Density of norms (field theory)

carpetpool · 2018-12-28, 00:09

In this post here, I asked for the conditional probability for an integer N being prime given that all prime q dividing n are congruent to 1 modulo 2*p (for some prime p). As a result, I also got the answer of how many integers N not exceeding x can be written as a product of primes only congruent to 1 modulo 2*p. This is asymptotically D(x) = c*x*(log(x))^(1/(p-1) - 1) for some constant c, which seems to be decreasing significantly as p increases.

How many integers N not exceeding x can

(I) be written as a product of primes only congruent to 1 modulo 2*p
and

(II) in addition to (I), N can be expressed as the norm for some integral element f in the ring of integers in K=Q(zeta(p)) where K is the field of p-th roots of unity (the p-th cyclotomic field) ?

The condition for (II) can be restated as there is at least one ideal of norm N that is principal in K.

I am hoping for a precise answer (as in my last thread) in an attempt to solve another problem related to this. Again, any information is helpful, and thanks for help.

science_man_88 · 2018-12-28, 00:19

Quote:

Originally Posted by carpetpool

In this post here, I asked for the conditional probability for an integer N being prime given that all prime q dividing n are congruent to 1 modulo 2*p (for some prime p). As a result, I also got the answer of how many integers N not exceeding x can be written as a product of primes only congruent to 1 modulo 2*p. This is asymptotically D(x) = c*x*(log(x))^(1/(p-1) - 1) for some constant c, which seems to be decreasing significantly as p increases.

How many integers N not exceeding x can

(I) be written as a product of primes only congruent to 1 modulo 2*p
and

(II) in addition to (I), N can be expressed as the norm for some integral element f in the ring of integers in K=Q(zeta(p)) where K is the field of p-th roots of unity (the p-th cyclotomic field) ?

The condition for (II) can be restated as there is at least one ideal of norm N that is principal in K.

I am hoping for a precise answer (as in my last thread) in an attempt to solve another problem related to this. Again, any information is helpful, and thanks for help.

(2kp+1)(2jp+1)=4jkp^2+2(k+j)p+1 = 2(2jkp+k+j)p+1 so as many as the natural numbers up to X/(2p) of form 2jkp+k+j for some natural numbers k and j.

carpetpool · 2018-12-28, 19:05

It seems to be that the answer is 0, although I don't know for sure because many of my previous posts seem to be getting lack of attention due to the little known information and research of these topics. However, I did find this article. Still, any other explanations of it are welcome.

2018-12-28, 00:09	#1
carpetpool "Sam" Nov 2016 2²×3⁴ Posts	Density of norms (field theory) In this post here, I asked for the conditional probability for an integer N being prime given that all prime q dividing n are congruent to 1 modulo 2p (for some prime p). As a result, I also got the answer of how many integers N not exceeding x can be written as a product of primes only congruent to 1 modulo 2p. This is asymptotically D(x) = cx(log(x))^(1/(p-1) - 1) for some constant c, which seems to be decreasing significantly as p increases. How many integers N not exceeding x can (I) be written as a product of primes only congruent to 1 modulo 2*p and (II) in addition to (I), N can be expressed as the norm for some integral element f in the ring of integers in K=Q(zeta(p)) where K is the field of p-th roots of unity (the p-th cyclotomic field) ? The condition for (II) can be restated as there is at least one ideal of norm N that is principal in K. I am hoping for a precise answer (as in my last thread) in an attempt to solve another problem related to this. Again, any information is helpful, and thanks for help.

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2018-12-28, 19:05	#3
carpetpool "Sam" Nov 2016 144₁₆ Posts	It seems to be that the answer is 0, although I don't know for sure because many of my previous posts seem to be getting lack of attention due to the little known information and research of these topics. However, I did find this article. Still, any other explanations of it are welcome.