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Old 2017-10-03, 06:16   #10
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Nov 2016

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Originally Posted by Dr Sardonicus View Post
In the field of n-th roots of unity K = Q(\zeta_{n}), the prime numbers p congruent to 1 (mod n) are precisely those which are norms of ideals in the ring of integers OK. Invoking the "assumption of ignorance" that no ideal class is favored over any other, the "obvious" answer is that 1/h of these primes are norms of principal ideals (that is, are norms of algebraic integers in K), where h is the class number (standard notation).

And in fact this is correct. It may be viewed as a generalization of the equal distribution of primes in arithmetic progressions. See THE ASYMPTOTIC DISTRIBUTION OF PRIME IDEALS IN IDEAL CLASSES

For the field k = Q(\zeta23) with h = 3 (constructed in advance with bnfinit()), I ran a script to get the actual count for primes up to 100,000. As you can see, it's pretty close to 1/3 of all primes congruent to 1 mod 46.

I would like to announce one result involving the field K = Q(\zeta_{23}).


For a prime p = 1 (mod 23), p is a norm of a principal ideal if and only if there are 3 solutions to x^3-x-1 = 0 (mod p).

To see this is the same list as the one quoted in Dr. Sardonicus's first post, I modified a PARI/GP script which includes a bnf construction of any number field, and k.

normU(k,w,n,m) =
bnf = bnfinit(w);
print("Up to 100000 there are ",j," primes congruent to 1 mod k and ",l," are norms of principal ideals")

here w is any polynomial, and the script finds primes n < p < m, and p = 1 (mod k) such that p is the norm of a principal ideal in the field that w defines.

The list above is the same as the one Dr. Sardonicus's first post, and also the primes p = 1 (mod 23) which satisfy x^3-x-1 = 0 (mod p).

In the field K = Q(\zeta_{29}), what is the polynomial w for which if p = 1 (mod 29), and p is the norm of a principal ideal, then p satisfies w = 0 (mod p), if it is not principal, then p does not satisfy w = 0 (mod p)?

Update: For K23, this works for any polynomial w having a discriminant D of -23. In fact replacing x^3-x-1 with x^2+x+6, x^2+3*x+8, x^3+3*x^2+2*x-1,... would all give the same result.

Thanks for help.

Last fiddled with by carpetpool on 2017-12-11 at 06:00
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