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!

  • 1
    $\begingroup$ Your counterexample shows that more argument is needed at that point. For example, $c_1+c_2$ is ruled put because from $N(t\alpha)=t^n N(\alpha)$, we know that our polynomial must be homogeneous of degree $n$. (This alone doesn't help yet either) $\endgroup$ Commented Feb 12, 2020 at 7:41
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    $\begingroup$ Author here, confirming this is indeed my mistake (rather than OP's). Thank you for pointing it out. I'll try to fix when I get a chance although this may not be for a while. But I will quickly at least delete the wrong proof now and leave a placeholder indicating it should be filled in later. $\endgroup$
    – Evan Chen
    Commented Feb 12, 2020 at 17:25

1 Answer 1


I think you are right and the proof is faulty.

  • As you pointed out, the fact that the norm $N(\alpha)$ of a number $\alpha\in K$ is an integer does not lead to a bound on the denominators of the coefficients $c_i$. As an example let $K=\Bbb{Q}(i)$, $n\in\Bbb{Z}_{>0}$ and let $$\alpha=\alpha(n):=\left(\frac{2+i}{2-i}\right)^n=\frac{(2+i)^{2n}}{5^n}.$$ We have $N(\alpha(n))=1$ for all $n$, as $(2+i)/(2-i)$ and all its powers lie on the unit circle. Yet the numerators of the coefficients of $\alpha(n)$ are both $5^n$ – something we can make as large as we wish. Of course, the numbers $\alpha(n)$ are not algebraic integers. In fact, the claim of that theorem is true, but the author's attempt to prove it with norm (of $\alpha$) alone is doomed to fail (barring something extra in the preceeding results, I don't have the time to check).
  • Another problem in that proof is that after we have derived an upper bound $N$ on the denominators it does not follow that $$\mathcal{O}_K\subseteq\frac1N\bigoplus_i \Bbb{Z}\alpha_i.$$ After all, if we know that a denominator is bounded by $4$, we might have $\alpha_1/3\in\mathcal{O}_K$, but this is not an element of $\frac14\bigoplus_i\Bbb{Z}\alpha_i$. This problem is easier to fix though. For once we have a bound $N$ on the absolute values of the denominators, then it does follow that $$\mathcal{O}_K\subseteq\frac1{N!}\bigoplus_i \Bbb{Z}\alpha_i.$$ The rest of the argument then works as explained.
  • To get that upper bound $N$ for the denominators I think they should use the trace. If $\alpha$ is an algebraic integer, then all the traces $tr(\alpha\alpha_i)$ are rational integers. This leads to a linear system of equations, satisfied by the $c_i$s, with integer coefficients. The determinant of that system thus gives a bound to the denominators (Cramer's rule and all that). There are scores of details to check. Consult introductory books on algebraic numbers for them.
  • $\begingroup$ I think what is especially important about the fix in the last point is that you (in contrast to the original argument) make sufficient use of the multiplicative structure of $\mathcal{O}_K$ $\endgroup$ Commented Feb 12, 2020 at 17:59

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