Re: - Representing units by a quadratic form

On Apr 13, 6:10 am, ballade1 <balla...@xxxxxxxxx>
wrote:
Let F be a field, and q a quadratic form (in n
variables) over F. Denote by F* the group of nonzero
elements of F.

We say that an element d of F* is represented by q
if
there are x_1, .., x_n in F such that q(x_1, ..,
x_n) = d.

Define

D(q) = {d in F* : q represents d}

Suppose now q has matrix

[1 0]
[0 a]

for nonzero a, i.e. q(x, y)= x^2 + ay^2.

In this case, why does D(q) form a subgroup of F*
under multiplication?

Clearly 1 = q(1, 0); and if d, e are represented by
q, then

q(x, y) = d and q(u, v) = e,

so that de = q(x, y) q(u, v);

but how do we write this in the form q(r, s)?

I find

(x^2 + ay^2)(u^2 + av^2)

= (xu)^2 + a[(xv)^2 + (yu)^2] + a^2 (yv)^2

but this doesn't quite yet have the right ' form' .

Also, if d = q(x, y) is a unit in F, then how can
we express

1/q(x, y) as q(t, w) ?

Thank you in advance.

Formally introduce a square root k = sqrt(-a) in your
field
and notice:

( x + k y )(u + k v ) = (x u - a y v ) + k (x v +
v + y u )

It suggests we look at

q(x u - a y v, x v + y u )
= ( x u - a y v )^2 + a ( x v + y u )^2
= x^2 ( u^2 + a v^2 ) + a y^2 ( u^2 + a v^2 )
= q(x, y) q(u, v)

Since proof of this formula does not involve k
directly.
It works in your original field. You can apply
similar
trick to the inverse to get the obvious

1/Q(u,v) = Q( u/Q(u,v), -v/Q(u,v) ) = Q(u/Q(u,v),
v), v/Q(u,v)).

when Q(u,v) is a unit.




How does one formally introduce a square root in a field?
Rather, why does k (above) exist in F? This is what I had thought about doing at first, but was convinced it wasn't quite legal.
.

Relevant Pages

  • Re: - Representing units by a quadratic form
    ... It works in your original field. ... trick to the inverse to get the obvious ... How does one formally introduce a square root in a ...
    (sci.math)
  • Re: - Representing units by a quadratic form
    ... why does Dform a subgroup of F* under multiplication? ... Formally introduce a square root k = sqrtin your field ... It works in your original field. ... trick to the inverse to get the obvious ...
    (sci.math)
  • Re: Order question
    ... We will show that every element in the subgroup of order q is a QR, ... by constructing a square root for each element. ... A low hamming weight q would be advantageous for allowing quick tests ... q values are in fact safe then they would be useful for DL protocols. ...
    (sci.crypt)
  • Re: XROOT and complex arguments
    ... You're right, TSIMP does the trick! ... (I still wonder why the square root of i is automatically simplified ... while the cube root is not...) ...
    (comp.sys.hp48)