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Titus Piezas III science forum Guru Wannabe
Joined: 10 Mar 2005
Posts: 102

Posted: Wed Jul 19, 2006 12:02 pm Post subject:
Re: A conjecture on class numbers



Wonderful! So glad to know. Here's two more, on the jfunction (Note:
Discriminant d is understood to be d < 0):
1. Given a fundamental d (not a multiple of 3) with class number h(d),
then jfunction j(tau) = x^3, where x is an algebraic integer of degree
h(d). In other words, it is a perfect "cube".
2. If d = 3m with m not a square and h(d) a power of 2, then j(tau) =
v^2x^3 where v is a fundamental unit of m and x is an algebraic integer
of degree h(d).
Note: We will define the fundamental unit v here as p+q*Sqrt[m] such
that p^2mq^2 = +/1, hence uses either the + or  case of Sqrt[m].
Example with tau = (1+Sqrt[d])/2, d = 3*17, h(d) = 2:
J(tau) = (4+Sqrt[17])^2(48^3)(5+Sqrt[17])^3
Are both true?
Titus 

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victor_meldrew_666@yahoo. science forum beginner
Joined: 19 May 2006
Posts: 17

Posted: Tue Jul 18, 2006 7:18 pm Post subject:
Re: A conjecture on class numbers



titus_piezas@yahoo.com wrote:
Quote:  Let m,n be any element of {0,1,2,....}
1. Let F(n) be a squarefree number of form 8n+3. If F(n) is prime,
then 8F(n) has class number h(d) of form 4m+2. If composite, then of
form 4m.
Ex. F(n) = 11 with h(8*11) = 2;
F(n) = 35 with h(8*35) = 4.
2. Let F(n) be a squarefree number of form 8n+5. If F(n) is prime,
then both 4F(n) and 8F(n) have class numbers of form 4m+2. If
composite, then of form 4m.
Ex. F(n) = 13 with h(4*13) = 2, h(8*13) = 6;
F(n) = 85 with h(4*85) = 4, h(8*85) = 12.
True or not?

These are true, and quite wellknown.
Let m be a positive, odd, squarefree integer.
Genus theory tells us that the 2torsion subgroup
classgroup of the classgroup of Q(sqrt(2m)) has order
2^d where d is the number of prime factors of m.
In particular, when m is composite, the classgroup has
a subgroup of order 4 and so the classnumber is a
multiple of 4.
Suppose then that m = p is prime. Then the classgroup
has only one class of order 2. This is the class of the
unique ideal P_2 of norm 2. The classnumber is divisible
by 4 iff there is an element of order 4 in the classgroup.
This would be represented by an ideal I with I^2 equivalent
to P_2. Thus I^2 P_2 would be a principal ideal
<a + b sqrt{2p}>. Taking norms gives
2 N(I)^2 = a^2 + 2p b^2.
Considering this equation modulo p shows that 2 must be
a quadratic residue modulo p. Thus if (2/p) = 1, that is
if p = 3 or 5 (mod , then the classgroup has no element
of order 4 and so the class number is even but not
a multiple of 4.
Victor Meldrew 

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Titus Piezas III science forum Guru Wannabe
Joined: 10 Mar 2005
Posts: 102

Posted: Fri Jul 07, 2006 5:29 am Post subject:
A conjecture on class numbers



Hello all,
Just two curious conjectures here:
Let m,n be any element of {0,1,2,....}
1. Let F(n) be a squarefree number of form 8n+3. If F(n) is prime,
then 8F(n) has class number h(d) of form 4m+2. If composite, then of
form 4m.
Ex. F(n) = 11 with h(8*11) = 2;
F(n) = 35 with h(8*35) = 4.
2. Let F(n) be a squarefree number of form 8n+5. If F(n) is prime,
then both 4F(n) and 8F(n) have class numbers of form 4m+2. If
composite, then of form 4m.
Ex. F(n) = 13 with h(4*13) = 2, h(8*13) = 6;
F(n) = 85 with h(4*85) = 4, h(8*85) = 12.
True or not?
P.S. Numbers 8n+1 and 8n+7 are not that wellbehaved.
Titus 

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