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Let $D$ be a square free integer. I am looking at primes representable as $x^2+Dy^2$, where $x,y\in\mathbb Z$. I wonder whether it is always true that this set of primes is the union of finitely many arithmetic progressions intersected with the set of all primes?

This looks like a classical questions but I could never find an answer. It is certainly true if $\mathbb Z[\sqrt D]$ is a principal ideal domain. I think I had a proof in the case of class number two but I cannot remember it. It is of course true if one looks at norms of ideals rather than norms of elements but this is not what I want.

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    $\begingroup$ This is the subject of this beautiful book Cox, David A. Primes of the form x2+ny2. Fermat, class field theory and complex multiplication. A Wiley-Interscience Publication. John Wiley & Sons, Inc $\endgroup$
    – Niels
    Commented Dec 20, 2011 at 11:48
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    $\begingroup$ Seems like a really nice but really hard to get book... Where is Dover when we need them... $\endgroup$
    – Igor Rivin
    Commented Dec 20, 2011 at 12:09
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    $\begingroup$ This is also partially addressed in Silverman's Advanced topic in elliptic curves. The answer is in general negative: it is true iff the ray class group of associated order is a 2-group. There are only finitely many such orders. $\endgroup$
    – Dror Speiser
    Commented Dec 20, 2011 at 15:32
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    $\begingroup$ Hard to get in Bonn? What's the world coming to? The answer to our question is positive if and only if the class group has exponent 2. The key word is genus theory for one direction, and norm limitation theorems from class field theory in the other. $\endgroup$
    – Franz Lemmermeyer
    Commented Dec 20, 2011 at 15:32
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    $\begingroup$ @Dror: finitely many if D is assumed to be positive. $\endgroup$
    – Franz Lemmermeyer
    Commented Dec 20, 2011 at 15:33

2 Answers 2

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EDIT: evidently the Cox book is now print-on-demand, at WILEY

There are two situations where a positive integral form represents all the primes (at least those that do not divide the discriminant) given by arithmetic progressions. On situation, appropriate for your $x^2 + n y^2,$ is when $n$ is an idoneal number, which is when there is only one class per genus. This includes the case of only one genus, but also such thngs as $x^2 + 5 y^2,$ where the other class of that discriminant is in a different genus, $2 x^2 + 2 x y + 3 y^2.$ Note that we also know exactly what primes are represented by the second form.

It is also good enough when a genus of two classes is simply one class and its opposite, as in $3 x^2 \pm 2 x y + 5 y^2.$ Either form represents all primes with certain values modulo 56. Note that this cannot happen with the principal genus.

In general, there are elaborate conditions on what primes are represented by the principal form. Typical for class number 3: (Gauss) a prime $p \equiv 1 \pmod 3$ is represented by $x^2 + 27 y^2 $ if and only if 2 is a cubic residue $\pmod p,$ which is to say that $z^3 -2$ factors into three distinct linear factors $\pmod p.$ A prime $p > 3$ with $(-44 | p) =1$ is represented by $x^2 + 11 y^2$ if and only if $z^3 + z^2 -z+1 $ factors into three distinct linear factors $\pmod p.$ For all the class number three discriminants, see K. S. Williams and R. H. Hudson, Acta Arithmetica, vol. 57 (1991) pages 131-153, Representation of primes by the principal form of the discriminant $-D$ when the classnumber $h(-D)$ is 3

For classnumber 4: (Gauss) a prime is represented by $x^2 + 64 y^2$ if and only if 2 has four distinct fourth roots $\pmod p,$ which is to say that $z^4 -2$ factors into four distinct linear factors $\pmod p.$ Note that $p \equiv 1 \pmod 8$ anyway. A prime $p > 3$ with $(-56 | p) =1$ is represented by $x^2 + 14 y^2$ if and only if $z^4 + 2z^2 -7 $ factors into four distinct linear factors $\pmod p.$ For all the class number four discriminants,see K. S. Williams and D.Liu, Tamkang Journal of Mathematics, vol. 25 (1994) pages 321-334, Representation of primes by the principal form of negative discriminant $\Delta$ when $h(\Delta)$ is 4

For a table with all discriminants from $-3$ to $-451,$ all class numbers mixed together, see pages 539-542 in Advanced Topics in Computational Number Theory by Henri Cohen, 1999. The table is called "Discriminants and Hilbert class fields of imaginary quadratic fields." The polynomial given tells you, once a prime is represented by the principal genus of a discriminant, what polynomial must factor completely for the prime to be represented by the principal form itself. This table gives the polynomials $f_n(x)$ promised in Theorem 9.2 on page 180 of Primes of the form $x^2 + n y^2$ by David A. Cox, AMAZON

If you cannot find the two articles or the wonderful Cohen table, email me. My websites are currently down, some of it is usually posted there. The thing about Cohen's table is that it is difficult to find a polynomial that works, it is incredibly difficult to find one with small coefficients, compare some articles by Kaltofen and Yui.

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This is discussed in http://websites.math.leidenuniv.nl/algebra/Stevenhagen-Primes.pdf

and

http://www.dms.umontreal.ca/~andrew/Courses/Chapter4.pdf

The latter seems to be less machinery laden (both are interesting, however), and proposition 4.1 gives the condition (which, by reciprocity, appears to imply the affirmative answer to your question).

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    $\begingroup$ Proposition 4.1 tells when a number is represented by \emph{some} quadratic form of a given determinant. As far as I understand, this is equivalent to being a norm of an \emph{ideal} rather than a norm of an element. Thus this is not I want I'll look more carefully at the papers though. $\endgroup$
    – Roman Fedorov
    Commented Dec 20, 2011 at 11:38
  • $\begingroup$ @Roman: You are right (this is the right answer when the class number is 1, but not in general), so the Stevenhagen thing is most likely more relevant. $\endgroup$
    – Igor Rivin
    Commented Dec 20, 2011 at 12:00

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