Yes, of course p^j, p^k, p^(jk). Ok, so in the case where p=2,
jk=32, j could be 2,4,8,16 with k 16,8,4,2 respectively ?
The matrix representation should be a *block matrix* representation,
with an "outer" representation of 2x2, 4x4, 8x8, 16x16, and an
"inner" representation of 16x16, 8x8, 4x4, 2x2, respectively.
I assume that it is possible to do this incrementally, with each
step being a factor of 2 -- i.e., a 2x2 array of smaller blocks?
So we automatically get a Strassen-like recursive matrix multiplication
algorithm for these block matrices ?
-----Original Message-----
>From: rcs(a)xmission.com
>Sent: Jul 24, 2017 10:53 PM
>To: math-fun(a)mailman.xmission.com
>Cc: rcs(a)xmission.com
>Subject: Re: [math-fun] Matrices to represent extension field elements
>
>Addressing only the third-to-last paragraph, about extensions & towers:
>If j and k are relatively prime positive numbers, you can develop the
>fields GF[p^j], GF[p^k], and GF[p^(jk)]. The last contains the first two
>as subfields. If you select irreducible polynomials (mod p) of degrees
>j and k, you can adjoin roots of both polynomials to GF[p] and get the
>product degree field GF[p^(jk)]. It works, I've written the code, it runs.
>
>If j divides k, then GF[p^j] is embeddable as a subfield of GF[p^k];
>if you've selected a representation of GF[p^k], there's a unique subset
>that makes up GF[p^j]. The embedding is unique as to the set of elements,
>but not wrt the actual mapping, because of automorphisms of GF[p^j].
>
>If you choose an N, and consider its divisibility box, with 1 at the
>lower left front corner, and N at the opposite corner, each node J of the box
>has a field GF[p^J], and the subfields have the same relationship as
>the box nodes.
>
>Rich
>
>--------
>Quoting hbaker1 <hbaker1(a)pipeline.com>:
>> A recent post mentioned that (square) matrices were a kind of
>> "universal" representation for fields, in that a matrix A whose
>> characteristic polynomial is p[x] can represent a root of this
>> polynomial in the extension field.
>>
>> The problem is that such a matrix doesn't represent a *single* root of
>> p[x], but simultaneously represents *all* of the roots of p[x].
>> (Hint: diagonalize the matrix so that the main diagonal consists of
>> the eigenvalues, i.e., all the roots of p[x].)
>>
>> Suppose we have the simplest situation: a symmetric matrix with all
>> distinct real roots. Since the matrix itself represents *all* of
>> these roots simultaneously, it can't participate, e.g., in ordering
>> comparisons with other numbers which may lie in the middle of this set
>> of roots.
>>
>> So there needs to be some way to distinguish exactly which root we're
>> talking about when given the matrix A, before we can realistically
>> utilize this matrix as a representation of one of the real roots of
>> the polynomial.
>>
>> ---
>>
>> In the usual case of real matrices with complex eigenvalues, the
>> coefficients of the polynomial, the entries of the matrix, and the
>> roots themselves are usually more-or-less continuous functions of one
>> another. So if we vary a polynomial coefficient by a little, one or
>> more of the roots will vary in a more-or-less predictable way.
>>
>> If we're going to extend finite fields in a p-adic way, then
>> presumably there will be an analogous "continuity" between the
>> polynomial coefficients and the roots.
>>
>> It is my understanding that all of the fields GF(p^k) with a given
>> prime p and a given positive power k are isomorphic to one another.
>> So does this mean that instead of extending by p^k, one could instead
>> extend first by p^i, and subsequently by p^j, such that i+j=k ?
>>
>> Perhaps one could build (always?) some sort of "tower" of extensions
>> whereby each field is embedded in the next ?
>>
>> Under these conditions, could the polynomial coefficients be "close"
>> to those of the extensions, and the polynomial roots be "close" to
>> those of the extension, where "close" is measured in some p-adic way?
