Denominator of Algebraic Number

A priori, this so called denominator may not exist. As such, the remainder of this note is devoted to proving that indeed, as above, it is well defined.

Proof

Let $\mathbb{K}$ be a number field of degree $n$ over $\mathbb{Q}$. $\mathbb{K}$ has an integral basis $\{{\alpha }_1,\dots ,{\alpha }_n\}$ and as such, every element $\alpha \in \mathbb{K}$ can be expressed as

where $m_1,\dots ,m_n,k_1,\dots ,k_n\in \mathbb{Z}$ and $k_1,\dots ,k_n\ne 0$. That is, everything is a $\mathbb{Q}$ linear combination of the basis. Then, letting $k=\mathrm{lcm}(k_1,\dots ,k_n)$, we have that

is a $\mathbb{Z}$ linear combination of the integral basis and hence an algebraic integer. That is, $k\alpha \in {\mathcal{O}}_{\mathbb{K}}$.

Proof

Of course every algebraic number $\alpha$ is an element of the number field $\mathbb{Q}(\alpha )$ and thus there exists a positive integer $k$ such that $k\alpha \in {\mathcal{O}}_{\mathbb{Q}(\alpha )}$ which is a subset of the set of all algebraic integers.