Algebraic Number Theory (II): Number Fields

25 Jan 2019

• algebraic number theory

A number field $K$ is a subfield of $\bb C$ which is a finite extension of $\bb Q$. These are the main objects of study of algebraic number theory. In this post, we look at some basic examples and properties, and define the trace and norm functions in number fields.

Quadratic fields

For our first example, consider the quadratic fields $\bb Q[\sqrt m]$ for any integer $m\neq0,1$. Clearly $\bb Q[\sqrt m]=\bb Q[\sqrt{d^2m}]$ for any nonzero integer $d$; hence we may assume that $m$ is squarefree. We call the fields $\bb Q[\sqrt m]$ for $m>1$ (resp. $m<0$) the real quadratic fields (resp. imaginary quadratic fields).

When $m\neq0,1$ is squarefree, the minimal polynomial of $\sqrt m$ over $\bb Q$ is $X^2-m$, which is irreducible over $\bb Q$ (since we can check that $\sqrt m$ is not a rational number). Hence $[\bb Q[\sqrt m]:\bb Q]=2$.

Proposition: Every quadratic field is normal over $\bb Q$.

Proof: Note that the conjugates of $\sqrt m$ are $\pm\sqrt m$, both of which lie in $\bb Q[\sqrt m]$. $\qed$

Conversely, we show that these $\bb Q[\sqrt m]$ completely parametrise number fields of degree $2$ over $\bb Q$.

Proposition: Every number field of degree $2$ over $\bb Q$ is one of the quadratic fields.

Proof: If $K=\bb Q[\alpha]$ has degree $2$, then $\alpha$ is the root of some quadratic polynomial $aX^2+bX+c$ with coefficients in $\bb Z$; then the quadratic formula gives

$$\alpha=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\in\bb Q[\sqrt{b^2-4ac}].\ \qed$$

Proposition: The fields $\bb Q[\sqrt m]$, for $m\neq0,1$ squarefree, are pairwise non-isomorphic.

Proof: Let $m,n\neq0,1$ be distinct squarefree integers, and suppose that there exists a field isomorphism $\varphi:\bb Q[\sqrt m]\to\bb Q[\sqrt n]$. This is an embedding of $\bb Q[\sqrt m]$ into $\bb C$, so by normality of $\bb Q[\sqrt m]/\bb Q$ we have $\bb Q[\sqrt m]=\bb Q[\sqrt n]$. In particular, $\sqrt m\in\bb Q[\sqrt n]$.

(Alternatively, without using normality, note that $\varphi(\sqrt m)=\pm\sqrt m\in\bb Q[\sqrt n]$ implies $\sqrt m\in\bb Q[\sqrt n]$.)

Then there exists $a,b\in\bb Q$ such that

$$\begin{aligned} \sqrt m&=a+b\sqrt n\\ \therefore m&=a^2+b^2n+2ab\sqrt n\\ \therefore\sqrt n&=\frac{m-a^2-b^2n}{2ab}, \end{aligned}$$

contradicting $\sqrt n\not\in\bb Q$. $\qed$

Trace and norm

Let $K$ be a number field, with $[K/\bb Q]=n$. Then by elementary field theory, there are $n$ embeddings $\sigma_1,\ldots,\sigma_n$ of $K$ in $\bb C$. Now for $\alpha\in K$, define the functions

$$\begin{aligned} T^K(\alpha)&=\sigma_1(\alpha)+\cdots+\sigma_n(\alpha),\\ N^K(\alpha)&=\sigma_1(\alpha)\cdots\sigma_n(\alpha), \end{aligned}$$

called respectively the trace and norm of $\alpha$ with respect to the field $K$. We will usually write $T^K=T$ and $N^K=N$ if the field $K$ is clear from context.

Since every $\sigma_k$ fixes $\bb Q$ pointwise, it is routine to check the following:

Proposition: Let $K$ be a number field with $[K:\bb Q]=n$.

1. For $r\in\bb Q$, we have $T(r)=nr$ and $N(r)=r^n$.

2. For $\alpha,\beta\in K$ and $r\in\bb Q$, we have

   $$\begin{aligned} T(\alpha+\beta)&=T(\alpha)+T(\beta),&T(r\alpha)&=rT(\alpha),\\ N(\alpha\beta)&=N(\alpha)N(\beta),&N(r\alpha)&=r^nN(\alpha).\ \qed \end{aligned}$$

For a concrete example, let us evaluate the trace and norm over quadratic fields.

Proposition: Let $K=\bb Q[\sqrt m]$, where $m\neq0,1$ is squarefree. Then for any $a,b\in\bb Q$, we have

$$\begin{aligned} T(a+b\sqrt m)&=2a,\\ N(a+b\sqrt m)&=a^2-mb^2. \end{aligned}$$

Proof: Note that each conjugate of $\sqrt m$ over $\bb Q$ determines an embedding of $K$ into $\bb C$. Hence the two embeddings of $K$ are given by

$$\sigma_\pm(a+b\sqrt m)=a\pm b\sqrt m.$$

Hence the two possible images of $a+b\sqrt m$ have sum $2a$ and product $a^2-mb^2$. $\qed$

For example, for $m=-1$, we recover the norm function over the Gaussian integers.

As a special case, we have the following interpretation for $K=\bb Q[\alpha]$:

Proposition: Let $\alpha\in\bb C$ be algebraic. Then $T^{\bb Q[\alpha]}(\alpha)$ and $N^{\bb Q[\alpha]}(\alpha)$ are respectively the sum and product of the conjugates of $\alpha$ over $\bb Q$.

Proof: If $\alpha$ has degree $d$ over $\bb Q$, then the $d$ embeddings of $\bb Q[\alpha]$ send $\alpha$ precisely to each of its $d$ conjugates. $\qed$

Corollary: $T^{\bb Q[\alpha]}(\alpha)$ and $N^{\bb Q[\alpha]}(\alpha)$ are rational.

Proof: By Vieta’s formulas, the minimal polynomial of $\alpha$ over $\bb Q$ is

$$X^d-T^{\bb Q[\alpha]}(\alpha)X^{d-1}+\cdots+(-1)^dN^{\bb Q[\alpha]}(\alpha),$$

and we are done since the coefficients are rational. $\qed$

In general, we have the following result:

Proposition: Let $K$ be a number field with $[K:\bb Q]=n$, and let $\alpha\in K$ be an algebraic number with degree $d$ over $\bb Q$. Then

$$\begin{aligned} T^K(\alpha)&=\frac ndT^{\bb Q[\alpha]}(\alpha),\\ N^K(\alpha)&=N^{\bb Q[\alpha]}(\alpha)^{n/d}. \end{aligned}$$

Proof: This follows from the fact that every embedding of $\bb Q[\alpha]$ in $\bb C$ extends to exactly $\frac nd$ embeddings of $K$ in $\bb C$. $\qed$

Corollary: For any number field $K$ and any $\alpha\in K$, we have $T^K(\alpha)$ and $N^K(\alpha)$ are rational. $\qed$

Relative trace and norm

Let $K,L$ be number fields such that $K\subseteq L$, with $[L:K]=n$. Let $\sigma_1,\ldots,\sigma_n$ be the embeddings of $L$ in $\bb C$ which fix $K$ pointwise. Now for $\alpha\in L$, define the functions

$$\begin{aligned} T^L_K(\alpha)&=\sigma_1(\alpha)+\cdots+\sigma_n(\alpha),\\ N^L_K(\alpha)&=\sigma_1(\alpha)\cdots\sigma_n(\alpha), \end{aligned}$$

called respectively the relative trace and relative norm of $\alpha$ with respect to $L/K$. This is a generalisation of the trace and norm defined previously, with $T^K=T^K_{\bb Q}$ and $L^K=L^K_{\bb Q}$. In this context, we have the following analogous results:

Proposition: Let $K,L$ be number fields such that $K\subseteq L$, with $[L:K]=n$.

1. For $\delta\in K$, we have $T(\delta)=n\delta$ and $N(\delta)=\delta^n$.

2. For $\alpha,\beta\in L$ and $\delta\in K$, we have

   $$\begin{aligned} T(\alpha+\beta)&=T(\alpha)+T(\beta),&T(\delta\alpha)&=\delta T(\alpha),\\ N(\alpha\beta)&=N(\alpha)N(\beta),&N(\delta\alpha)&=\delta^nN(\alpha).\ \qed \end{aligned}$$

Proposition: Let $\alpha\in\bb C$ be algebraic, and let $K$ be a number field. Then $T^{K[\alpha]}_ K(\alpha)$ and $N^{K[\alpha]}_ K(\alpha)$ are respectively the sum and product of the conjugates of $\alpha$ over $K$. $\qed$

Corollary: $T^{K[\alpha]}_ K(\alpha)$ and $N^{K[\alpha]}_ K(\alpha)$ are in $K$. $\qed$

Proposition: Let $K,L$ be number fields such that $K\subseteq L$, with $[L:K]=n$, and let $\alpha\in L$ be an algebraic number with degree $d$ over $K$. Then

$$\begin{aligned} T^L_K(\alpha)&=\frac ndT^{K[\alpha]}_ K(\alpha),\\ N^L_K(\alpha)&=N^{K[\alpha]}_ K(\alpha)^{n/d}.\ \qed \end{aligned}$$

Corollary: For any number fields $K,L$ such that $K\subseteq L$, and any $\alpha\in L$, we have $T^L_K(\alpha)$ and $N^L_K(\alpha)$ are in $K$. $\qed$

Finally, we prove the transitivity of the relative trace and relative norm.

Proposition: Let $K,L,M$ be number fields with $K\subseteq L\subseteq M$. Then for all $\alpha\in M$, we have

$$\begin{aligned} T^L_K(T^M_L(\alpha))&=T^M_K(\alpha),\\ N^L_K(N^M_L(\alpha))&=N^M_K(\alpha). \end{aligned}$$

Proof: Let $\sigma_1,\ldots,\sigma_n$ be the embeddings of $L$ in $\bb C$ fixing $K$ pointwise, and let $\tau_1,\ldots,\tau_m$ be the embeddings of $M$ in $\bb C$ fixing $L$ pointwise.

We want to compose the $\sigma_i$ with the $\tau_j$. As such, we fix a normal extension $N/\bb Q$ such that $M\subseteq N$, and choose an extension of each $\sigma_i$ and $\tau_j$ to an automorphism of $N$, which we will also denote by $\sigma_i$ and $\tau_j$. Now

$$\begin{aligned} T^L_K(T^M_L(\alpha))&=\sum_i\sigma_i\left(\sum_j\tau_j(\alpha)\right)=\sum_{i,j}\sigma_i\circ\tau_j(\alpha),\\ N^L_K(N^M_L(\alpha))&=\prod_i\sigma_i\left(\prod_j\tau_j(\alpha)\right)=\prod_{i,j}\sigma_i\circ\tau_j(\alpha). \end{aligned}$$

Now consider the $mn$ functions $\sigma_i\circ\tau_j\rvert_M$. Each of these is an embedding of $M$ in $\bb C$ fixing $K$ pointwise. Furthermore, if any two of them are equal, say $\sigma_i\circ\tau_j=\sigma_{i’}\circ\tau_{j’}$, then

$$\sigma_i|_ L=\sigma_i\circ\tau_j|_ L=\sigma_{i'}\circ\tau_{j'}|_ L=\sigma_{i'}|_ L,$$

so $i=i’$, hence $j=j’$. Thus the $mn$ functions $\sigma_i\circ\tau_j\rvert_M$ are pairwise distinct, so they form the complete list of embeddings of $M$ in $\bb C$ fixing $K$ pointwise (since $mn=[M:L][L:K]=[M:K]$). Therefore

$$\begin{aligned} \sum_{i,j}\sigma_i\circ\tau_j(\alpha)&=T^M_K(\alpha),\\ \prod_{i,j}\sigma_i\circ\tau_j(\alpha)&=N^M_K(\alpha), \end{aligned}$$

and we are done. $\qed$