Norms and Traces

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 number field generated by $ a$, which is the smallest number field that contains $ a$. If $ a\in \overline{\mathbf{Q}}$ then $ a$ has a minimal polynomial $ f(x)\in\mathbf{Q}[x]$, and the of $ a$ are the roots of $ f$. For example the element $ \sqrt{2}$ has minimal polynomial $ x^2-2$ and the Galois conjugates of $ \sqrt{2}$ are $ \pm \sqrt{2}$.

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 5.2.1 (Norm and Trace)   The and of $ a$ from $ L$ to $ K$ are

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

It is standard 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)}{\mathrm{Det}}(\ell_a)$, and the coefficient of $ x^{\deg(f)-1}$ is $ -\tr (\ell_a)$.

Proposition 5.2.2   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 (since it is the polynomial in $ K[x]$ of least degree that is satisfied by $ a$). Since $ f$ is irreducible, $ [K(a):K]=\deg(f)$, and $ a$ satisfies a polynomial if and only if $ \ell_a$ does, 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$ by Exercise 9). $ \qedsymbol$

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

Corollary 5.2.3   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 $ 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 $ \tau_1\sigma',\ldots,\tau_d\sigma'$ 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 5.2.2.

Proposition 5.2.4   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 5.1.8 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 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 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$. 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 5.2.5   The ring of integers $ \O _K$ of a number field is Noetherian.

Proof. By Proposition 5.2.4, 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 the Hilbert Basis Theorem, $ \O _K$ is Noetherian. $ \qedsymbol$

William Stein 2004-05-06