Elementary proof of $\mathcal{O}_K$ is a free abelian group of rank $n$ Let $K$ be a number field of degree $n$. It is known that $\mathcal{O}_K$ is a free abelian group of rank $n$. Most of the proofs start with defining a non-degenarete form and after some arguments, conclude the result. Also, the result can be shown via geometric methods like: The ring of integers of a number field is finitely generated..
So, I wonder whether there is an elementary proof that only uses basic abstract algebra? Alternative proofs are also appreciated.
 A: I ambitiously claim that there is no proof that doesn't use properties of the trace pairing (or equivalent) - and here's a sketch of why.
One half is easy: we need to show that there is a linear map $\mathbb{Z}^n\hookrightarrow \mathcal{O}_K$. Well, an integral basis $a_1, \dots, a_n$ for $K/\mathbb{Q}$ clearly does this for us: the sum $a_1 \mathbb{Z} + \dots + a_n \mathbb{Z}$ is a direct sum (this follows easily from the definition of a integral basis), so has rank $n$, and it lives inside $\mathcal{O}_K$ (by definition of an integral basis). The definitions were set up to make this work.
For the converse: we need to show that there is some linear map $\mathcal{O}_K \hookrightarrow \mathbb{Z}^n$. Well, if this exists, then we can certainly project onto any of its $n$ coordinates, which will give us a nontrivial linear map $\mathcal{O}_K\to \mathbb{Z}$. Now, what functions are there that do this?
(Important note: we already know - because some clever person has told us so - that the maps $t_i: x\mapsto \mathrm{Tr}_{K/\mathbb{Q}}(a_ix)$ work. In fact, the map $(t_1, \dots, t_n): \mathcal{O}_K\to \mathbb{Z}^n$ is injective, which is another way of saying that the $t_i$ are all $\mathbb{Z}$-linearly independent. So we happen to know, a fortiori, that $\mathcal{O}_K \cong \mathbb{Z}^n$.
But that's not the question you asked. What you asked is: couldn't we have chosen something simpler than the maps $t_i$? And we're going to use the above knowledge to show that the answer is "no".)
We know that we need linear maps from $\mathcal{O}_K$ to $\mathbb{Z}$. Let's look at the space of all such maps we could have chosen - I'll it denote by $\mathrm{Hom}(\mathcal{O}_K, \mathbb{Z})$, and note that it's a $\mathbb{Z}$-module. Our question is: are there nontrivial elements of $\mathrm{Hom}(\mathcal{O}_K, \mathbb{Z})$ that are somehow "nicer" than the maps $t_i$?
But, if $\mathcal{O}_K\cong \mathbb{Z}^n$ (as we secretly know by the important note above), then $\mathrm{Hom}(\mathcal{O}_K, \mathbb{Z})$ is just isomorphic to $\mathrm{Hom}(\mathbb{Z}^n, \mathbb{Z})$. This is the dual of $\mathbb{Z}^n$, which is again isomorphic to $\mathbb{Z}^n$. In particular, $\mathrm{Hom}(\mathcal{O}_K, \mathbb{Z})$ is a $\mathbb{Z}$-module of rank $n$. But in our important note above, we showed that the maps $t_1, \dots, t_n$ were $n$ linearly independent elements of this module. In other words, there's nothing in $\mathrm{Hom}(\mathcal{O}_K, \mathbb{Z})$ that isn't just a linear combination of the $t_i$.
So no, in conclusion: you need to use properties of the $t_i$ because you need some maps, and (up to linear combination) they're the only ones available.
