Let $K = \mathbb{Q}(\sqrt{41})$, $\omega = \frac{1}{2}(1+\sqrt{41})$, and $\mathcal{O}_K= \mathbb{Z}[\omega ]$ be the ring of integers of $K$. Let $\alpha = 27 +10 \omega$ be the fundamental unit of $\mathcal{O}_K$. The characteristic polynomial of $\alpha $ is $f(x) = x^2 -64 x - 1$. Using $\text{disc}(f) = \text{disc}(K) \times [\mathcal{O}_K : \mathbb{Z}[\alpha ]]^2$, we find $\text{disc}(f) = 64^2 + 4 = 41\times 10^2$ so $[\mathcal{O}_K : \mathbb{Z}[\alpha ]] = 10 < 41$.

Now let $K$ be an extension of the rational numbers, $\mathcal{O}_K$ be the ring of integers of $K$. What is known on the existence of a unit $\alpha$ of $\mathcal{O}_K$ such that the index $[\mathcal{O}_L : \mathbb{Z}[\alpha ]] < N$, where $L = \mathbb{Q}(\alpha )$ and $N$ depends on the discriminant of the field $K$, for example if $N = \text{disc}(K)$ or the square root of this?

  • 2
    $\begingroup$ The index $[{\mathbf Q}(\alpha):{\mathbf Z}[\alpha]]$ is not what you mean: it's infinite! (Think about $[{\mathbf Q}:{\mathbf Z}]$.) Probably what you meant is the index $[{\mathcal O}_K:{\mathbf Z}[\alpha]]$, which is finite for real quadratic $K$. Most number fields don't have one fundamental unit. $\endgroup$
    – KCd
    Jun 30, 2013 at 2:28
  • $\begingroup$ Oh yeah :) need to edit.. thanks! $\endgroup$ Jun 30, 2013 at 2:31
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    $\begingroup$ But you still need to put a constraint on $K$: a general number field doesn't have a fundamental unit. And even if a number field has a fundamental unit, that unit might not generate the number field (and thus the ring it generates doesn't have finite index in ${\mathcal O}_K$). An example is $K = {\mathbf Q}(\zeta_5)$, for which the unit group of ${\mathcal O}_K$ is $\pm\mu_5((1+\sqrt{5})/2)^{\mathbf Z}$, so $\alpha := (1+\sqrt{5})/2$ is a fundamental unit but $[{\mathcal O}_K:{\mathbf Z}[\alpha]]$ is infinite. $\endgroup$
    – KCd
    Jun 30, 2013 at 2:38

1 Answer 1


If $K$ is a number field of signature $(r_1,r_2)$, the rank of the abelian group generated by units is $r_1 + r_2 -1$, so there is not always a "fundamental unit" in general. Let me assume, therefore, that you are talking about a real quadratic field. (The same analysis as below should also apply to imaginary cubic fields, the other case where the unit rank is one.) The ring of integers of $K$ is given by $$\mathbf{Z}[\alpha], \ \text{where} \ \beta = \frac{\Delta + \sqrt{\Delta}}{2} ,$$ and where $\Delta$ is the discriminant of $K$. The fundamental unit has the shape $$\alpha = a + b \beta.$$ The index of $\mathbf{Z}[\alpha]$ in $\mathcal{O}_K = \mathbf{Z}[\beta]$ is equal to $b$. We should be a little careful, because we can multiply the unit by $\pm 1$ or take its inverse. But the inverse of a fundamental unit is, up to sign, its conjugate, which is equal to $$\overline{\alpha} = (a + b d) - b \beta,$$ so the number $b$ remains unchanged. You want a bound for $b$. Fix an embedding of $K$ into $\mathbf{R}$, and assume that $\alpha > 1$. Then $|\overline{\alpha}| < 1$, and so, in terms of size,

$$\alpha \sim \alpha - \overline{\alpha} = 2 b \beta - b d = b \sqrt{\Delta}$$

That is, the size of $b$ is roughly the size of $\alpha/\sqrt{\Delta}$.

From the Brauer-Siegel theorem (which comes from the class number formula and estimates for $L$-values), we have the estimate

$$h_K \log |\alpha| \ll \Delta^{1/2 + \epsilon}_K,$$

where the implied constant depends on $\epsilon$ and is ineffective. This is the same formula which (when applied to imaginary quadratic fields) shows that there are only finitely many imaginary quadratic fields of class number one. Moreover, this bound is "optimal" in that one also has the converse inequality $h_K \log |\alpha| \gg \Delta^{1/2 - \epsilon}_K$. The integer $h_K$ appearing here is the class number, and it's hard to estimate in general, but at least one knows that $h_K \ge 1$. In fact, Gauss conjectured that $h_K = 1$ infinitely often. Hence we end up with the estimate:

$$\text{index} \ll \exp\left(\Delta^{1/2 + \epsilon}\right),$$

which is as good as you will be able to get. Indeed, assuming Gauss' conjecture, the index will be this size (at least with $1/2 - \epsilon$) infinitely often.

That being said, the index will also be $1$ infinitely often, just by taking $\sqrt{d}$ with $d = n^2 + 1$, where $\alpha = n + \sqrt{n^2+1}$ generates the ring of integers when $n$ is odd. In general, the index can be small only when the class number is large.


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