Algebraic Number Theory, Lecture 2

Bryden R. Cais

September 29, 2002

1 Traces, Norms, and Discriminants

Let A ⊂ B be rings, and suppose that B is a finitely generated free A module, that is B ≃ A^{n}. For an element b ∈ B, we can define

Tr_{B/A}(b) := Tr(m_{b}),

where m_{b} : B → B is the matrix of multiplication by b. One can check that this is independent of a choice of basis. We also define

N_{B/A}(b) := det(m_{b}).

If A ⊂ B ⊂ C are modules (each free and finitely generated over A) then a straightforward matrix argument gives

Tr_{C/A} = Tr_{C/B} ∘ Tr_{B/A}
N_{C/A} = N_{C/B} ∘ N_{B/A} .

Lemma 1. Suppose that L/K is a finite separable extension of fields of degree d and that σ_{1}, …, σ_{d} are the d distinct embeddings of L into K̄ fixing K. Then

Tr_{L/K} = ∑_{i=1}^{d} σ_{i}  (1)

N_{L/K} = ∏_{i=1}^{d} σ_{i}.  (2)

Proof. Let α ∈ L have minimal polynomial

min_{α}(x) = x^{n} + a_{n−1}x^{n−1} + … + a_{0} ∈ K[x]

and set F = K(α). It follows at once from the definition of the characteristic polynomial that min_{α}(x) is the characteristic polynomial of multiplication by α, and consequently that

Tr_{L/K}(α) = −a_{n−1}  (3)
N_{L/K}(α) = (−1)^{n}a_{0}.  (4)

Now we know that

F ≃ K[x]/(min_{α}(x)).

Therefore, if α = α_{1}, α_{2}, …, α_{n} are the n distinct roots of min_{α}(x) in K̄, then the n distinct embeddings of F in K̄ are given by

x ↦ α_{i}.

It now follows from (3) and (4) that

Tr_{L/K}(α) = ∑_{i=1}^{n} α_{i}
N_{L/K}(α) = ∏_{i=1}^{n} α_{i}.

We are therefore done if F = L. If not, it is not hard to see that

∑_{i=1}^{d} σ_{i}(α) = ∑_{j=1}^{n} ∑_{σ_{i}(α)=α_{j}} α_{j} = ∑_{j=1}^{n} deg(L/F)α_{j} = deg(L/K) Tr_{F/K}(α).

Since α ∈ F, multiplication by α as a linear map of L as an F vector space is just diagonal; therefore, the corresponding matrix for L as a K vector space is block diagonal. The claim for Tr L/K(α) follows from this, and the proof for N_{L/K}(α) is similar so we omit it.

The following properties of Tr_{L/K} and N_{L/K} follow easily from the definitions:

1. Tr_{L/K}(α + β) = Tr_{L/K}(α) + Tr_{L/K}(β).

2. If a ∈ K then Tr_{L/K}(aα) = a Tr_{L/K}(α).

3. If α ∈ K then Tr_{L/K}(α) = deg(L/K).

4. N_{L/K}(αβ) = N_{L/K}(α) N_{L/K}(β).

5. If a ∈ K then N_{L/K}(aα) = a^{deg(L/K)} N_{L/K}(α).

6. We have that ⟨x, y⟩ ↦ Tr_{L/K}(xy) is a K bilinear pairing.

Now suppose that A ⊂ B are rings with B ≃ A^{n} and let β_{1}, …, β_{n} ∈ B. Definition 1. We define the discriminant of β_{1}, …, b_{n} to be

disc(β_{1}, …, β_{n}) = Δ(β_{1}, …, β_{n}) = det(Tr_{B/A}(β_{i}β_{j})).

Observe that if M ∈ N_{n×n}(A) and if (γ_{1}, …, γ_{n}) = (β_{1}, …, β_{n})M, then since

Tr_{B/A}(γ_{1}γ_{j}) = M^{T} (Tr_{B/A}(β_{i}β_{j}))M,

we have

disc(γ_{1}, …, γ_{n}) = (det(M))^{2} disc(β_{1}, …, β_{n}).

Lemma 2. Suppose that B ≃ A^{n}.

1. If {β_{1}, …, β_{n}} and {γ_{1}, …, γ_{n}} are two A bases for B then

disc(γ_{1}, …, γ_{n}) = u^{2} disc(β_{1}, …, β_{n})

for some unit u ∈ A^{∗}.

2. If {β_{1}, …, β_{n}} is an A basis for B and

(disc(γ_{1}, …, γ_{n})) = (disc(β_{1}, …, β_{n})) ⊂ A

then {γ_{1}, …, γ_{n}} is also an A basis for B.

Proof. This is clear.

Example 1. Let D ≠ 0, 1 be a square free integer and take A = ℤ, ℤ[\sqrt{D}]. Then clearly \{1, \sqrt{D}\} is a ℤ basis of B and we have

disc(1, \sqrt{D}) = det (\begin{matrix}Tr_{B/A}(1) & Tr_{B/A}(\sqrt{D}) \\ Tr_{B/A}(\sqrt{D}) & Tr_{B/A}(D) \end{matrix})

= det (\begin{matrix}2 & 0 \\ 0 & 2D \end{matrix})

= 4D.

As another example, if D ≡ 1 mod 4 then it is not difficult to see that α = \frac{1+\sqrt{D}}{2} is an algebraic integer. Letting B = ℤ ⊕ αℤ, we find that

disc(1, α) = det (\begin{matrix}Tr_{B/A}(1) & Tr_{B/A}(α) \\ Tr_{B/A}(α) & α^{2} \end{matrix})

= det (\begin{matrix}2 & 1 \\ 1 & \frac{1+D}{2} \end{matrix} )

= D.

2 Integrality

Definition 2. An element α ∈ B is integral over A if it satisfies f(α) = 0 for some monic f ∈ A[x]. If every element of B is integral over A, we say that B is integral over A.

Proposition 1. An element α ∈ B is integral over A if and only if A[α] ⊂ B is a finitely generated A module.

Proof. We have in any case a surjective map

A[x]/(f(x)) ↠ A[α].

If α is integral, so that f is monic, then A[x]/(f) is a finitely generated A module, with basis 1, x, …, x^{n−1}, where n = deg(f). Conversely, if A[α] is a finitely generated A module, let ω_{i} for i = 1, …, m be a basis for A[α] over A. Then

αω_{i} = ∑_{j=1}^{m} b_{ij}ω_{j}

for each i, where the b_{ij} ∈ A. We then see that α is a root if the characteristic polynomial of the matrix given by the b_{ij}, which is clearly monic. Lemma 3. 1. If B is a finitely generated A module then every element of B is integral over A.

2. The A-integral elements of B form a subring of B. Proof. If α, β are integral over A, then A[α, β] is finitely generated since A[α, β] = A[α][β] and β is integral over A[α] since it is integral over A. Since α + β, αβ ∈ A[α, β], it follows that A[α + β], A[αβ] are finitely generated A modules, from which the lemma follows. Definition 3. The integral closure of A ⊂ B is the ring of all elements of B that are integral over A. Definition 4. The ring A is said to be integrally closed in B if it is its own integral closure in B. Lemma 4. The integral closure of A in B is integrally closed. Proof. Let C be the integral closure of A in B and suppose that α ∈ B is integral over C. Then α^{n} + c_{n−1}α^{n−1} + ⋯ + c_{0} = 0 for some c_{i} ∈ C. Then

A[c_{0}, …, c_{n−1}, α]

is obviously finitely generated over A and hence α is integral over A. Definition 5. A domain A is integrally closed if it is integrally closed in its field of fractions. Example 2. ℤ is integrally closed since it is integrally closed in ℚ. Lemma 5. Suppose that A is an integrally domain with field of fractions K and that L/K is a finite separable extension of fields. Then α ∈ L is integral over A if and only if min_{α}(x) ∈ K[x] is in A[x]. Proof. One direction is obvious. So suppose that α is integral over A. Let p(x) ∈ A[x] be any monic polynomial having α as a root. Then for any embedding σ : L ↪ K̄ fixing K we see that σ(α) is a root of p(x). It follows that σ(α) is integral and hence, since the σ(α) are precisely the roots of min_{α}(x), that the coefficients of min_{α}(x) (which are elementary symmetric polynomials in the σ(α)) are integral over A. On the other hand, these coefficients are fixed by each σ and therefore lie in K. Using the fact that A is integrally closed shows that min_{α}(x) ∈ A[x], as required. Definition 6. Let K be a number field. The ring of integers, O_{k} is the integral closure of ℤ in K. Lemma 6. If α ∈ O_{K} then

Tr_{K/ℚ}(α), N_{K/ℚ}(α) ∈ ℤ.

More generally, we have Proof. The trace and norm are coefficients of min_{α}(x) ∈ ℤ[x] by Lemma 5. Lemma 7. If L/K is a finite extension of number fields and α ∈ O_{L} then

Tr_{L/K}(α), N_{L/K}(α) ∈ O_{K}.