20171004, 15:03  #12 
Feb 2017
Nowhere
3887_{10} Posts 
The question of determining the Hilbert Class Field was quite instructive to me in the use of PariGP. As a result, I can give defining polynomials for three quadratic extensions of the field k of 29th roots of unity, whose join is the Hilbert Class Field k^{(1)} of k. The extension k^{(1)}/k is elementary Abelian of degree 8; that is, the Galois group is the direct product of three cyclic groups of order 2.
x^2 + Mod(3*t^27  t^26  2*t^24  2*t^23  t^21  3*t^20  t^18  t^17  t^16  t^15  2*t^14  2*t^11  t^10  t^8  2*t^7  2*t^4  t^3  t^2  t  1, t^28 + t^27 + t^26 + t^25 + t^24 + t^23 + t^22 + t^21 + t^20 + t^19 + t^18 + t^17 + t^16 + t^15 + t^14 + t^13 + t^12 + t^11 + t^10 + t^9 + t^8 + t^7 + t^6 + t^5 + t^4 + t^3 + t^2 + t + 1) x^2 + Mod(2*t^27  2*t^25 + t^24  t^23 + t^22  2*t^21  2*t^19  t^17  t^16  t^15  2*t^14  t^13  t^12 + t^11  t^10  t^9  t^8  t^7 + t^6  t^5  t^4  2*t^3  t^2  t  1, t^28 + t^27 + t^26 + t^25 + t^24 + t^23 + t^22 + t^21 + t^20 + t^19 + t^18 + t^17 + t^16 + t^15 + t^14 + t^13 + t^12 + t^11 + t^10 + t^9 + t^8 + t^7 + t^6 + t^5 + t^4 + t^3 + t^2 + t + 1) x^2 + Mod(t^27 + t^26 + 2*t^25 + t^24 + t^23 + t^21 + t^20 + t^19 + t^17 + t^16 + 2*t^15 + t^14 + t^13 + t^11 + t^10 + 3*t^9 + 2*t^8 + 2*t^7 + 2*t^4 + 2*t^3 + 3*t^2 + t + 1, t^28 + t^27 + t^26 + t^25 + t^24 + t^23 + t^22 + t^21 + t^20 + t^19 + t^18 + t^17 + t^16 + t^15 + t^14 + t^13 + t^12 + t^11 + t^10 + t^9 + t^8 + t^7 + t^6 + t^5 + t^4 + t^3 + t^2 + t + 1) It was not difficult to find a degree8 polynomial for k^{(1)}/k. From it, I did produce a degree224 polynomial in Z[x] defining k^{(1)} as an extension of Q, but due to the high degree and large coefficients, I was unable to do any computations with it. 
20171005, 03:41  #13 
"Sam"
Nov 2016
140_{16} Posts 
Okay, Thank you for the effort in finding these? Also I would like to know what the degree 8 polynomial in k the field of the 29th roots of unity, that defines the Hilbert Class Field L of k? Also, what is this polynomials discriminant?

20171005, 04:17  #14 
"Sam"
Nov 2016
101000000_{2} Posts 
A second announcement I would like to present, is the "creation" of new ideal classes in a certain field.
So suppose we take the field k of the 23rd roots of unity, which has class number 3. The primes p in k which are norms of principal ideals are the list in my previous three posts. In my previous two posts, I commented on the fact that in k, a prime p is a norm of a principal ideal if and only if there are solutions to x^3x1 = 0 (mod p). However the statement that in k, p is principal if and only if there are solutions to x^3x1 = 0 (mod p) is false because p = 6533 = 47*139 is a norm of a principal ideal in the field k of the 23rd roots of unity, but NOT in k = Q(sqrt(23)). Let primes p, 'p, and q = 1 (mod 23). If P = p*'p, P is principal if and only if p and 'p are both not norms of principal ideals, or both are norms of principal ideals. P*q is always principal. (20:59) gp > normU(23,x^3x1,1,100000) Up to 300000 there are 429 primes congruent to 1 mod k and 141 are norms of principal ideals (20:59) gp > print(v) (20:59) gp> We also have elements of the form a*x^21+b*x^20+c*x^19....t*x^2+u*x+v for instance any element has a norm in k (21:07) gp > norm(Mod(x^3+x^22*x1,polcyclo(23))) %1613 = 121356143 I.e. all the elements in k are mapped to a norm of a principal ideal in k, the field of the 23rd roots of unity. For each prime p a norm of principal ideal, there are exactly 22 elements which correspond to norm of p. Now, to "create" a different ideal class, we will define a prime p = 1 (mod 23) as principal if and only if there are solutions to x^33*x1 = 0 (mod p), and a prime with no solutions as nonprincipal. Again, the condition on P, is still true. (20:59) gp > normU(23,x^33*x1,1,100000) Up to 300000 there are 429 primes congruent to 1 mod k and 136 are norms of principal ideals (20:59) gp > print(v) [] Although x^33*x1 does not define the Hilbert Class Field L of k, in the construction of this new "nonreal" cyclotomic field c, it does. In c, 47*139 = 6533 is still a norm of a principal ideal, and there should exist the same number of elements as in k. However, 599*829 = 496571 is not a norm of a principal in c and there are no elements with norm 496571. The thing I am unsure about is using the new definition for a norm of a principal ideal in c, how do we compute the norm of an element in c? For instance what elements w in c correspond to norm(Mod(w,c)) = 1657 Since 1657 is prime, there should be exactly 22 elements in c whose norm is 1657. (Note 1657 is NOT principal in k, and there are no elements in the real field of the 23rd roots of unity.) Last thing, which number field does c define? It is indeed a cyclotomic field of order 23, however the ideal classes are different compared to those in k. Thanks for reading this, anyone into number fields, norms, elements, creation of different number fields, please help. Thanks. Last fiddled with by carpetpool on 20171211 at 06:00 
20171005, 04:53  #15  
"Sam"
Nov 2016
2^{6}×5 Posts 
Quote:
In k, the polynomial is x^3x1 which defines the Hilbert Class Field L of k, In c, the polynomial is x^33*x1 which defines the Hilbert Class Field L of k, For example, in k the element x2 has norm 2^231 = 8388607 = 47*178481. In c what is the norm of x2? Not all numbers which are norms of principal ideals in k are the same in c, the ideal classes are different in both fields, but produce the same number of solutions. Last fiddled with by carpetpool on 20171211 at 06:01 

20171005, 15:48  #16  
Feb 2017
Nowhere
3887_{10} Posts 
Quote:
(BTW, you're using the same symbol, k, for two different fields here.) As to the solutions of the cubic mod p, I note that this only pertains to primes, and 47*139 isn't prime. 

20171006, 01:55  #17  
"Sam"
Nov 2016
2^{6}·5 Posts 
Quote:
Argument made simple, in the 23rd roots of unity, we see that 2*23+1 = 47 is the norm of an ideal, which is NOT principal, and no principal ideals have a norm of 47. There should exist infinitely many field extensions K/Q of the 23rd root of unity where 47 is the norm of a prinicpal ideal. How to compute norms of elements in a field extension, I am unsure of, and also finding all elements with norm p in a rational field extension of the nth roots of unity. 

20171006, 14:35  #18  
Feb 2017
Nowhere
13^{2}×23 Posts 
Quote:
Alas, in many cases, the Hilbert class field tower does not terminate. In particular (see INFINITE HILBERT CLASS FIELD TOWERS FROM GALOIS REPRESENTATIONS, which cites René Schoof, Infinite class field towers of quadratic fields, J. Reine Angew. Math. 372 (1986), 209–220), this is the case for Q(\zeta_{877}). In fact, it is true for "almost all" cyclotomic fields, as shown in INFINITE HILBERT CLASS FIELD TOWERS OVER CYCLOTOMIC FIELDS 

20171007, 15:03  #19 
Dec 2012
The Netherlands
10111110001_{2} Posts 
René's homepage (with many of his papers) can be found here: http://www.mat.uniroma2.it/~schoof/

20171008, 12:13  #20  
"Sam"
Nov 2016
2^{6}×5 Posts 
Quote:
Okay, so is Q(\zeta_{n}), n = 877 the first case of a cyclotomic field not having its Hilbert class field terminate (i.e. counter example)? Second, The field K of nth roots of unity for prime n > 19, has a class number h > 1. There should be a field extension K/Q of the nth roots of unity which there is an element (in that field extension) with norm of p for any prime p = 1 (mod n). Does this stand true? Last, can the processes described here of computing the norm of an element in a cyclotomic field be applied to computing the norm of an element in a field extension? 

20171008, 14:13  #21  
Feb 2017
Nowhere
13^{2}×23 Posts 
Quote:
Quote:
If you mean L/K with an element from L to K of norm p, of course there is. If L is the extension of K obtained by adjoining an nth root r of p (where K/Q is the field of nth roots of unity, and n is greater than 2), then the norm from L to K of r is p. Quote:
Last fiddled with by Dr Sardonicus on 20171008 at 14:15 

20171008, 19:45  #22  
Feb 2017
Nowhere
13^{2}·23 Posts 
Quote:
I shouldn't have excluded the value n = 2. If n is even, you need an nth root of p rather than of p. This also assumes x^n + p or x^n  p is irreducible over the field of nth roots of unity. Alas, that isn't always true (e.g. p = 2, n = 8). If n is even, any extension defined by polynomial with constant term p which is irreducible in K[x] will fill the bill. if n is odd, an irreducible polynomial with constant term p instead of p will do the job. 

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