The theorem and proof in question are from Evan Chen's An Infinitely Large Napkin. I've tried to include the relevant part here, but if you wish to view it in context please see Theorem 47.2.12 from the PDF here.
Theorem: Let $K$ be a number field of degree $n$.
Then its ring of integers $O_K$ is a free $\mathbb{Z}$-module of rank $n$,
i.e. $O_K \cong \mathbb{Z}^{\oplus n}$ as an abelian group.
In other words, $O_K$ has a $\mathbb{Z}$-basis of $n$ elements as
$$ O_K = \left\{ c_1\alpha_1 + \dots
+ c_{n-1}\alpha_{n-1} + c_n\alpha_n \mid c_i \in \mathbb{Z} \right\} $$
where $\alpha_i$ are algebraic integers in $O_K$.
Proof: Pick a $\mathbb{Q}$-basis of $\alpha_1$, …, $\alpha_n$ of $K$ and WLOG the $\alpha_i$ are in $O_K$ by scaling.
Consider $\alpha \in O_K$,
and write $\alpha = c_1\alpha_1 + \dots + c_n\alpha_n$.
We will try to bound the denominators of $c_i$.
Look at $N(\alpha) = N(c_1\alpha_1 + \dots + c_n\alpha_n)$.
If we do a giant norm computation, we find that $N(\alpha)$
is a polynomial in the $c_i$ with fixed coefficients.
(For example, $N(c_1 + c_2\sqrt 2) = c_1^2 – 2c_2^2$, say.)
But $N(\alpha)$ is an integer, so the denominators of the $c_i$
have to be bounded by some very large integer $N$.
Thus
$$ \bigoplus_i \mathbb{Z} \cdot \alpha_i \subseteq O_K
\subseteq \frac 1N \bigoplus_i \mathbb{Z} \cdot \alpha_i. $$
The latter inclusion shows that $O_K$ is a subgroup
of a free group, and hence it is itself free.
On the other hand, the first inclusion shows it's rank $n$.
My issue is with the conclusion that the denominators of the $c_i$ are bounded. Consider for instance the polynomial $c_1 + c_2$. Surely this can take integer values with arbitrarily small denominators, for instance with $\frac{1}{x} + \frac{x-1}{x}$?
Is this a flaw in the proof, and if so, is the proof rescuable (for instance, using some more properties of the polynomial) or entirely wrong? (While the source is an informal document, I've yet to find any other major errors and am not exactly a math expert, so at the moment I'm leaning towards the proof being correct and me doing something obviously stupid.) Thanks!
Best Answer
I think you are right and the proof is faulty.