considering all embeddings K ,→ C, solve for the xi using Cramer’s rule, and in
this way show that each xi is an algebraic integer in L divided by a bounded denominator.
Note given embeddings σK : K ,→ C and σL : L ,→ C, there is exactly one embedding
σKL : KL ,→ C such that restricts to σK and σL. It is clearly unique if it exists. To
show existence, write K = Q(α) ∼= Q(x)/(f (x)) by PET, and note that the characteristic polynomial of f does not change upon passing to L because of the degree assumption. Hence KL = L(α) = L(x)/(f (x)), and in extending σL to σKL, we are allowed to send α = x to
Fix an embedding σ : L ,→ C, and let σ1, . . . , σm be all embeddings K ,→ C. Then
applying σk to 13.2 we obtain the system of equations m
X
i=1
σk(αi)xi = σk(γ), 1 ≤ k ≤ m.
By Cramer’s rule, letting D = det[(σk(αi))k,i] we get Dxi = Di where Di has the ith column
of D replaced by (σk(αi))mk=1. Note that D and Di are both algebraic integers. Using
disc(OK/Z) = D2 (Proposition 3.4), we get
disc(OK/Z)xi = DDi.
Hence disc(OK/Z)xi is an algebraic integer (in OL). Since the βj are an integral basis for
OL, this forces r | disc(OK/Z). Similarly, r | disc(OL/Z), as needed.
Now we prove the second part. Choose (α1, . . . , αm) a basis for K/Q and (β1, . . . , βn) a
basis for L/Q. Then (αjβk)1≤j≤m,1≤k≤n is a basis for KL/Q. For γ ∈ KL, let (γ)j,k denote
the coordinate of αjβk in γ. Then the mn × mn matrix
[Tr(αi1βi2αi01βi02)] = " X 1≤j≤m,1≤k≤n (αi1βi2αi01βi02αjβk)j,k # = " X 1≤j≤m,1≤k≤n (αi1αi01αj)j(βi2βi02βk)k # = " X 1≤j≤m X 1≤k≤n (αi1αi01αj)j(βi2βi02βk)k # = [Tr(αi1αi01)] ⊗ [Tr(βi2βi02)]. Taking determinants and using
det(A ⊗ B) = det(A)ndet(B)m, A ∈ Mm×m, B ∈ Mn×n
we get
disc(KL/Q) = disc(K/Q)[L:Q]disc(L/Q)[M :Q].
§5 Problems
1. Suppose that f ∈ Z[x] is irreducible and has a root of absolute value at least 32. Prove
that if α is a root of f then f (α3+ 1) 6= 0.
2. Let a1, . . . , an be algebraic integers with degrees d1, . . . , dn. Let a01, . . . , a 0
n be the con-
jugates of a1, . . . , an with greatest absolute value. Let c1, . . . , cn be integers. Prove
that if the LHS of the following expression is not zero, then |c1a1+ . . . + cnan| ≥ 1 |c1a01| + · · · + |cna0n| d1d2···dn−1 .
Number Theory, §13.5. For example, |c1+ c2 √ 2 + c3 √ 3| ≥ 1 |c1| + |2c2| + |2c3| 3 .
3. Let p be a prime and consider k pth roots of unity whose sum is not 0. Prove that the absolute value of their sum is at least kp−21 .
Chapter 14
Ideal factorization
§1 Discrete Valuation Rings
Definition 1.1: Let K be a field. A discrete valuation on K is a surjective function v : K× → Z such that for every x, y ∈ K×,
1. π is a group homeomorphism: v(xy) = v(x)v(y). 2. v(x + y) ≥ min(v(x), v(y)).
We set v(0) = ∞.
A discrete valuation ring (over Z) is a local integral domain R (not a field), whose fraction field has a discrete valuation v.
An element t with v(t) = 1 is a uniformizing parameter.
Proposition 1.2: Suppose R is a DVR with fraction field K. Let v be the valuation on K. 1. The units are exactly the elements with 0 valuation:
R×= v−1(0).
2. Its maximal idea is the set of elements with positive valuation. m= {x : v(x) > 0} .
3. R is a PID with ideals mn= {x : v(x) ≥ n} = (tn) for n ∈ N.
4. R is a UFD; any element can be written uniquely in the form utn where u is a unit.
Lemma 1.3: Let A be a local domain with maximal ideal m principal and nonzero. If T
n≥0mn = 0 then A is a DVR.
Theorem 1.4: Let (A, m) be a Noetherian local domain. The following conditions are equivalent.
2. A is a normal domain of dimension 1. (Dimension 1 means that the longest chain of prime ideals is 2: p0 ⊆ p1.) (Since A is local this means it has only two prime ideals.)
3. A is a normal domain of depth 1. (There is a nonzero x ∈ A with m ∈ Ass(A/hxi).) 4. A is a regular local ring of dimension 1. (Regular means its maximal ideal is generated
by a number of elements equal to its dimension. So here it means m is principal.) 5. m is principal and nonzero.
Proof. Note (5) =⇒ (1) uses Krull Intersection Theorem: For R a Noetherian ring, a an ideal, and M a finitely generated module (esp. when M = R), then there exists x ∈ a such that (1 + x) ∞ \ n=0 anM = 0.
§2 Dedekind Domains
Definition 2.1: A Dedekind domain is a normal Noetherian integral domain A such that every nonzero prime ideal is maximal.
Proposition 2.2: A local integral domain is Dedekind iff it is a DVR.
Proposition 2.3: For every nonzero prime ideal p in a Dedekind domain A, the localization Ap is a DVR. (Locally, Dedekind domains are DVR’s.)
(The converse, i.e. if Ap is a DVR for every p, then A is Dedekind, holds using Serre’s
criterion.)
Theorem 2.4 (Unique factorization of prime ideals): Let A be a Dedekind domain. Every proper nonzero ideal of A can be written uniquely as a product of prime ideals.
Proof. Let a be a proper nonzero ideal of A.
1. If A is Noetherian, then every ideal a ⊆ A contains a product b = Q prk
k of nonzero
prime ideals: Otherwise, choose a maximal counterexample a (possible since A is Noetherian). Since a is not prime, there exist x, y 6∈ a such that xy ∈ a. By the maximality assumption both a + (x) and a + (y) contain a product of prime ideals, and so does a ⊇ (a + (x))(a + (y)).
2. By the Chinese Remainder Theorem
A/b ∼=Y
k
A/prk
k
Number Theory, §14.2.
3. If p is a maximal ideal in a ring A, and q = pAp, then the natural map A/pm →
(A/pm)
p = Ap/qm is an isomorphism. (Indeed, it is injective because p is prime and
surjective because any s ∈ A − p is invertible modulo pm, on account of (s) + pm= A.) Thus Y k A/prk k ∼= Y k Apk/q rk k .
(This is where we use the fact that nonzero prime ideals are maximal.)
4. Combining the above, we get a one-to-one correspondence between ideals in A contain- ing b, and ideals in Q
kApk/q
rk
k . All ideals in the last ring are in the form
Q kq sk k /q rk k , so a is of the formQ kq sk
k . Moreover, different prime ideals containing b correspond to
different Q
kq sk
k /q rk
k , which are different for different sk, giving uniqueness.
Corollary 2.5: Let A be a Dedekind domain. 1. If a =Q pp rk k and b = Q pp sk
k are ideals in A and p is a nonzero prime ideal then
a⊇ b ⇐⇒ rk ≥ sk for all k
⇐⇒ aAp ⊇ bAp for all p.
2. If a ⊃ b 6= 0 are ideals in A then a = b + (a) for some a ∈ A. In particular, if b ∈ a then there exists a ∈ A such that a = (a, b); i.e. each ideal is generated by at most two elements.
3. (Inverses) Let a 6= 0 be an ideal of A. There exists a nonzero ideal a∗ such that aa∗ is principal.
(a) We can choose a∗ so aa∗ = (a) for given a ∈ a.
(b) Alternatively we can choose a∗ to be relatively prime to a given ideal c 6= 0. Proof. 1. The forward direction was shown in the course of the theorem. The reverse
directions are easy.
2. Choose any a ∈ a\{0}. By unique factorization, we can write (a) = pu1
1 · · · purr
a= pv1
1 · · · pvrr
for primes p1, . . . , pr and uj ≥ vj ≥ 0. Now choose bj ∈ p vj
j \p vj+1
j . By the Chinese
remainder theorem we can choose b such that b ≡ bj (mod p vj+1
j ) for all j. Since
ordpj(bj) = vj, by item 1, the highest power of pj dividing (b) is vj. The highest power of pj dividing (a) is uj ≥ vj, so the highest power of pj dividing (a, b) is vj. Now for
a prime q 6∈ {p1, . . . , pr}, we have a 6∈ q (else q would divide a), so q does not divide
(a, b). We conclude (a, b) = pv1 1 · · · p vr r , as needed.
3. (a) follows from item 1; for (b), use item 2 and 3(a) to write a = ac + (a) = ac + aa∗ = a(c + a∗).
Theorem 2.6: Assume AKLB, and K/L is finite separable. If A is a Dedekind domain, then so is B. In particular, taking A = Z and K = Q, every ring of integers in a finite separable extension of Q is Dedekind.
Proof.
1. B is noetherian: By Proposition 13.4.1, B is a finitely generated A-module, hence a Noetherian A-module, hence Noetherian as a ring.
2. B is integrally closed by Proposition 13.1.11(2).
3. Every nonzero prime ideal q of B is maximal: Take a nonzero β ∈ q and let its minimal polynomial be xn+ an−1xn−1+ · · · + an. Then an= −βn− · · · − a1β ∈ βB ∩ A ⊆ q ∩ A.
This shows q ∩ A 6= 0; since A is Dedekind and q ∩ A is prime, q ∩ A is maximal and A/q is a field. Since B is integral over A, B/q is integral over A/q.
Lemma 2.7: An integral domain B containing a field k and algebraic over k is a field. Proof. Let β ∈ B be nonzero. Then k[β] is a finite dimensional vector space and the multiplication-by-β map mβ : k[β] → k[β] is injective, hence surjective. Thus there
exists β0 so ββ0 = 1, i.e. β has an inverse.
The lemma shows B/q is a field. Hence q is maximal.
Alternatively, this follows directly from “lying-over” and “going up” for integral exten- sions.
Theorem 2.8: Suppose K is a finite extension of Q. Then unique factorization of ideals holds in OK.
Proof. Combine Theorem 2.4 and Theorem 2.6.
§3 Primary decomposition*
[ADD: Commutative algebra generalization, and a new proof of unique ideal factorization]
§4 Ideal class group
Number Theory, §14.5.
Definition 4.1: A fractional ideal of A is a nonzero A-submodule of K such that da ∈ A for some d ∈ A.
A principal fractional ideal is one of the form
(b) := bA := {ba|a ∈ A}. The product of two fractional ideals is
ab=nXaibi|ai ∈ a, bi ∈ b
o .
Note that given a nonzero A-submodule of K, it is finitely generated iff it is a fractional ideal. (Take common denominators of the generators.)
We can extend unique factorization to fractional ideals, in the same way that we can extend unique factorization from Z to Q.
Theorem 4.2: The set Id(A) of fractional ideals is a free abelian group on the set of prime ideals. Thus each fraction ideal can be uniquely written in the form
a=Y
p
prp.
Proof. Freeness follows from unique factorization (Theorem 2.4) and existence of inverses follows from Corollary 2.5(3a).
Now we are ready for the following definition.
Definition 4.3: Let P (A) be the group of principal ideals of A. The ideal class group C(A) is Id(A)/P (A). Its order is the class number.
The ideal class group and class number of K are defined as the ideal class group and class number of OK.
Note that we have an exact sequence
0 → P (A) → I(A) → C(A) → 0.
The class number is 1 iff all A is a PID. Thus in some sense it measures how far A is from being a PID.
Alternatively there is an exact sequence
1 → OK× → K×→ IK → CK → 1
where the map K× → K is given by a 7→ (a).
Theorem 4.4 (Approximation Theorem): Let x1, . . . , xm ∈ A, and p1, . . . , pm be distinct
prime ideals. For any x ∈ N, there is x ∈ A such that ordpi(x − xi) > n for all i.
§5 Factorization in extensions
Assume AKLB, with A Dedekind and L/K finite separable. A prime ideal p ⊂ A will factor in B:
pB = Pe1
1 · · · P eg
g .
We say ei is the ramification index of Pi. For P | p, we write e(P/p) for the ramification
index and f (P/p) for the residue class degree [B/P : A/p]. 1. If ek> 1 for some k, p is ramified in B.
(a) If g = 1 and e1 > 1, p is totally ramified.
(b) When |A/p| = pn, p prime, and p - [B/P : A/p], then p is tamely ramified.
2. If ei = fi = 1 for all i, p splits completely.
3. If pB stays prime, p is inert.
Lemma 5.1: A prime ideal P divides p iff P ∩ K = p.
Theorem 5.2 (Degree equation): Let m = [L : K] and suppose pB = Pe1
1 · · · P eg g . Then g X i=1 eifi = m.
If L/K is Galois, then all the ei are equal and all the fi are equal. Letting e and f denote
these common values,
ef g = m.
Proof. We show both sides of the equation equal dimA/p(B/pB).
For the LHS, by the Chinese Remainder Theorem B/pB ∼=Qgi=1B/Pei
i so dimA/p(B/pB) = g X i=1 dimA/p(B/Peii). (14.1)
Consider the filtration
B ⊃ Pi ⊃ · · · ⊃ Peii.
There are no ideals between any two consecutive ideals by Corollary 2.5 (the first iff), so