## Number Fields

An (algebraic) number field is a subfield of $$\mathbb{C}$$ 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 $$I = \langle 2, 1+\sqrt{-3} \rangle$$. Then $$I \ne \langle 2\rangle$$, but $$I^2 = 2I$$. Furthermore, $$I$$ is the unique prime ideal containing $$\langle 2\rangle$$ and hence $$\langle 2\rangle$$ is not the product of prime ideals.

Example: Let $$\omega = e^{2 \pi i / m}$$ for an odd 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 $$f = g h$$ 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 $$m g, n h \in \mathbb{Z}[x]$$, so the coefficients of $$m g, n h$$ have no common factor. (For example, if $$m g$$ did have a common factor $$d$$, we could replace $$m$$ by $$m/d$$; we know $$d | m$$ because the leading coefficient of $$m g$$ is $$m$$ since $$g$$ is monic.) Now if $$m n > 1$$, then let $$p$$ be some prime dividing $$m n$$. Consider the equation $$m n f = (m g)(n h) = 0 (mod p)$$. Since $$\mathbb{Z}_p[x]$$ is an integral domain, this means $$(m g) = 0$$ or $$(n h)=0$$, implying that $$p$$ divides all the coefficients of $$m g$$ or $$n h$$, 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

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

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

$x^2 - 2r x + r^2 - m s^2$

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

Theorem: Let $$a \in \mathbb{C}$$. 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 $$\mathbb{C}$$ that has a finitely generated additive group.

4. $$a A \subset A$$ for some finitely generated additive subgroup $$A \subset \mathbb{C}$$.

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

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

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

$\begin{pmatrix} a a_1 \\ \vdots \\ a a_n \end{pmatrix} = M \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix}$

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

$(a I - M) \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} = 0$

But since the vector $$(a_1 ... a_n)$$ is nonzero, we must have that $$det(a I - M) = 0$$. 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 $$a+b, a b$$

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 $$\mathbb{C}$$ form a ring, which we shall denote by $$\mathbb{A}$$. 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 $$\mathbb{Z}_K$$.

Ben Lynn blynn@cs.stanford.edu 💡