Talk:Complex multiplication

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I am expecting multiplication in the complex field at this(Complex multiplication) title. --MarSch 12:52, 3 Jun 2005 (UTC)

How's that? Dmharvey Talk 16:57, 3 Jun 2005 (UTC)

better than nothing --MarSch 18:31, 3 Jun 2005 (UTC)

Apparently there are many n for which is very near an integer, for example n = 58 and 652. How come? - Fredrik | talk 21:39, 16 September 2005 (UTC)

I'm attempting to figure that out right now... I have a very good understanding of undergraduate mathematics, and a beginning understanding of set theory, group theory, analysis, et cetera. I'm finding this article somewhat difficult to follow, but I really, really want to know the answer to the question you asked, so I'm going to try. --Monguin61 04:12, 15 December 2005 (UTC)

[edit] Trying to follow explanation

I guess this is evident to anyone who knows what theyre doing, so maybe the article doesn't need the notation explained, but I was wondering if anyone could tell me what

actually means? I assume the Z denotes a group of some sort, and the stuff inside the brackets is a parameter defining the group, but what type of group exactly? Or am I totally off? --Monguin61 04:24, 15 December 2005 (UTC)

Z[α] is the subring of the complex numbers generated by the complex number α if you want it in simple terms. Charles Matthews 08:53, 15 December 2005 (UTC)

Z is the set of all integers. The thing in square brackets is something not belonging to Z, being "adjoined", so that we get the smallest set contaning all integers and containing the thing in square brackets and closed under addition, subtraction, and multiplication. See Herstein's advance undergraduate text Topics in Algebra. Michael Hardy 00:15, 16 December 2005 (UTC)

I will add a link explaining this notation. But I cannot either follow the "explanation", I mean, some more details (even if unprecise and handwaving), or at least an "external link" seem necessary, because readers come from "far away" to this page (referred to in pi and other places), hungry for this explanation (why dist( exp(...), Z) is so small )). — MFH:Talk 14:14, 16 March 2006 (UTC)

[edit] Trying to follow ...

I followed this article up to this point:

This is a typical elliptic curve with complex multiplication, in the sense that over the complex number field they are all found as such quotients, in which some order in the ring of integers in an imaginary quadratic field takes the place of the Gaussian integers.

Is this trying to say that every elliptic curve can be obtained as C/cZ[F] where c is some complex number, and F is an imaginary quadratic field? Is this furthermore implying that such a Z[F] is a unique factorization domain? Last but not least, is there an implication that for any f in F, that exp (f\pi) == close approximation to integer?

Anyway, I barely understand what I just asked; this is all new ot me. I'll try to hunt down book references for this topic, but if anyone has one handy, let me know? linas 01:56, 19 January 2006 (UTC)

Serge Lang, Elliptic Functions might suit you. Every elliptic curve is C/Λ for a lattice Λ, but the cases where Λ has anything to do with a quadratic field are very special. The associated τ must be one of a countable set of algebraic numbers, for example. Charles Matthews 08:50, 19 January 2006 (UTC)

OK, I understand the bit about elliptic curves; I found something on the web that skteches the proof, viz. along the lines that the j-invariant is an algebraic integer. If the ring of integers is a UFD, then j is a "rational integer". By then truncating the fourier expansion for j, (the q-series expansion for j) (justified because because q is so small), one obtains the Ramanujan number \exp \pi \sqrt{163}.

Now for the trick question: are there similar results for an "irrational" quadratic "field" (well, ok, ring, then) of the form

I have a near-identity that clearly has a square-root of pi sitting next to an integer, and am trying to explain this ident. I'm looking for anything related to such a field... linas 18:38, 20 January 2006 (UTC)

Simple answer: no. The ring is just a polynomial ring, and it doesn't embed in the complex plane as a lattice. Charles Matthews 19:31, 20 January 2006 (UTC)

Since sqrt(pi) is not algebraic, I would even dare to say that this is isomorphic to Q[X]. — MFH:Talk 02:01, 1 April 2006 (UTC)