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 2017-09-17, 04:10 #1 carpetpool     "Sam" Nov 2016 26×5 Posts Prime numbers norms of modulo cyclotomic polynomials For any given prime n > 19, and a random prime p = 2*k*n+1, what is the probability there will exist n-1 elements (polynomials) u such that norm(Mod(u,polycylo(n)) = p? (Here polycyclo(n) is the nth cyclotomic polynomial) For example, in polycyclo(n), n = 23, p = 599, we see that there exists elements u with norm(Mod(u,polcyclo(23)) = 599, but for prime p = 967 (which is of the form 2*k*23+1), there does not exist any elements u with norm(Mod(u,polcyclo(23)) = 967. Of the primes p = 1 (mod 2*23), for which there does exist elements with norm(Mod(u,polcyclo(23)) = p, they are seemingly random, as there is no special / additional arithmetical or modular condition on them besides having p = 1 (mod 2*23). The same is true with polcyclo(29), except there being fewer primes p = 1 (mod 2*29) for which there are elements u such that norm(Mod(u,polcyclo(29)) = p. As p = 4931 this is true for but not p = 5279 (both primes of the form 2*29*k+1). There is a similar concept for all polcyclo(n) with class number > 1, (although this question is for prime n) where some primes p = 1 (mod n), there will exist elements u and norm(Mod(u,polcyclo(n)) = p, while others will not. Are the primes 'randomly' split up somehow? Is there, and what is the general probability for a prime p = 1 (mod n) that there will be norm solutions in the field polycyclo(n), given polycyclo(n) has class number c? For instance, what is the probablity (for each of the corresponding cyclotomic fields below), that there will be an element u with norm(Mod(u,polcyclo(n)) = p, and prime p = k*n+1 (2*k*n+1 since all of these n are odd primes). p = 45520819, and norm(Mod(u,polcyclo(23)) = p p = 2913344873, and norm(Mod(u,polcyclo(29)) = p p = 11661616457, and norm(Mod(u,polcyclo(31)) = p p = 746726311549, and norm(Mod(u,polcyclo(37)) = p ... I would like to apply this for all polcyclo(n), if possible. Thanks for help.