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

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

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

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

and

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

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

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

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

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

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

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

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. $\qedsymbol$