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 \(m \gt 0\), and the imaginary quadratic fields are those with \(m \lt 0\). 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 \(m n \gt 1\), 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 💡