Rational Linear Combinations of Integral Basis

Theorem

If \(\{\alpha_1, \dots, \alpha_n\}\) is an integral basis of \(\mathcal{O}_\mathbb{K}\), then \(\{\alpha_1, \dots, \alpha_n\}\) is a basis for \(\mathbb{K}\) over \(\mathbb{Q}\). Moreover, if

\[ x = m_1\alpha_1 + \dots + m_n\alpha_n \in \mathbb{K}\]

then \(x \in \mathcal{O}_\mathbb{K}\) if and only if all \(m_i\) are in \(\mathbb{Z}\).

Proof

Let \(\{\alpha_1, \dots, \alpha_n\}\) is an integral basis for \(\mathcal{O}_\mathbb{K}\), and as such, they are \(\mathbb{Z}\) linearly independent. Now let \(m_1, \dots, m_n, k_1, \dots, k_n \in \mathbb{Z}\) with \(k_1, \dots, k_n \neq 0\) and consider the \(\mathbb{Q}\) linear combination

\[ \frac{m_1}{k_1}\alpha_1 + \frac{m_2}{k_2}\alpha_2 + \dots + \frac{m_n}{k_n}\alpha_n.\]

Now we have that

\[\begin{align*} & \frac{m_1}{k_1}\alpha_1 + \frac{m_2}{k_2}\alpha_2 + \dots + \frac{m_n}{k_n}\alpha_n = 0 \\ \implies& \frac{m_1k_1\dots k_n}{k_1}\alpha_1 + \frac{m_2k_1\dots k_n}{k_2}\alpha_2 + \dots + \frac{m_nk_1\dots k_n}{k_n}\alpha_n = 0 \\ \implies& m_1k_2\dots k_n\alpha_1 + m_2k_1k_3\dots k_n\alpha_2 + \dots + m_nk_1\dots k_{n - 1}\alpha_n = 0 \\ \end{align*}\]

which means that \(m_i k_1 \dots k_{i - 1} k_{i + 1} \dots k_n = 0\) for each \(i\) since \(\{\alpha_1, \dots, \alpha_n\}\) are \(\mathbb{Z}\) linearly independent. However \(k_j \neq 0\) for all \(j\), and so \(m_i = 0\) for all \(i\). This means that \(\frac{m_i}{k_i} = 0\) for all \(i\) and thus \(\{\alpha_1, \dots, \alpha_n\}\) is \(\mathbb{Q}\) linearly independent in \(\mathbb{K}\). Since this is a set of \(n\) vectors in an \(n\) dimensional space, linear independence implies it is a basis.

It is important to note that a converse result does not hold. That is, given a field basis of \(\mathbb{K}\) comprised entirely of integral elements, this does not necessarily have to be an integral basis.

Example

\(\mathbb{Q}(\sqrt{5})\) has field basis \(\{1, \sqrt{5}\}\) but this is not an integral basis.

See ring of integers of quadratic field for details. In short, \(5 \equiv 1 \pmod 4\) and so \(\mathcal{O}_{\mathbb{Q}(\sqrt{5})} \neq \mathbb{Z}[\sqrt{5}]\).

As such, we need more sophisticated methods, namely the integral basis algorithm.