Algebraic Integers¶

Consider \(\mathbb{Z} \hookrightarrow \mathbb{Q} \hookrightarrow \mathbb{F}\) as a field extension with \(n = [\mathbb{F} : \mathbb{Q}] = \operatorname{dim}_{\mathbb{Q}} \mathbb{F} < \infty\). The question is that, do we have \(\mathbb{Z} \subset \mathcal{O}_{\mathbb{F}} \subset \mathbb{F}\) such that:

\(\mathbb{F}\) is a field of fractions of \(\mathcal{O}_{\mathbb{F}}\).
\(\mathcal{O}_{\mathbb{F}}\) is a ring such that \(\mathbb{Z} \subset \mathcal{O}_{\mathbb{F}}\).
\(\mathcal{O}_{\mathbb{F}}\) is a free module over \(\mathbb{Z}\) isomorphic to \(\mathbb{Z}^n\).

i.e. to ﬁnd the bottom-left question mark on the following diagram:

Due to the simplicity of the structure of \(\mathbb{Z}\), the third requirement is simply determining the ring structure of \(\mathcal{O}_{\mathbb{F}}\) in an easier way.

Integrality & Algebraicity¶

Note that \(\mathbb{Z} = \{x \in \mathbb{Q} \mid x - a = 0, a \in \mathbb{Z}\}\), we can define a similar concept:

\[ \{\alpha \in \mathbb{F} \mid f(\alpha) = 0, f \text{ is a monic polynomial with degree } n, f(x) \in \mathbb{Z}[x]\} \]

Theorem

Let \(R\) be a unital commutative ring and \(A \subset R\) a subring. Take \(x \in R\), then the following are equivalent:

There exists \(a_0, a_1, \ldots, a_{n-1} \in A\) such that \(x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 = 0\);
The ring \(A[x]\) is a finitely generated \(A\)-module;
There is a subring \(B \subset R\) such that \(A[x] \subset B \subset R\) and \(B\) is a finitely generated.

Proof

\(1 \Rightarrow 2 \Rightarrow 3\): Trivial.
\(3 \Rightarrow 1\): By \(3\), we have

\[ B = Ay_1 + Ay_2 + \cdots + Ay_n \]

Hence

\[ x_iy_i \]