Can an integral basis be extended?

Question

Let $K$ and $L$ be finite degree number fields with $K\subseteq L$. Then clearly $O_K\subseteq O_L$. Can an integral basis for $O_K$ always be extended to an integral basis for $O_L$? I was wondering because I was trying to prove that $\Delta_K\vert \Delta_L$ where $\Delta_K$ and $\Delta_L$ denotes the fields' discriminants over $\mathbb{Q}$ and was thinking it might help if an integral basis for $O_K$ could be extended to one for $O_L$.

Answer 1

I think it's true and the following proof should go through.

Lemma. (Basis Extension Theorem for Modules) Let $\def\O{\mathcal{O}}\def\Z{\mathbb{Z}}M$ be an $R$-module and $N \subseteq M$ be a submodule such that:

• $N$ is free
• $M/N$ is free

Then $M$ is free and any basis of $N$ can be extended to a basis of $M$.

Proof. Recall/show that any short exact sequence $0 \to N \to M \to F \to 0$ of $R$-modules with $F$ free splits. In our case that yields $M \cong N \oplus M/N$. $\quad\square$

Now let $L/K$ be an extension of number fields. By the Lemma, it suffices to show that:

• $\O_K$ is free (as a $\Z$-module)
• $\O_L/\O_K$ is free (as a $\Z$-module)

$\O_K$ is free by the (algebraic number) theory, of course. (Or apply this proof here first to $K/\mathbb{Q}$!) To show that $\O_L/\O_K$ is free, it suffices to show that it's torsion-free (because we work over $\Z$, which is a PID). But $\O_L/\O_K$ is torsion-free because $\O_K$ is integrally closed. Indeed, assume $x \in \O_L$ satisfies $nx \in \O_K$ for some $n \in \Z_{>0}$. Then $x \in \operatorname{Frac}(\O_K) = K$ and $x$ is integral over $\Z$ because $x \in \O_L$. Since $\O_K$ is integrally closed, this implies $x \in \O_K$.