Advanced Algebra II Jun. 1, 2007

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Definition

An integral domain D is a Dedekind domain if 1. D is a Noetherian ring;

2. D is integrally closed;

3. all non-zero prime ideals are maximal ideals.

That is, it’s an 1-dimensional Noetherian integrally closed domain.

Let K be its field of quotients.

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Example

Z is a Dedekind domain.

Let Q(√

d )/Q be a quadratic extension. And let O be the algebraic integers in Q(√

d ). Then O is a Dedekind domain.

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Definition

Let a C D be an ideal. A fractional ideal b is a finitely generated D-submodule of K , that is

b= ca := {x ∈ K |x = ca, c ∈ K^∗, a ∈ a}.

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In fact, every finitely generated D-module M in K can be

generated by elements {^a_b¹, ...,^a_bⁿ} for some b. Let a := (a₁, ..., a_n) and c = b⁻¹, then M = ca.

We can define the product on the set of fractional ideals in a natural way. There is an identity element, D in the set of fractional ideals. And in particular, if a C D, then ab ⊂ b. For a ∈ K^∗, we write

(a) = aD

and any fractional ideal of this form is called principal. Clearly (a)(b) = (ab).

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We can define the product on the set of fractional ideals in a natural way. There is an identity element, D in the set of fractional ideals. And in particular, if a C D, then ab ⊂ b.

For a ∈ K^∗, we write

(a) = aD

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We can define the product on the set of fractional ideals in a natural way. There is an identity element, D in the set of fractional ideals. And in particular, if a C D, then ab ⊂ b.

For a ∈ K^∗, we write

(a) = aD

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Theorem

Every non-zero ideal a C D can be written as a product of prime ideals of D,

a= p1· · · p_n.

Moreover, this representation is unique up to the order of prime ideals.

Theorem

Every fractional ideal of a Dedekind domain is invertible.

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A fractional ideal a is said to be invertible if there is a fractional ideal b such that ab = D We need the following Lemmas.

For a fractional ideal a, we consider R_a:= {x ∈ K |x a ⊂ a} and a⁻¹:= {x ∈ K |x a ⊂ D}.

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A fractional ideal a is said to be invertible if there is a fractional ideal b such that ab = D We need the following Lemmas.

For a fractional ideal a, we consider R_a:= {x ∈ K |x a ⊂ a} and a⁻¹:= {x ∈ K |x a ⊂ D}.

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Lemma

If D is Noetherian and integrally closed, then R_a = D.

Now let S := {a|R_a ( a⁻¹}. S is non-empty if D is not a field and let a = (a) for some a ∈ D − D^∗.

Since D is Noetherian, we have a maximal element m in S . We claim that m is prime and invertible.

Given a ∈ D − m and b ∈ D such that ab ∈ m. Note that there is x ∈ K − D such that x m ⊂ D. We first consider m + (a) C D. Then x (m + (a)) ⊂ D + x (a). Since m is maximal, we must have x (a) = (xa) 6⊂ D.

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Lemma

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Lemma

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Lemma

Given a ∈ D − m and b ∈ D such that ab ∈ m. Note that there is x ∈ K − D such that x m ⊂ D. We first consider m + (a) C D.

Then x (m + (a)) ⊂ D + x (a). Since m is maximal, we must have x (a) = (xa) 6⊂ D.

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Next we consider m + (b). Now ax (m + (b)) ⊂ ax m + (abx ) ⊂ D.

Hence m + (b) ∈ S and m + (B) = m by maximality. We thus have b ∈ m.

Suppose now that D is Dedekind, hence prime ideal is maximal. So we have m ⊂ mm⁻¹⊂ D. But m 6= mm⁻¹, otherwise

,m⁻¹⊂ R_m= D which is absurd. It follows that m⁻¹m= D.

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Next we consider m + (b). Now ax (m + (b)) ⊂ ax m + (abx ) ⊂ D.

Hence m + (b) ∈ S and m + (B) = m by maximality. We thus have b ∈ m.

Suppose now that D is Dedekind, hence prime ideal is maximal.

So we have m ⊂ mm⁻¹⊂ D. But m 6= mm⁻¹, otherwise ,m⁻¹⊂ R_m= D which is absurd. It follows that m⁻¹m= D.

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Lemma

Every prime ideal of a Dedekind domain is invertible.

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[Proof of theorems]

We have seen that if m is a mximal element in S , then m is invertible and prime.

We claim that a non-zero ideal a C D is invertible if and only if a= m1· · · m_r, with m_i being maximal elements in S .

One direction is clear. So that a is proper invertible, then

D ( a⁻¹. Hence a ∈ S and there is maximal m1 ∈ S containing a.

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[Proof of theorems]

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[Proof of theorems]

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Clearly, we have a ⊂ am⁻¹₁ ⊂ D. Notice that a 6= am⁻¹₁ , otherwise m⁻¹₁ ⊂ R_a = D.

If am⁻¹₁ = D, then a = am⁻¹₁ m₁ = Dm1= m1, we are done. If am⁻¹₁ ( D, then am⁻¹₁ is again invertible ( with inverse a⁻¹m₁). Proceed this process, and by Noetherian condition, we end up with D= am⁻¹₁ m⁻¹₂ ...m⁻¹_r and this proves the claim.

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If am⁻¹₁ = D, then a = am⁻¹₁ m₁ = Dm1= m1, we are done.

If am⁻¹₁ ( D, then am⁻¹₁ is again invertible ( with inverse a⁻¹m₁). Proceed this process, and by Noetherian condition, we end up with D= am⁻¹₁ m⁻¹₂ ...m⁻¹_r and this proves the claim.

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If am⁻¹₁ ( D, then am⁻¹₁ is again invertible ( with inverse a⁻¹m₁).

Proceed this process, and by Noetherian condition, we end up with D= am⁻¹₁ m⁻¹₂ ...m⁻¹_r and this proves the claim.

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If am⁻¹₁ ( D, then am⁻¹₁ is again invertible ( with inverse a⁻¹m₁).

Proceed this process, and by Noetherian condition, we end up with D= am⁻¹₁ m⁻¹₂ ...m⁻¹_r and this proves the claim.

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For any prime ideal p C D, pick any a ∈ p, we consider (a) which is invertible.

Hence we have m1m₂...mr = (a) ⊂ p. Thus mj ⊂ p for some j . Since every prime is maximal, we have m_j = p. It follows that p is invertible.

Finally, for any a C D, a ⊂ p1 for some prime ideal. Since prime is invertible, we have a ⊂ ap⁻¹₁ ⊂ D. Clearly a 6= ap⁻¹₁ , otherwise p⁻¹⊂ R_a = D. By Noetherian condition, we will end up with ap⁻¹₁ p⁻¹₂ ...p⁻¹_r = D. It follows that a = p1· · · p_r.

It remains to show the uniqueness. If p₁· · · p_r = q₁· · · q_s. Then p₁⊃ q₁· · · q_s. It follows that p₁ ⊃ q_j for some j . Since prime is maximal, we have p1= qj. By multiplying p⁻¹₁ and prove by induction, we are done.

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For any prime ideal p C D, pick any a ∈ p, we consider (a) which is invertible. Hence we have m1m₂...mr = (a) ⊂ p. Thus mj ⊂ p for some j . Since every prime is maximal, we have m_j = p. It follows that p is invertible.

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Finally, for any a C D, a ⊂ p1 for some prime ideal. Since prime is invertible, we have a ⊂ ap⁻¹₁ ⊂ D.

Clearly a 6= ap⁻¹₁ , otherwise p⁻¹⊂ R_a = D. By Noetherian condition, we will end up with ap⁻¹₁ p⁻¹₂ ...p⁻¹_r = D. It follows that a = p1· · · p_r.

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Finally, for any a C D, a ⊂ p1 for some prime ideal. Since prime is invertible, we have a ⊂ ap⁻¹₁ ⊂ D. Clearly a 6= ap⁻¹₁ , otherwise p⁻¹⊂ R_a = D.

By Noetherian condition, we will end up with ap⁻¹₁ p⁻¹₂ ...p⁻¹_r = D. It follows that a = p1· · · p_r.

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For a fractional ideal b, it’s of the form ca for some a C D.

Suppose that a = p1· · · p_r, then c⁻¹p⁻¹₁ · · · p⁻¹_r b= D. So b is invertible.

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Exercise

c⁻¹p⁻¹₁ · · · p⁻¹_r = b⁻¹.

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Let I_D be the set of fractional ideals. It’s naturally an abelian group via the multiplication of fractional ideals. There is a group homomorphism ϕ : K^∗ → I_D by ϕ(x ) = (x ), the fractional ideal generated by x . Let Cl (D) be the cokernel, which is called the ideal classes of D.

Proposition

Dis PID if and only if Cl (D) = 1.

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Proof.

If Cl (D) = 1, then every ideal a C D is a fractional ideal, i.e.

a= cD for some c ∈ K . We may write c = ^b_a and c ∈ a ⊂ D implies that b = at for some t ∈ D. It follows that c ∈ D and thus a= (c) is a principal ideal.

On the other hand, if D is PID, then every primes ideal p is principal, hence so is every fractional ideal.

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It’s easy to see that for all prime ideal p C D, we can define v_P : I_D → Z in a natural way. This gives rise to:

Corollary

I_D is a free abelian group. In fact, I_D ∼= ⊕_pZ.

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Let us consider the kernel of ϕ. For x ∈ K^∗, ϕ(x ) = 1 means that (x ) = D. In other words, x ∈ D and indeed x is a unit in D.

Hence we have kernel equals to D^∗. By abuse the language, we call D^∗ the ”group of unit of K ” and denoted UK usually. It’s thus essential to study units and classgroups.

We close this section by considering extensions. Let L/K be a finite separable extension and let O be the integral closure of D in L.

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Theorem

Let L/K be a finite separable extension of fields and keep the notation as above. Then

1. O is a finitely generated D-module which span L over K . 2. O is a Dedekind domain.

3. Every prime ideal P C O lies above a primes ideal p of D. For every prime ideal p C D, there exists at least one, and at most [L : K ], primes lying over p.

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Proof.

For any D-module X of L, we consider

X^D:= {x ∈ L|tr (xy ) ∈ D, ∀y ∈ X }.

It’s easy to see that O ⊂ O^D.

Pick X a free-module with a basis {x1, ..., xn} so that this is also a basis of L/K . We may assume that x_i ∈ O by eliminating the enumerators.

Since trace is non-singular, one has dual basis {y1, ..., yn} for X^D. Summarize, we have D-modules.

X ⊂ O ⊂ O^D ⊂ X^D.

Since D is Noetherian and X^D is finitely generated, so is O. This proves (1).

For (2), it’s easy.

Now consider p C D a prime ideal. It’s clear that pO ( O. So O/pO is a finite dimensional algebra over D/p. Hence there are finitely many maximal ideals. These maximal ideals corresponds to maximal ideals lying over p.

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Use this theorem to the case of algebraic number fields L/Q with algebraic numbers O. There is a free Z-basis ω1, ..., ωnof O. so that it spans L over Q.

We define the discriminant

det(tr_L/_A(ω_iω_j)) = det(ω^σ_i )².

Note that this is a rational algebraic integer, hence an integer. One can also prove that this is independent of choice of basis. Thus we call it dL.

Proposition d_L≡ 0, 1 mod 4.

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Use this theorem to the case of algebraic number fields L/Q with algebraic numbers O. There is a free Z-basis ω1, ..., ω_nof O. so that it spans L over Q.

We define the discriminant

det(tr_L/_q(ωiωj)) = det(ω^σ_i )². This follows from







tr (ω1ω1) tr (ω1ω2) . . . tr (ω1ωn)

... ...

tr (ω_nω₁) tr (ω_nω₂) . . . tr (ω_nω_n)





=







ω₁^σ¹ ω₁^σ² . . . ω^σ₁ⁿ

... ...

ω_n^σ¹ ω_n^σ² . . . ω^σ₁ⁿ













ω^σ₁¹ ω₂^σ¹ . . . ω^σ_n¹

... ...

ω^σ₁ⁿ ω₂^σⁿ . . . ω^σ_nⁿ





. Note that this is a rational algebraic integer, hence an integer. One can also prove that this is independent of choice of basis. Thus we call it dL.

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Proposition dL≡ 0, 1 mod 4.

Let t be the number of real embeddings, then there are n − t imaginary embeddings. It’s clear that n − t = 2s since there is complex conjugation. Then one sees that

Exercise

sgn(d_L) = (−1)^s.

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Let D be a Dedekind domain and let a be a non-zero ideal.

Suppose that D/a has finite element. We define Na := |D/a|.

Proposition Nab = NaNb.

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In fact, if K is an algebraic number field, then N(aD) = |N_{K /}_Qa|

for all a ∈ K^∗. Proposition

Every ideal class contains an D-ideal. And two D-ideals are equivalent if and only a1a₁ = a2a₂ for some non-zero a1, a2 ∈ D.

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Theorem

For a number field K , the ideal class group C_K is bounded by κ :=Qn

j =1(Pn

i =1|v_i^σ^j|), hence finite.

Example

Let K = Q(ω). dK = −3. {1, ω} is a basis. κ = 4. We look for ideals with norm ≤ 4. They are principal, hence e get a PID.

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Let D be a Dedekind domain and K be its field of quotients. Let p be a prime ideal. Then we have v_p : K^∗ → Z such that:

v_p(xy ) = v_p(x )v_p(y ). and v_p(x + y ) ≥ inf (v_p(x ), v_p(y )).

Also this is surjective. We may extend the definition to K by assign v_p(0) = ∞.

It follows that v (1) = 0.

For a given K with a function v : K → Z ∪ {∞} with above property is called a valuation. For a valuation, let

R := {x ∈ K |v (x ) ≥ 0} and m := {x ∈ K |v (x ) ≥ 1}. Then R is called the valuation ring. It’s easy to check that m is a maximal ideal, called the valuation ideal.

We can show that R is a local PID.