We have an elementary sharp lower bound for the regulator of a real quadratic field as a function of the discriminant

$$R\geq \tfrac{1}{2}(\sqrt{d-4}+\sqrt{d})$$

It is sharp because the equality holds infinitely often for $d=x^2+4$.

The problem of finding a good upper seems much more complicated, but there's still is a very nice (and relatively easy) bound for $\mathbb{Q}(\sqrt{D})$ depending only on $D$.

Loo-Keng Hua proved that $L(1,\chi)<1+\tfrac{1}{2} \log D$, so using the trivial estimate $h\geq 1$ and substituting on Dirichlet's class number formula we get

$$R<\sqrt{D}(\tfrac{1}{2}\log D+1)$$

The way this bound is presented (indirectly) in this survey suggests that it might be the best one currently known for all real quadratic fields (well, for $D>5$). Given how old Hua result is (1942), this seems unlikely, but I haven't been able to find a better one so far.

I am aware of much better estimates which work for sufficiently large $D$. For example it follows from work of Lavrik that

$$R<(0.263+\mathcal{o}(1))\sqrt{D}\log D$$

What is the best known bound for $R$ which holds for all real quadratic fields and depends only on $D$? (or for $D>k$ with the $k$ explicit and "small")

I'm also interested in what the true bound is expected to be.

  • $\begingroup$ Perhaps it would be better to write $o(1)$ instead of $\delta$ in the last display. Because what you mean is: the bound holds for $D$ sufficiently large in terms of $\delta$, and this is exactly what the $o(1)$ notation stands for. $\endgroup$
    – GH from MO
    Oct 23, 2016 at 17:18
  • $\begingroup$ @GHfromMO Thanks for the suggestion, I've edited to change the notation. $\endgroup$
    – Myshkin
    Oct 23, 2016 at 17:50
  • 1
    $\begingroup$ "Much better" depends on the application. The Lavrik bound is still within a constant factor (indeed, within a factor of $2$) of the bound using $h=1$ and Dirichlet, so if that's the best known then for many purposes it's essentially equivalent to the bound with a coefficient of $1/2$ in place of $0.263$. The tough question is whether the regulator is $o(\sqrt{D} \log D)$ for large $D$, and if so then by how much. $\endgroup$ Oct 23, 2016 at 18:50

1 Answer 1


Stephane Louboutin has several papers on getting explicit bounds for $L(1,\chi)$, for $\chi$ a character $\pmod q$. They're all of the strength of $1/2 \log q + $ an explicit constant. Some of his results include information on $\chi(2)$ which is sometimes helpful.

The best theoretical upper bound is due to P.J. Stephens: it gives $$ L(1,\chi) \le \frac{1}{4} \Big( 2- \frac{2}{\sqrt{e}} + o(1)\Big) \log q = (0.1967\ldots +o(1)) \log q, $$ for a quadratic character $\chi \pmod q$ (see for example Upper bounds for $|L(1,\chi)|$). Here the $1/4$ is from Burgess and the $2-2/\sqrt{e}$ comes from Vinogradov's trick for the least quadratic non-residue. To go beyond $1/2$ in the bound for $L(1,\chi)$ explicitly, one would have to work with either explicit Burgess type bounds or with explicit versions of Vinogradov's trick. I don't think anyone has carried that out -- nothing new here, but just needs elbow grease.

On GRH Theorem 1.5 from Lamzouri-Li-Soundararajan will give you explicit bounds for $L(1,\chi)$. These show that the regulator is bounded by something of size $\sqrt{d} \log \log d$, which is likely the correct order of magnitude (but this is unknown).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.