In an unfinished (and as of now unpublished) article intended for the encyclopedia of mathematics, Arnold Scholz wrote:

"Classifying extensions according to the Galois group of their normal closure provides us with a new point of view. Not only the minimal discriminants but also the mean values of the ideal densities differ considerably, and have the following values for discriminants with large prime factors:

  • $\sqrt{\zeta(2)}$ for quadratic extensions;
  • $\sqrt[3]{\zeta(3)^2}$ and $\sqrt{\zeta(2)}\sqrt[3]{\zeta(3)}$ for a cubic extension according as it is cyclic or noncyclic;
  • $\sqrt{\zeta(2)}\sqrt{\zeta(4)}$ and $\sqrt{\zeta(2)}^3$ for cyclic and biquadratic quartic extensions, respectively."

I'd like to know what Scholz is talking about here. Ideal density might be some limit of the form "number of ideals with norm $\le x$" / $x$, and mean value should denote some average over number fields. But what exactly is Scholz doing here?

Edit. Apparently (this is suggested by some remarks he made elsewhere), Scholz called the expression $$ \prod_p \phi(p^n)/\Phi_K(p) $$ the ideal density of a number field $K$, where $\phi$ and $\Phi_K$ denote Euler's phi function in the rationals and in $K$, respectively, and where $n$ denotes the degree of $K$. This expression occurs in the product formula for the zeta function. I still don't know where to go from here.

As for Robin's remark on the density of fields ordered by discriminants, Scholz claimed, in a letter to Hasse dated Sept. 27, 1938, the following: The Dirichlet series $$ G(s) = \sum_{Gal(K)=G} D_K^{-s}, $$ where the sum is over all quartic fields whose normal closure has Galois group $G$, have abscissa of convergence $\alpha(D)=1$, $\alpha(Z) = \alpha(V) = \frac{1}{2}$ and probably $\alpha(S)=1$, $\alpha(A)=\frac{1}{2}$, where $D$, $Z$, $V$, $A$, $S$ denote the dihedral, cyclic, four, alternating and symmetric group. Moreover, $$ \lim_{s \to 1/2} \frac{Z(s)}{V(s)} = 0, $$ where $Z(s)$ and $V(s)$ are the Dirichlet series defined above for $G=Z$ and $G=V$. This is all correct, as we know now, but how could Scholz have discovered (and, for $G = D$, $Z$, $V$, proved) these results in the 1930s?

  • $\begingroup$ I've retagged this with the generic nt.number-theory tag (more people will see it that way). $\endgroup$ Mar 21 '10 at 15:58
  • 2
    $\begingroup$ ideal density is residue at $1$ of $\zeta_k$; if you take geometric mean of this over fields $k$, I think you get quantities as stated. No idea if this is what he meant. $\endgroup$
    – moonface
    Mar 21 '10 at 18:23
  • $\begingroup$ @moonface: I just checked in sage and got: $(\prod_{0 < -D < X} h(D))^{1/r(X)} \sim c\sqrt{X}$ , where $r(X)$ is the number of fundamental negative discriminants greater than $−X$ , and the sum ranges only on those as well. The experimental data says that $c$ is about 0.231, not close to 6. Did I misunderstand? $\endgroup$ Mar 21 '10 at 19:18
  • $\begingroup$ Two things: The density of ideals (as measured by norms) is not the same as the class number; rather, it's proportional to $h(D)/\sqrt{D}$ in the imaginary quadratic case. Secondly, the phrase "with large prime factors" makes a significant difference. I should say I didn't test this experimentally, but playing with it on paper it seems like the answers are as Scholz mentioned. $\endgroup$
    – moonface
    Mar 21 '10 at 19:43
  • $\begingroup$ You should multiply by $\pi$ (that's the constant of proportionality). Then it looks a lot more like $\sqrt{\zeta(2)}$. $\endgroup$
    – moonface
    Mar 21 '10 at 21:06

For explaining Scholz's ideas on ideal densities, consider the Dedekind zeta function of a number field $K$. If we collect all Euler factors for the prime ideals above a prime number $p$, then a short calculation shows that this product equals $$ E_p = \frac{1 - \frac1p}{(1 -p^{-f_1}) \cdots (1 - p^{-f_g})}. $$ If $\Phi$ denotes Euler's phi function in $K$ and $\phi$ the usual phi function for integers, then this equation may be rewritten in the form $$ \lim_{s \to 1} \frac{\zeta_K(s)}{\zeta(s)} = \frac{2^t\pi^shR}{v \sqrt{|D|}} = \prod_p \frac{\phi(p^n)}{\Phi(p)}. $$ The mean ideal density of a family $F$ of number fields $K_1$, $K_2$, $K_3$, \ldots is defined as $$ {\rm MID}(F) = \lim_{n \to \infty} \Big( \lim_{s \to 1} \frac{\zeta_{K_1}(s)}{\zeta(s)} \cdots \lim_{s \to 1} \frac{\zeta_{K_n}(s)}{\zeta(s)} \Big)^{\frac 1n} $$ if this limit exists.

Let us now compute the mean ideal density of quadratic number fields ordered by discriminants. Here $$ \Phi_K(p) = \begin{cases} (p-1)^2 & \text{ if $p$ splits}, \\ p(p-1) & \text{ if $p$ ramifies}, \\ p^2-1 & \text{ if $p$ is inert}, \end{cases} $$ hence the product of the Euler factors is $$ E_p = \frac{\phi(p^2)}{\Phi_K(p)} = \begin{cases} \frac{p}{p-1} > 1 & \text{ if $p$ splits}, \\ \qquad \quad 1 & \text{ if $p$ ramifies}, \\ \frac{p}{p+1} < 1 & \text{ if $p$ is inert} \end{cases} $$ and we find $$ \lim_{s \to 1} \frac{\zeta_K(s)}{\zeta(s)} = \prod_p \frac{p}{p-1} \prod_q \frac{q}{q+1} , $$ where $p$ runs through the split primes and $q$ through those that are inert.

Using some handwaving we expect that half the primes split and the other half remains inert, so heuristically we should get $$ {\rm MID}(F) = \lim_{n \to \infty} \Big( \prod_{p_1} \frac{p_1}{p_1-1} \prod_{q_1} \frac{q_1}{q_1+1} \cdots \prod_{p_n} \frac{p_n}{p_n-1} \prod_{q_n} \frac{q_n}{q_n+1}\Big)^{\frac1n}, $$ where $p_j$ are primes splitting in $K_j$ etc. Changing limits we find $$ \lim_{s \to 1} \lim_{n \to \infty} \sqrt[n]{\frac{\zeta_1(s)}{\zeta(s)} \cdots \frac{\zeta_n(s)}{\zeta(s)}} = \Big(\prod_p \frac{p}{p-1} \prod_q \frac{q}{q+1} \Big)^{\frac12} ,$$ where the products now are over all primes $p$ and $q$. Now $$ \Big(\prod_p \frac{p}{p-1} \prod_p \frac{p}{p+1} \Big)^{\frac12} = \Big(\prod_p \frac{p^2}{p^2-1} \Big)^{\frac12} = \Big(\prod_p \frac1{1-\frac1{p^2}} \Big)^{\frac12} = \sqrt{\zeta(2)} $$ in agreement with Scholz's claim.

This prediction does not agree well with numerical experiments, and a closer examination reveals that this is due to our omitting the ramified primes from our considerations. In fact, for a given prime number $p$ about $\frac{X}{p+1}$ of the discriminants below $X$ are divisible by $p$, and in the remaining fields half the primes split and the rest are inert. If we take this behaviour into consideration it turns out that we expect the mean ideal density for quadratic number fields to be $$ {\rm MID(F)} = \sqrt[3]{\frac{2^2}{2^2-1}} \sqrt{\zeta(2)} , $$ which agrees very well with numerical experiments.

"Computing" the mean ideal densities for other families of number fields is straightforward. I do not know how difficult it is to make these calculations rigorous, or what to do with them.


This may be a reference to the the Davenport-Heilbronn theorem on the distribution of discriminants of number fields. See math.stanford.edu/~fthorne/davenport-heilbronn.pdf for a nice exposition. Strictly speaking the Davenport-Heilbronn theorem is the case of cubic extensions. Quadratic extensions are easy. I believe general quartic (and quintic) extensions were proved recently by Bhargava; I don't know who first proved the special cases of quartics that Scholz lists.

  • $\begingroup$ Scholz died in 1942. Scholz knew special cases of these results, but it is not known how he proved them. And I do not see the connection with "ideal densities". $\endgroup$ Mar 21 '10 at 16:48
  • $\begingroup$ Neither do I. I'm no longer sure there is any connection with the D-H theorem ... apart from the appearance of $\zeta(3)$. $\endgroup$ Mar 21 '10 at 17:57

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