In article <··········@fsuj01.rz.uni-jena.de> ···@rz.uni-jena.de (Ralf Muschall) writes:
In article <·······················@192.0.2.1> ······@netcom.com (Henry Baker) writes:
:: I forgot to say that for algebraic numbers (roughly solutions of polynomials
:: with integer coefficients -- e.g., sqrt(2)), there is a very straightforward
:: way to calculate with them -- by representing such an algebraic number by a
:: square matrix whose characteristic polynomial is the minimal polynomial which
:: has the given algebraic number as a root.
:: The Cayley-Hamilton theorem tells you that one of these matrices satisfies
:: its own characteristic polynomial, so you can now calculate with the set
:: of square matrices generated by one of these particular matrices.
A layman's question:
Wouldn't this approach map all zeros of the polynomial to the same
[set of] matri[x|ces]?
I.e., sqrt(2) and (-sqrt(2)) would be indistinguishable.
This might not be a problem in pure algebra, where they
*are* indistinguishable, but in other sciences, the users might
want to know whether a number is +1.414 or -1.414.
Ralf
1) If we represent sqrt(2) as [[0,2],[1,0]], we have -sqrt(2) = -[[0,2],[1,0]],
so the two roots of X^2 - 2 ARE different,
BUT
2) to compare the elements of our field {[[a,2b],[b,a]] : a, b rational} by
size, we need to embed it into the real numbers, and there are TWO embeddings,
one mapping [[0,2],[1,0]] to 1.414..., the other mapping it to -1.414...
To make things unique one has to decide about the representation of sqrt(2) as
a matrix ( [[0,2],[1,0]] or -[[0,2],[1,0]]), and about the embedding.
Stefan
In article <····················@euclid.math.nat.tu-bs.de>,
Stefan Loewe <······@euclid.math.nat.tu-bs.de> wrote:
:In article <··········@fsuj01.rz.uni-jena.de> ···@rz.uni-jena.de (Ralf Muschall) writes:
: In article <·······················@192.0.2.1> ······@netcom.com (Henry Baker) writes:
: :: I forgot to say that for algebraic numbers (roughly solutions of polynomials
stuff deleted....
: A layman's question:
:
: Wouldn't this approach map all zeros of the polynomial to the same
: [set of] matri[x|ces]?
:
: I.e., sqrt(2) and (-sqrt(2)) would be indistinguishable.
Not really. For any finite extension of the rationals (say Q(a)) where a
is an algebraic number whose minimal polynomial is f, a and conj(a) are
distinguished by the Galois group, i.e. the action of the Galois group
Gal(a/Q) sends a to -a.
To distinguish a from -a in practice, (say on a computer), you simply
must specify which embedding to choose, i.e. how to embed a in C. This
can be done by simply giving the argument of a in the complex plane,
along with its magnitude.
--
Bob Silverman
The MathWorks Inc.
24 Prime Park Way
Natick, MA