Number Fields

An (algebraic) number field is a subfield of $ℂ$ whose degree over $\mathbb{Q}$ is finite.

It turns out that number fields are Dedekind domains thus all their ideals factor uniquely into prime ideals. An example of a ring where this is not true is $\mathbb{Z}[\sqrt{-3}]$: take the ideal $I=⟨2,1+\sqrt{-3}⟩$. Then $I\ne ⟨2⟩$, but $I^2=2I$. Furthermore, $I$ is the unique prime ideal containing $⟨2⟩$ and hence $⟨2⟩$ is not the product of prime ideals.

Example: Let $\omega =e^{2\pi i/m}$ for an integer $m$. Then $\mathbb{Q}[\omega ]$ is the $m$th cyclotomic field. The $m$th and $2m$th cyclotomic fields are identical because we have $\omega =-{\omega }^{m+1}=-({\omega }^2{)}^{(m+1)/2}\in \mathbb{Q}[{\omega }^2]$.

Example: The quadratic fields are the number fields $\mathbb{Q}[\sqrt{m}]$ for squarefree $m$ (and these are exactly the number fields with degree 2 over $\mathbb{Q}$). The real quadratic fields are the $\mathbb{Q}[\sqrt{m}]$ with $m>0$, and the imaginary quadratic fields are those with $m<0$. Notice the $m=-1,-3$ cases are also cyclotomic fields. By considering the equation $\sqrt{m}=a+b\sqrt{n}$ it can be seen that the fields $\mathbb{Q}[\sqrt{m}]$ are pairwise nonisomorphic.

An algebraic integer is a complex number that is a root of a monic polynomial in $\mathbb{Z}[x]$.

Theorem: The minimal polynomial of an algebraic integer over $\mathbb{Q}$ has coefficients in $\mathbb{Z}$.

Proof: This follows from the following lemma.

Lemma: Let $f\in \mathbb{Z}[x]$ be monic. Then if $f=gh$ for monic $g,h\in \mathbb{Q}[x]$ then $g,h\in \mathbb{Z}[x]$.

Proof: (This is a special case of Gauss' Lemma.) Let $m,n$ be the smallest positive integers such that $mg,nh\in \mathbb{Z}[x]$, so the coefficients of $mg,nh$ have no common factor. (For example, if $mg$ did have a common factor $d$, we could replace $m$ by $m/d$; we know $d\mid m$ because the leading coefficient of $mg$ is $m$ since $g$ is monic.) Now if $mn>1$, then let $p$ be some prime dividing $mn$. Consider the equation $mnf=(mg)(nh)=0(\mathrm{mod}p)$. Since ${\mathbb{Z}}_p[x]$ is an integral domain, this means $(mg)=0$ or $(nh)=0$, implying that $p$ divides all the coefficients of $mg$ or $nh$, which is a contradiction. Thus $m=n=1$.

Corollary: The only algebraic integers in $\mathbb{Q}$ are the ordinary integers.

Corollary: Let $m$ be a squarefree integer. The set of algebraic integers in quadratic field $\mathbb{Q}[\sqrt{m}]$ is

$$\begin{array}{cc}\{a+b\sqrt{m}:a,b\in \mathbb{Z}\}& \mathrm{if}m=2,3(\mathrm{mod}4);\\ \{\frac{a+b\sqrt{m}}{2}:a,b\in \mathbb{Z},a=b(\mathrm{mod}2)\}& \mathrm{if}m=1(\mathrm{mod}4).\end{array}$$

Proof: Let $\alpha =r+s\sqrt{m}$ for rationals $r,s$. If $s\ne 0$ then the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is

$$x^2-2rx+r^2-m{s}^2$$

Thus $\alpha$ is integral if and only if $2r$ and $r^2-m{s}^2$ are integers, implying the results above.

Theorem: Let $a\in ℂ$. The following are equivalent.

1. $a$ is an algebraic integer.

2. $\mathbb{Z}[a]$ is finitely generated as an additive group.

3. $a$ belongs to a subring of $ℂ$ that has a finitely generated additive group.

4. $aA\subset A$ for some finitely generated additive subgroup $A\subset ℂ$.

Proof: (1) $\Rightarrow$ (2): If $n$ is the degree of $a$ over $\mathbb{Q}$ then $\mathbb{Z}[a]$ is generated by $1,a,...,a^{n-1}$.

(2) $\Rightarrow$ (3) $\Rightarrow$ (4) is clear.

(4) $\Rightarrow$ (1): Let $a_1,...,a_n$ generate $A$. We may write

$$\left(\begin{array}{c}a{a}_1\\ ⋮\\ a{a}_n\end{array}\right)=M\left(\begin{array}{c}{a}_1\\ ⋮\\ a_n\end{array}\right)$$

where $M$ is some $n\times n$ matrix with integer entries. Thus

$$(aI-M)\left(\begin{array}{c}{a}_1\\ ⋮\\ a_n\end{array}\right)=0$$

But since the vector $(a_1...a_n)$ is nonzero, we must have that $\det (aI-M)=0$. This is a monic equation in $a$ (of degree $n$), so $a$ must be an algebraic integer.

Corollary: If $a,b$ are integral, so are $a+b,ab$

Proof: Since $\mathbb{Z}[a],\mathbb{Z}[b]$ are finitely generated, we have that $\mathbb{Z}[a,b]$ is finitely generated.

Thus the set of algebraic integers in $ℂ$ form a ring, which we shall denote by $𝔸$. We call a the subring of algebraic integers of $K$ the number ring corresponding to $K$.

In other texts the ring of algebraic integers is denoted ${\mathbb{Z}}_{\bar{\mathbb{Q}}}$, and the number ring corresponding to a field $K$ is denoted ${\mathbb{Z}}_K$.