Re: [math-fun] Where do roots live in GF(2^k) ?
Thanks, Dan. The algebraic closure of the infinite sequence GF(p),GF(p^2),GF(p^3),... as described in the Wikipedia article is pretty ugly -- at least as described. I was hoping for something as elegant as the complex numbers with their usual topology. Perhaps I need to check back in about 300 years? At 10:08 AM 7/21/2017, Dan Asimov wrote:
Every field K Â if it does not already contain all the roots of polynomials in K[x] Â is a proper subfield of a larger field that is algebraically closed.
E.g., see <https://en.wikipedia.org/wiki/Finite_field#Algebraic_closure>.
I don't know how such an algebraic closure is usually visualized.
But I believe that, for any prime p, the Galois group of [the algebraic closure of the finite field F(p)] over F(p) is isomorphic to the same thing for any other prime.
ÂDan
From: Henry Baker <hbaker1@pipelline.com> Jul 21, 2017 9:06 AM: -----
OK, if I extend the rationals with the root alpha of an irreducible polynomial p[x], I can plot alpha on the complex plane; indeed, I can plot *all* of the roots of p[x] on the complex plane. So all of these "extension roots" live in the complex plane.
Is there an analogous (single) place/field where all extension roots of GF(p) live -- i.e., a larger field which includes all of the extension fields of GF(p) ?
There seems to be a problem, since there are many (isomorphic) ways to extend GF(p); perhaps these are all different in this larger field?
* Henry Baker <hbaker1@pipeline.com> [Jul 22. 2017 14:26]:
Thanks, Dan.
[...]
I was hoping for something as elegant as the complex numbers with their usual topology.
Now, long after my (naive, failed) attempts to somehow nicely visualize things like polynomial roots over GF(p^k), I strongly suspect the problem really is in that "topology" thing. I'd be very grateful if someone could elaborate. Best regards, jj
Perhaps I need to check back in about 300 years?
[...]
The algebraic closure of the rationals consists of the algebraic numbers, and is a countable field. The complex numbers are an uncountable field. In between the two lie the algebraic closures of the various transcendental extensions of the rationals. -- Gene On Friday, July 21, 2017, 5:46:55 PM PDT, Henry Baker <hbaker1@pipeline.com> wrote: Thanks, Dan. The algebraic closure of the infinite sequence GF(p),GF(p^2),GF(p^3),... as described in the Wikipedia article is pretty ugly -- at least as described. I was hoping for something as elegant as the complex numbers with their usual topology. Perhaps I need to check back in about 300 years? At 10:08 AM 7/21/2017, Dan Asimov wrote:
Every field K — if it does not already contain all the roots of polynomials in K[x] — is a proper subfield of a larger field that is algebraically closed.
E.g., see <https://en.wikipedia.org/wiki/Finite_field#Algebraic_closure>.
I don't know how such an algebraic closure is usually visualized.
But I believe that, for any prime p, the Galois group of [the algebraic closure of the finite field F(p)] over F(p) is isomorphic to the same thing for any other prime.
—Dan
From: Henry Baker <hbaker1@pipelline.com> Jul 21, 2017 9:06 AM: -----
OK, if I extend the rationals with the root alpha of an irreducible polynomial p[x], I can plot alpha on the complex plane; indeed, I can plot *all* of the roots of p[x] on the complex plane. So all of these "extension roots" live in the complex plane.
Is there an analogous (single) place/field where all extension roots of GF(p) live -- i.e., a larger field which includes all of the extension fields of GF(p) ?
There seems to be a problem, since there are many (isomorphic) ways to extend GF(p); perhaps these are all different in this larger field?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Yes, you are technically correct. However, I was thinking about the naive freshman engineering class where we plot roots on the complex plane to determine -- e.g., the stability of some linear system. It's a lot shorter conversation if we plot them in the usual (uncountable) complex plane. Now that finite fields are so important in crypto and coding theory, where are these freshmen (& bright high schoolers) going to find their roots? At 08:31 AM 7/22/2017, Eugene Salamin via math-fun wrote:
The algebraic closure of the rationals consists of the algebraic numbers, and is a countable field.
The complex numbers are an uncountable field.
In between the two lie the algebraic closures of the various transcendental extensions of the rationals.
-- Gene
On Friday, July 21, 2017, 5:46:55 PM PDT, Henry Baker <hbaker1@pipeline.com> wrote:
Thanks, Dan.
The algebraic closure of the infinite sequence GF(p),GF(p^2),GF(p^3),... as described in the Wikipedia article is pretty ugly -- at least as described.
I was hoping for something as elegant as the complex numbers with their usual topology.
Perhaps I need to check back in about 300 years?
At 10:08 AM 7/21/2017, Dan Asimov wrote:
Every field K -- if it does not alreeady contain all the roots of polynomials in K[x] -- iss a proper subfield of a larger field that is algebraically closed.
E.g., see <https://en.wikipedia.org/wiki/Finite_field#Algebraic_closure>.
I don't know how such an algebraic closure is usually visualized.
But I believe that, for any prime p, the Galois group of [the algebraic closure of the finite field F(p)] over F(p) is isomorphic to the same thing for any other prime.
--Dan
From: Henry Baker <hbaker1@pipelline.com> Jul 21, 2017 9:06 AM: -----
OK, if I extend the rationals with the root alpha of an irreducible polynomial p[x], I can plot alpha on the complex plane; indeed, I can plot *all* of the roots of p[x] on the complex plane. So all of these "extension roots" live in the complex plane.
Is there an analogous (single) place/field where all extension roots of GF(p) live -- i.e., a larger field which includes all of the extension fields of GF(p) ?
There seems to be a problem, since there are many (isomorphic) ways to extend GF(p); perhaps these are all different in this larger field?
You might look at p-adic numbers. For example, sqrt(2) mod 7^N is (radix 7) +-...13 . Although some folks are uncomfortable about 2 being closer to 345 than to 9 or 51. I think you can get some milage mapping sqrt(6) (mod 7^N) to an "imaginary" 7-adic, adjoining i = sqrt(-1). You could also adjoin (rational) fractional powers to allow sqrt(7). I have no clue whether this is a complete representation system for algebraics (mod 7^N). The three cube roots of 1 seem straightforward, with sevens&units digits 01, 42, and 24. cbrt(2) (mod 7^N) looks like a problem. Even if we extend by all J^(1/K), we still face nested radicals, and non-radical algebraics. An interesting puzzle to see how far this can be pushed. Rich ----------- Quoting Henry Baker <hbaker1@pipeline.com>:
Yes, you are technically correct.
However, I was thinking about the naive freshman engineering class where we plot roots on the complex plane to determine -- e.g., the stability of some linear system. It's a lot shorter conversation if we plot them in the usual (uncountable) complex plane.
Now that finite fields are so important in crypto and coding theory, where are these freshmen (& bright high schoolers) going to find their roots?
At 08:31 AM 7/22/2017, Eugene Salamin via math-fun wrote:
The algebraic closure of the rationals consists of the algebraic numbers, and is a countable field.
The complex numbers are an uncountable field.
In between the two lie the algebraic closures of the various transcendental extensions of the rationals.
-- Gene
On Friday, July 21, 2017, 5:46:55 PM PDT, Henry Baker <hbaker1@pipeline.com> wrote:
Thanks, Dan.
The algebraic closure of the infinite sequence GF(p),GF(p^2),GF(p^3),... as described in the Wikipedia article is pretty ugly -- at least as described.
I was hoping for something as elegant as the complex numbers with their usual topology.
Perhaps I need to check back in about 300 years?
At 10:08 AM 7/21/2017, Dan Asimov wrote:
Every field K -- if it does not alreeady contain all the roots of polynomials in K[x] -- iss a proper subfield of a larger field that is algebraically closed.
E.g., see <https://en.wikipedia.org/wiki/Finite_field#Algebraic_closure>.
I don't know how such an algebraic closure is usually visualized.
But I believe that, for any prime p, the Galois group of [the algebraic closure of the finite field F(p)] over F(p) is isomorphic to the same thing for any other prime.
--Dan
From: Henry Baker <hbaker1@pipelline.com> Jul 21, 2017 9:06 AM: -----
OK, if I extend the rationals with the root alpha of an irreducible polynomial p[x], I can plot alpha on the complex plane; indeed, I can plot *all* of the roots of p[x] on the complex plane. So all of these "extension roots" live in the complex plane.
Is there an analogous (single) place/field where all extension roots of GF(p) live -- i.e., a larger field which includes all of the extension fields of GF(p) ?
There seems to be a problem, since there are many (isomorphic) ways to extend GF(p); perhaps these are all different in this larger field?
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Eugene Salamin -
Henry Baker -
Joerg Arndt -
rcs＠xmission.com