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A conjecture on class numbers
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Titus Piezas III
science forum Guru Wannabe


Joined: 10 Mar 2005
Posts: 102

PostPosted: Fri Jul 07, 2006 5:29 am    Post subject: A conjecture on class numbers Reply with quote

Hello all,

Just two curious conjectures here:

Let m,n be any element of {0,1,2,....}

1. Let F(n) be a square-free 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 square-free 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 well-behaved.

--Titus
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victor_meldrew_666@yahoo.
science forum beginner


Joined: 19 May 2006
Posts: 17

PostPosted: Tue Jul 18, 2006 7:18 pm    Post subject: Re: A conjecture on class numbers Reply with quote

titus_piezas@yahoo.com wrote:

Quote:
Let m,n be any element of {0,1,2,....}

1. Let F(n) be a square-free 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 square-free 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 well-known.

Let m be a positive, odd, squarefree integer.
Genus theory tells us that the 2-torsion 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 Cool, 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

PostPosted: Wed Jul 19, 2006 12:02 pm    Post subject: Re: A conjecture on class numbers Reply with quote

Wonderful! So glad to know. Here's two more, on the j-function (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 j-function 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^2-mq^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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