Rational Linear Combinations of Integral Basis

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 $k_1,\dots ,k_n\ne 0$ and consider the $\mathbb{Q}$ linear combination

Now we have that

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 $k_j\ne 0$ 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.

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

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