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Old 2018-12-24, 08:37   #1
carpetpool
 
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"Sam"
Nov 2016

32810 Posts
Post 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
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