Primes Produced by Quadratic Forms

[David Harden]

>btw, even with respect to non-exceptional primes, the statement >like
>"primes of the form x^2+n*y^2 are those of the form U*k+V" >seems to be
>possible only if the class number h(-4n) = degree(f_n(t)) does not
>exceed 2. For degree(f_n(t)) > 2, there seems to be no direct
>relationship between the solveblity of f_n(t) modulo p and the
>determinant of f_n(t).
>However, I don't have a proof that the case h(-4n)>2 cannot >result in
>a simple description of the primes represented by x^2+n*y^2.

I have a counterexample: Q(sqrt(-30)) has class number 4.

4 inequivalent ideals in Z[sqrt(-30)] are (1), (2,sqrt(-30)), (3,sqrt(-30)) and (5,sqrt(-30)). Therefore these represent all ideal classes.

Let p be a prime such that (-30/p)=1. Then there are exactly two ideals (complex conjugates of each other, and inverses of each other in the ideal class group; since the ideal class group has exponent two, this means they are similar to each other) of norm p in Z[sqrt(-30)].

If they are principal, then one is generated by x + y*sqrt(-30) and it has norm x^2 + 30y^2.

If they are similar to (2,sqrt(-30)), then one is generated by (x + y*sqrt(-30), -15y + x*sqrt(-30)/2). The integrality of the generators forces x to be even (write x=2z), so the ideal is expressible as (2z + y*sqrt(-30), -15y + z*sqrt(-30)). The norm of this ideal is the absolute value of the determinant 
| 2z        y |
| -15y    z  | = 2z^2 + 15y^2, which is clearly nonnegative.

If they are similar to (3,sqrt(-30)), then one is generated by (x + y*sqrt(-30), -10y + x*sqrt(-30)/3). The integrality of the generators forces x to be a multiple of 3 (write x=3z), so the ideal is expressible as (3z + y*sqrt(-30), -10y + z*sqrt(-30)). The norm of this ideal is the absolute value of the determinant
| 3z       y |
| -10y    z | = 3z^2 + 10y^2, which is clearly nonnegative.

If they are similar to (5,sqrt(-30)), then one is generated by (x + y*sqrt(-30), -6y + x*sqrt(-30)/5). The integrality of the generators forces x to be a multiple of 5 (write x= 5z), so the
ideal is expressible as (5z + y*sqrt(-30), -6y + z*sqrt(-30)).
The norm of this ideal is the absolute value of the determinant
| 5z      y |
| -6y    z  | = 5z^2 + 6y^2, which is clearly nonnegative.

Thus, every prime p such that (-30/p)=1 is produced by one of the quadratic forms x^2 + 30y^2, 2x^2 + 15y^2, 3x^2 + 10y^2 or 5x^2 + 6y^2. It is easy to see (using quadratic reciprocity and its friends) that the primes p such that (-30/p)=1 are the primes where p == 1, 11, 13, 17, 23, 29, 31, 37, 43, 47, 49, 59, 67, 79, 101, or 113 (mod 120). 

The first form can, looking modulo 120, only produce primes p == 1, 31, 49 or 79 (mod 120).
The second form can, looking modulo 120, only produce primes p == 17, 47, 113 or 23 (mod 120).
The third form can, looking modulo 120, only produce primes p == 13, 43, 37 or 67 (mod 120).
The fourth form can, looking modulo 120, only produce primes p == 11, 101, 59 or 29 (mod 120).

Since each congruence class modulo 120 of primes p such that (-30/p)=1 is represented only once above, it follows that those congruential conditions are not only necessary but sufficient for expressibility in any of those forms.

---- David