 mersenneforum.org Norm equation of field
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 Register FAQ Search Today's Posts Mark Forums Read 2018-12-24, 08:37 #1 carpetpool   "Sam" Nov 2016 5·67 Posts Norm equation of field In any algebraic number field K, let e be an arbitrary element of K and N(e) is the norm of the element e in the field K. (I): It is well known that there exists an integer m such that for all primes p (which completely split in field K), there is an element e in K such that N(e) = m*p is solvable. Thus, there exists a "smallest" m such that this is true. In the rest of these results, m is the smallest such number. (II): For example, in the field K=Q(sqrt(-23)), K has class number 3 and its elements e are of the form a*r+b where r^2+r+6=0. N(a*r+b) = a^2 + a*b + 6*b^2 = p is not solvable for all primes p (which are not inert in K). However, N(a*r+b) = a^2 + a*b + 6*b^2 = 6*p is (non trivially) solvable for all primes p not inert in K, or if p is a quadratic residue mod 23, thus 6 is the "smallest" m as suggested in (I). (III): I decided to repeat example (II) using the base field K=Q(sqrt(-p)); for all odd primes p = 3 mod 4 and p > 23 and the elements of the form e = (a*r+b) where r^2+r+((p+1)/4)=0. Here are the results: p = 31, m = 10 p = 43, m = 1 p = 47, m = 24 p = 59, m = 15 p = 67, m = 1 p = 71, m = 24 p = 79, m = 40 p = 83, m = 21 p = 103, m = 56 p = 127, m = 88 It is obvious that when m = 1, N(e) = 1*p = p is solvable for all p, indicating that K has class number 1. (IV): I also attempted to compute "m" for some cyclotomic fields (the field of n-th roots of unity), K= Q(e^((2*pi*i)/n)) for prime n. K= Q(e^((2*pi*i)/n)) for n = 2, 3, 5, 7, 11, 13, 17, 19 m = 1 because K has class number one and N(e)=m*p is solvable for all primes p = 1 mod n. K= Q(e^((2*pi*i)/23)) m = 47^2, as N(e)=m*p is solvable for all primes p = 1 mod 23. K= Q(e^((2*pi*i)/29)) m = 59^2, as N(e)=m*p is solvable for all primes p = 1 mod 29. K= Q(e^((2*pi*i)/31)) m = 2^10, as N(e)=m*p is solvable for all primes p = 1 mod 31. K= Q(e^((2*pi*i)/37)) m = 149*223, as N(e)=m*p is solvable for all primes p = 1 mod 37. K= Q(e^((2*pi*i)/41)) m = 83^2*739, as N(e)=m*p is solvable for all primes p = 1 mod 41. K= Q(e^((2*pi*i)/43)) m = 173^2*431, as N(e)=m*p is solvable for all primes p = 1 mod 43. K= Q(e^((2*pi*i)/47)) m = 283^2*659, as N(e)=m*p is solvable for all primes p = 1 mod 47. And that's as far as I've gotten. Is there a general method or program to find a smallest m, given the information and criteria in (I)? Also, any verification of results I found would be useful too. Thanks for help. Last fiddled with by carpetpool on 2018-12-24 at 08:39  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post canny Abstract Algebra & Algebraic Number Theory 1 2018-05-22 07:17 chris2be8 Msieve 1 2015-09-13 02:24 paul0 Factoring 2 2015-01-19 11:52 paul0 Programming 6 2015-01-16 15:12 bonju Factoring 9 2005-08-26 13:29

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