Here I am storing various basic facts about Number Fields that are useful in other notes. I hope this becomes more complete as time goes on.

Number Fields

Recall that a number field $K$ is a finite extension of $\mathbb Q$. While we often think of number fields as $\mathbb Q(\alpha)$ for some algebraic number embedded in $\mathbb C$ it is useful to recall the general (unembedded) construction. $\mathbb Q[x]$ is the ring of polynomials with rational coefficients in the indeterminant $x$. If $f(x) \in \mathbb Q[x]$ is irreducible, then $f(x) \mathbb Q[x]$, the ideal formed from all rational polynomials divisible by $f(x)$, is a maximal ideal in $\mathbb Q[x]$. It follows that $K = \mathbb Q[x]/f(x) \mathbb Q[x]$ is a commutative ring with all non-zero elements invertible—that is a field.

In this construction, the elements of $K$ are cosets of the form $g(x) + f(x) \mathbb Q[x]$. If $g(x)$ and $h(x)$ generate the same coset, then we will write $g(x) \equiv h(x)$ (or $g(x) \equiv h(x) \bmod f(x)$ if more clarity is necessary). In this situation $f(x) | (g(x) – h(x))$.

Given the coefficients of $f(x)$, the arithmetic in $K$ is easy to perform. Suppose for $a_0, \ldots, a_{d-1}$ are the rational coefficients to $$f(x) = x^d + \sum_{n=0}^{d-1} a_n x^n,$$ then, $$ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}.$$ Now suppose $g(x) + f(x) \mathbb Q[x]$ is an arbitrary coset. By replacing monomials $x^n$ in $g(x)$ when $n > d$ (serially, if necessary) using this congruence, we see that $g(x) \equiv h(x)$ for some $h(x) \in \mathbb Q[x]$ with $\deg(g) < d$. The polynomial $h(x)$ is equivalent to the result of the Division Algorithm in $Q[x]$ for the remainder of $g(x)$ when divided by $f(x)$.

That is, as a group (in fact, as a vector space) $K$ is isomorphic to $\mathbb Q^d$ where the isomorphism is given by $$(b_0, \ldots, b_{d-1}) \mapsto x^d + \sum_{m=0}^{d-1} b_m x^m + f(x) \mathbb Q[x].$$ The only thing missing in this description is the multiplication. If we want to multiply two vectors $\mathbf b, \mathbf c \in \mathbb Q^d$, we set $g(x)$ to be the monic polynomial with coefficient vector $\mathbf b$ and $h(x)$ to be the polynomial with coefficient vector $\mathbf c$. We first multiply $g(x)$ and $h(x)$ as usual in $\mathbb Q[x]$, and then we use the equivalence $ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}$ to replace monomials in $g(x) h(x)$ (repeatedly if necessary) until we arrive at a polynomial $p(x)$ of degree $< d$. The coefficient vector of this polynomial in $\mathbb Q^d$ is the product of $\mathbf b$ and $\mathbf c$.

$K$, What is it Good for?

First, note that $\mathbb Q \hookrightarrow K$ by the map $r \mapsto r + f(x) \mathbb Q[x]$, and by definition (the fact that $K$ is a vector space of dimension $d$ over $\mathbb Q$) it is a number field of degree $d$ over $\mathbb Q$. This implies $\mathbb Q[x] \hookrightarrow K[x]$, and in particular, $f(x)$ has a life in $K[x]$. Because $f(x)$ is irreducible in $\mathbb Q[x]$ it has no zeroes in $\mathbb Q$. However, we will show that this is no longer the case in $K[x]$. And that is what $K$ is good for—producing a number field where $f(x)$ has a zero.

The element $x + f(x) \mathbb Q[x]$ is the root of $f(x)$ in $K$. To see this, we need only calculate $$f(x + f(x) \mathbb Q[x]) = f(x) + f(x) \mathbb Q[x] = 0 + f(x) \mathbb Q[x].$$

The element $x + f(x) \mathbb Q[x]$ is important as well because if we know how to multiply by this element, then we know how to multiply by arbitrary elements (which are, after all, simply linear combinations of its powers).

Multiplication by $x$ is a linear operator on $\mathbb Q[x]$, indeed $x( a g(x) + h(x) ) = a x g(x) + x h(x)$, and multiplication by $x + f(x) \mathbb Q[x]$ is a linear operator on $K$. We know $K$ is a vector space with basis $( x^n + f(x) \mathbb Q[x] : n=0,\ldots, d-1)$, so it makes sense to talk of the matrix of the multiplication operator, call it $T$, with respect to this basis. Note that, if we denote the standard basis of $\mathbb Q^d$ (with coordinates indexed from 0 to $d-1$ for consistency) by $\mathbf e_0, \ldots, \mathbf e_{d-1}$, then for $n < d-1$, $T \mathbf e_{n} = \mathbf e_{n+1}$. This corresponds to the multiplication $x x^{n} = x^{n+1}$ which remains true in $K$ if $n < d-1$. The final calculation, using the same equivalence that has gotten us so far $ x^d \equiv -a_0 – a_1 x – \cdots – a_{d-1} x^{d-1}$, shows that $T \mathbf e_{d-1} = -a_0 \mathbf e_0 – a_1 \mathbf e_1 – \cdots – a_{d-1} \mathbf e_{d-1}$. It follows that the matrix of $T$ with respect to the basis $(\mathbf e_n)$ is $$ \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & – a_1 \\ 0 & 1 & \cdots & 0 & – a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{d-1}\end{pmatrix}.$$

If this matrix looks familiar it is because it is the (Frobenius) companion matrix to $f(x)$ and the characteristic polynomial of this matrix (and hence the operator $T$) is $f(x)$. Indeed, the irreducibility of $f(x)$ implies that the minimal polynomial of $T$ is $f(x)$ as well.

But $f(x)$ has roots in $\mathbb C$. What about them?

The Fundamental Theorem of Algebra (ironically a theorem in analysis) guarantees that $f(x)$ has $d$ roots (counting multiplicity) in $\mathbb C$. How are they related to the root of $f(x)$ is $K$?

Let’s start with our favorite $\alpha \in \mathbb C$ such that $f(\alpha) = 0$. We know that $\alpha$ is either in $\mathbb R$ or it has a complex conjugate—more about that later. We can embed $K$ into $\mathbb C$ by sending $x + f(x) \mathbb Q[x] \rightarrow \alpha$. That is, $$ a_{d-1} x^{d-1} + \cdots + a_1 x + a_0 + f(x) \mathbb Q[x] \quad \mapsto \quad a^{d-1} \alpha^{d-1} + \cdots a_1 \alpha + a_0.$$ We denote this embedding by $\mathbb Q(\alpha) \subset \mathbb C$. Notice that if $\alpha \in \mathbb R$, then $\mathbb Q(\alpha) \subset \mathbb R$ and we call it a real embedding of $K$.

A count of the real and complex embeddings products the first of the classical invariants of a number field.

Invariants: Number of real and complex embeddings $r_1$ and $r_2$

Let us distinguish the real and complex roots of $f(x)$ by setting $\alpha_1, \ldots, \alpha_{r_1}$ to be the real roots and $\beta_1, \overline{\beta_1}, \ldots, \beta_{r_2}, \overline{\beta_{r_2}}$ be the non-real complex roots. Clearly $r_1 + r_2 = d$. Then the embeddings $\mathbb Q(\alpha_1), \ldots, \mathbb Q(\alpha_{r_1}) , \mathbb Q(\beta_1) , \ldots, \mathbb Q(\beta_{r_2})$ are called the archimedean embeddings of $K$.

The Norm and Trace

Here we wish to work in some generality and consider field extension $K | k$ where both are number fields. Little generality is lost by keeping the example $k = \mathbb Q$ at the front of your mind. However, as many properties of number fields ‘factor through’ intermediate fields (for instance $[ K : \mathbb Q] = [K : k] [k : \mathbb Q]$) it is useful to maintain some generality in notation etc.

We will also abandon our attempt to denote elements of $K | k$ as cosets in $k[x] / f(x) k[x]$, writing for instance $\alpha, \beta, \gamma, \ldots $ for generic field elements. Often we will implicitly identify $K$ with $k(\alpha)$ for some algebraic number $\alpha$ of degree $d$ over $k$. In this situation $\{1, \alpha, \ldots, \alpha^{d-1} \}$ is a basis for $K$, and the matrix of multiplication by $\alpha$ with respect to this basis is exactly the matrix of $T$ as before (the Frobenius companion matrix of the minimal polynomial of $K | k$).

More generally, given any $\gamma \in K$, we can make the linear operator “multiplication by $\gamma$” $T_{\gamma}$. If $\gamma$ is given as a $k$-linear combination of $\{1, \alpha, \ldots, \alpha^{d-1}\}$ then it is relatively easy to compute the matrix of $T_{\gamma}$ with respect to this basis. Note that this matrix has entries in $k$.

Norm and trace of $K|k$

The norm $N_{K|k} : K \rightarrow k$ and trace $\mathrm{Tr}_{K|k} : K \rightarrow k$ are respectively the determinant and trace of $T_{\gamma}$.

This definition is independent of basis, but can be computed explicitly in the basis $\{1, \alpha, \ldots, \alpha^{d-1}\}$.

If $\beta, \gamma \in K$, then $T_{\beta \gamma} =T_{\beta} \circ T_{\gamma}$ and $T_{\beta + \gamma} = T_{\beta} + T_{\gamma}$. The multiplicativity of the determinant and the additivity of the trace imply that $$N_{K|k}(\beta \gamma) = N_{K|k}(\beta) N_{K|k}(\gamma)$$ and $$\mathrm{Tr}_{K|k}(\beta + \gamma) = \mathrm{Tr}_{K|k}(\beta) + \mathrm{Tr}_{K|k}(\gamma).$$

The norm is a natural homomorphism from $K^{\times}$ onto $k^{\times}$ and the trace is a natural homomorphism from the additive group $(K, +)$ onto $(k,+)$.


Number fields
$f(x), g(x), h(x),$ etc.Polynomials, often in $\mathbb Q[x]$ or $k[x]$
$k, K, L$Number fields
$\alpha, \beta, \gamma,$ etcGeneric field elements. The field depends on context.
$[K : k]$, $d$The degree of a field extension. The fields depend on context.
$T_{\alpha}$The linear transformation on $K|k$ (fields context dependent) given by multiplication by $\alpha$.
$r_1, r_2$The number of real and complex embeddings (respectively) of $K$ (context dependent).
$N_{K|k}$, $\mathrm{Tr}_{K|k}$The Norm and Trace maps $K \rightarrow k$ given by $\alpha \mapsto \det( T_{\alpha})$ and $\alpha \mapsto \mathrm{Tr}( T_{\alpha})$.
$\mf o$, $\mf O$Rings of integers in $k$ and $K$.
$\mf a, \mf b, \mf A, \mf B,$ etc.Ideals in rings of integers. We often use lower case fraktur letters for ideals in $\mf o$ and capital fraktur letters for ideals in $\mf O$.
$\mf p, \mf q, \mf P, \mf Q$ Prime ideals in $\mf o$ and $\mf O$.
$\mathbb N \mf a, etc$ etcThe ideal norm $\mathbb N \mf a = [\mf o : \mf a]$.

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