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
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
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.

\(a\) is an algebraic integer.

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

\(a\) belongs to a subring of \(\mathbb{C}\) that has a finitely generated additive group.

\(a A \subset A\) 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 \(1, a, ..., a^{n1}\).
(2) \(\implies\) (3) \(\implies\) (4) is clear.
(4) \(\implies\) (1): Let \(a_1,...,a_n\) generate \(A\). We may write
where \(M\) is some \(n \times n\) matrix with integer entries. Thus
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 \(\mathbb{Z}_K\).