Norms and Traces

In this section we develop some basic properties of norms, traces, and discriminants, and give more properties of rings of integers in the general context of Dedekind domains.

Before discussing norms and traces we introduce some notation for field extensions. If $ K\subset L$ are number fields, we let $ [L:K]$ denote the dimension of $ L$ viewed as a $ K$-vector space. If $ K$ is a number field and $ a\in \overline{\mathbf{Q}}$, let $ K(a)$ be the extension of $ K$ generated by $ a$, which is the smallest number field that contains both $ K$ and $ a$. If $ a\in \overline{\mathbf{Q}}$ then $ a$ has a minimal polynomial $ f(x) \in \mathbf{Q}[x]$, and the Galois conjugates of $ a$ are the roots of $ f$. The are called the Galois conjugates because the are the orbit of $ a$ under the action of $ \Gal (\overline{\mathbf{Q}}/\mathbf{Q})$.

Example 2.4.1   The element $ \sqrt{2}$ has minimal polynomial $ x^2-2$ and the Galois conjugates are $ \sqrt{2}$ and $ -\sqrt{2}$. The cube root $ \sqrt[3]{2}$ has minimial polynomial $ x^3-2$ and three Galois conjugates $ \sqrt[3]{2}, \zeta_3\sqrt[3]{2}, \zeta_3^2\sqrt[3]{2}$, where $ \zeta_3$ is a cube root of unity.

We create the extension $ \mathbf{Q}(\zeta_3)(\sqrt[3]{2})$ in SAGE.

sage: L.<cuberoot2> = CyclotomicField(3).extension(x^3 - 2)
sage: cuberoot2^3

Then we list the Galois conjugates of $ \sqrt[3]{2}$.

sage: cuberoot2.galois_conjugates()
[cuberoot2, (-zeta3 - 1)*cuberoot2, zeta3*cuberoot2]

Note that $ \zeta_3^2 = -\zeta_3 - 1$:

sage: zeta3 = L.base_field().0
sage: zeta3^2
-zeta3 - 1

Suppose $ K\subset L$ is an inclusion of number fields and let $ a\in
L$. Then left multiplication by $ a$ defines a $ K$-linear transformation $ \ell_a:L\to L$. (The transformation $ \ell_a$ is $ K$-linear because $ L$ is commutative.)

Definition 2.4.2 (Norm and Trace)   The norm and trace of $ a$ from $ L$ to $ K$ are

$\displaystyle \Norm _{L/K}(a)=\det(\ell_a)$    and $\displaystyle \quad
\tr_{L/K}(a)=\tr (\ell_a).$

We know from linear algebra that determinants are multiplicative and traces are additive, so for $ a,b\in L$ we have

$\displaystyle \Norm _{L/K}(ab) = \Norm _{L/K}(a)\cdot \Norm _{L/K}(b)$


$\displaystyle \tr_{L/K}(a+b) = \tr_{L/K}(a) + \tr_{L/K}(b).$

Note that if $ f\in\mathbf{Q}[x]$ is the characteristic polynomial of $ \ell_a$, then the constant term of $ f$ is $ (-1)^{\deg(f)}\det(\ell_a)$, and the coefficient of $ x^{\deg(f)-1}$ is $ -\tr (\ell_a)$.

Proposition 2.4.3   Let $ a\in
L$ and let $ \sigma_1,\ldots, \sigma_d$, where $ d=[L:K]$, be the distinct field embeddings $ L\hookrightarrow \overline{\mathbf{Q}}$ that fix every element of $ K$. Then

$\displaystyle \Norm _{L/K}(a) = \prod_{i=1}^d \sigma_i(a)$    and $\displaystyle \quad
\tr_{L/K}(a) = \sum_{i=1}^d \sigma_i(a).

Proof. We prove the proposition by computing the characteristic polynomial $ F$ of $ a$. Let $ f\in K[x]$ be the minimal polynomial of $ a$ over $ K$, and note that $ f$ has distinct roots and is irreducible, since it is the polynomial in $ K[x]$ of least degree that is satisfied by $ a$ and $ K$ has characteristic 0. Since $ f$ is irreducible, we have $ K(a) = K[x]/(f)$, so $ [K(a):K]=\deg(f)$. Also $ a$ satisfies a polynomial if and only if $ \ell_a$ does, so the characteristic polynomial of $ \ell_a$ acting on $ K(a)$ is $ f$. Let $ b_1,\ldots,b_n$ be a basis for $ L$ over $ K(a)$ and note that $ 1,\ldots, a^m$ is a basis for $ K(a)/K$, where $ m=\deg(f)-1$. Then $ a^i b_j$ is a basis for $ L$ over $ K$, and left multiplication by $ a$ acts the same way on the span of $ b_j, a b_j, \ldots, a^m b_j$ as on the span of $ b_k, a b_k, \ldots, a^m b_k$, for any pair $ j,
k\leq n$. Thus the matrix of $ \ell_a$ on $ L$ is a block direct sum of copies of the matrix of $ \ell_a$ acting on $ K(a)$, so the characteristic polynomial of $ \ell_a$ on $ L$ is $ f^{[L:K(a)]}$. The proposition follows because the roots of $ f^{[L:K(a)]}$ are exactly the images $ \sigma_i(a)$, with multiplicity $ [L:K(a)]$ (since each embedding of $ K(a)$ into $ \overline{\mathbf{Q}}$ extends in exactly $ [L:K(a)]$ ways to $ L$). $ \qedsymbol$

It is important in Proposition 2.4.3 that the product and sum be over all the images $ \sigma_i(a)$, not over just the distinct images. For example, if $ a=1\in L$, then $ \Tr _{L/K}(a) = [L:K]$, whereas the sum of the distinct conjugates of $ a$ is $ 1$.

The following corollary asserts that the norm and trace behave well in towers.

Corollary 2.4.4   Suppose $ K\subset L \subset M$ is a tower of number fields, and let $ a\in M$. Then

$\displaystyle \Norm _{M/K}(a) = \Norm _{L/K}(\Norm _{M/L}(a))$    and $\displaystyle \quad
\tr_{M/K}(a) = \tr_{L/K}(\tr_{M/L}(a)).

Proof. For the first equation, both sides are the product of $ \sigma_i(a)$, where $ \sigma_i$ runs through the embeddings of $ M$ into $ \overline{\mathbf{Q}}$ that fix $ K$. To see this, suppose $ \sigma:L\to \overline{\mathbf{Q}}$ fixes $ K$. If $ \sigma'$ is an extension of $ \sigma$ to $ M$, and $ \tau_1,\ldots,
\tau_d$ are the embeddings of $ M$ into $ \overline{\mathbf{Q}}$ that fix $ L$, then $ \sigma'\tau_1,\ldots,\sigma'\tau_d$ are exactly the extensions of $ \sigma$ to $ M$. For the second statement, both sides are the sum of the $ \sigma_i(a)$. $ \qedsymbol$

The norm and trace down to  $ \mathbf {Q}$ of an algebraic integer $ a$ is an element of  $ \mathbf {Z}$, because the minimal polynomial of $ a$ has integer coefficients, and the characteristic polynomial of $ a$ is a power of the minimal polynomial, as we saw in the proof of Proposition 2.4.3.

Proposition 2.4.5   Let $ K$ be a number field. The ring of integers $ \O_K$ is a lattice in $ K$, i.e., $ \mathbf{Q}\O_K = K$ and $ \O_K$ is an abelian group of rank $ [K:\mathbf{Q}]$.

Proof. We saw in Lemma 2.3.15 that $ \mathbf{Q}\O_K = K$. Thus there exists a basis $ a_1,\ldots, a_n$ for $ K$, where each $ a_i$ is in $ \O_K$. Suppose that as $ x=\sum_{i=1}^n c_i a_i\in \O_K$ varies over all elements of $ \O_K$ the denominators of the coefficients $ c_i$ are arbitrarily large. Then subtracting off integer multiples of the $ a_i$, we see that as $ x=\sum_{i=1}^n c_i a_i\in \O_K$ varies over elements of $ \O_K$ with $ c_i$ between 0 and $ 1$, the denominators of the $ c_i$ are also arbitrarily large. This implies that there are infinitely many elements of $ \O_K$ in the bounded subset

$\displaystyle S = \left\{c_1 a_1 +\cdots + c_n a_n : c_i \in \mathbf{Q},  0\leq c_i \leq 1\right\}\subset K.$

Thus for any $ \varepsilon >0$, there are elements $ a,b\in \O_K$ such that the coefficients of $ a-b$ are all less than $ \varepsilon $ (otherwise the elements of $ \O_K$ would all be a ``distance'' of least  $ \varepsilon $ from each other, so only finitely many of them would fit in $ S$).

As mentioned above, the norms of elements of $ \O_K$ are integers. Since the norm of an element is the determinant of left multiplication by that element, the norm is a homogenous polynomial of degree $ n$ in the indeterminate coefficients $ c_i$, which is 0 only on the element 0. If the $ c_i$ get arbitrarily small for elements of $ \O_K$, then the values of the norm polynomial get arbitrarily small, which would imply that there are elements of $ \O_K$ with positive norm too small to be in $ \mathbf {Z}$, a contradiction. So the set $ S$ contains only finitely many elements of $ \O_K$. Thus the denominators of the $ c_i$ are bounded, so for some $ d$, we have that $ \O_K$ has finite index in $ A=\frac{1}{d}\mathbf{Z}a_1 + \cdots + \frac{1}{d}\mathbf{Z}a_n$. Since $ A$ is isomorphic to $ \mathbf{Z}^n$, it follows from the structure theorem for finitely generated abelian groups that $ \O_K$ is isomorphic as a $ \mathbf {Z}$-module to $ \mathbf{Z}^n$, as claimed. $ \qedsymbol$

Corollary 2.4.6   The ring of integers $ \O_K$ of a number field is noetherian.

Proof. By Proposition 2.4.5, the ring $ \O_K$ is finitely generated as a module over $ \mathbf {Z}$, so it is certainly finitely generated as a ring over $ \mathbf {Z}$. By Theorem 2.2.9, $ \O_K$ is noetherian. $ \qedsymbol$

William Stein 2012-09-24