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Snippet portion: From Iwaniec and Kowalski's Analytic Number Theory:

If the class number $h=h(D)$ is small, then there are only few prime ideals $\bf{p}$ of degree one with small norm. Indeed, if $p=\bf{p \bar{p}}$ with $(\bf{p},\bf{\bar{p}})=1$, then $\bf{p}^h$ is a principal ideal generated by $\frac{1}{2}(m+n\sqrt{D})$ with $n \ne 0$, when $p^h = \frac{1}{4}(m^2-n^2 D) \ge \frac{|D|}{4}$.

Therefore the least prime $p_1 = p_1(D)$ with $\chi_D(p_1)=1$ satisfies $p_1 \ge {(\frac{|D|}{4})}^{1/h}$.

Hence $\chi_D(n)$ agrees with $\mu(n)$ on all squarefree numbers $n \le {(\frac{|D|}{4})}^{1/h}$ with $(n,{(\frac{|D|}{4})}^{1/h})=1$. This property is not likely to hold in long segments (because $\chi_D$ is periodic while $\mu$ is not), therefore this suggests that h is rather large.

Question portion: Although the above argument would not work in a Real quadratic field ($D > 0$ so the last inequality in the first paragraph does not hold), it seems that if we replace the class number h with h times the regulator this should work.

Any ideas on how to actually show this?

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2 Answers

up vote 4 down vote accepted

The following is not a full answer, but perhaps gives you an idea of how to approach the result.

Let us consider the claim $$ p^{hR} \ge \Big(\frac{D}4\Big) $$ for the smallest noninert prime. I first show that the inequality holds whenever $p \ge 11$. In fact, we have $h \ge 1$ and $R \ge \frac12 \log D + O(1)$: the latter claim is clear since the fundamental unit $(t+u\sqrt{D})/2$ satisfies $u \ge 1$ and $t \approx u \sqrt{D}$. Since $\log p > 2$, the claimed inequality follows.

Thus we only have to look at small primes. If $R$ is large enough, the claim holds. If $R$ is as small as possible, the following happens. If $u = 1$, the discriminant must have a very special form. One possibility is $D = 4n^2 +1$. Here the fundamental unit is $\varepsilon = 2n+\sqrt{D}$ (unless $n = 1$), so $R \approx \frac12 \log D$. On the other hand, "Davenport's Lemma" shows that the minimal nontrivial norm is $n = N((2n+1+\sqrt{D})/2)$; see

  • Ankeny, Chowla & Hasse, On the class-number of the maximal real subfield of a cyclotomic field J. Reine Angew. Math. 217, 217-220 (1965);

    a different idea for a proof of this and similar results can be found in

  • Lemmermeyer & Pethö, Simplest cubic fields, Manuscr. Math. 88, No.1, 53-58 (1995).

Observe that we only have to check the claimed inequality for $n \le 7$.

Since the result holds for $R$ large and very small, I would guess that its proof is within reach. Good luck!

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This is a bit of a follow up to Franz's answer, but is also a bit different.

Dirichlet's class number formula tells us that $$hR = \frac{\mu(K)\sqrt{|D|}}{2^r(2\pi)^s}L(1,\chi)$$

A particular case of a famous result of Siegel says that for some positive constant $c$ $$L(1, \chi) > c|D|^{-1/4}$$

Combine to get that for some constant $c'$ $$hR > c'\sqrt[4]{|D|}$$

Notice that if $|D| > 2^{48}$ then $lg(|D|/4) < \sqrt[8]{|D|}$, so if $|D| > max\{2^{48},c'^{-8}\}$ then $$2^{hR} > 2^{c' \sqrt[4]{|D|}} > 2^{c'c'^{-1}\sqrt[8]{D}} > |D|/4$$ as required.

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The $c$ of Siegel is ineffective. Instead via $L$-functions you get $2^{hR}= 2^{2L(1\,chi)\sqrt d}$, so you need $(2\log 2)L(1,\chi)\sqrt D\ge\log (D/4)$ if I am correct. But this is not known effectively, as the best bounds are from the Gross-Zagier theorem and Goldfeld's work, and would be $\log(D)^{1-\epsilon}$ on the right. –  Junkie May 7 '10 at 4:16
    
True. But at least we know one exists. –  Dror Speiser May 7 '10 at 6:02
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