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| Al2009... |
 Posted: Sun Nov 01, 2009 7:20 pm |
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Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting field of the polynomial x^8 - x over Z_2.
It seems like this field has a basis consisting of 3 elements; {1, alpha, beta} satisfying alpha^2=-1. If this is correct, how can I find the remaining beta? If this is wrong, what are the elements of the splitting field of the polynomial x^8 - x over Z_2
Thanks. |
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| Achava Nakhash, the Loving Snake... |
| Posted: Sun Nov 01, 2009 8:08 pm |
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On Nov 1, 9:20 pm, Al2009 <conundr... at (no spam) naver.com> wrote:
[quote]Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting field of the polynomial x^8 - x over Z_2.
It seems like this field has a basis consisting of 3 elements; {1, alpha, beta} satisfying alpha^2=-1. If this is correct, how can I find the remaining beta? If this is wrong, what are the elements of the splitting field of the polynomial x^8 - x over Z_2
Thanks.1
[/quote]
The splitting field of x^8 - x over Z_2 is, indeed, the finite field
with 8 elements. One way to completely describe it would be to factor
(my own personal rant here - I believe that factorize is not correct
English and entered the mathematical lexicon under the influence of
the French, much as I used to hear about compacity instead of
compactness and couldn't stand either - end of rant). Anyway,
x^8 - x = x(x+1)(x^3+x^2+1)(x^3+x^2+1)
The two cubic factors are obviously irreducible (cubic with no roots),
so the splitting field has a basis of 1, x, x^2 where x is chosen
satisfy your choice of the cubic equation, and the basis is over Z_2,
so the elements are 1, x, x+1, x^2, x^2+x, x^2+1, x^2+x+1, and 0.
Addition is defined as polynomial addition over Z_2 and multiplication
is defined as polynomial multiplication where we reduce the product to
have degree 2 or less by using the chosen equation to eliminate all
terms with exponent greater than or equal to 3. The results will be
isomorphic whichever of the 2 polynomials you use for the definition.
HTH,
Achava |
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| AP... |
| Posted: Mon Nov 02, 2009 3:34 am |
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On Mon, 02 Nov 2009 00:20:03 EST, Al2009 <conundrums at (no spam) naver.com> wrote:
[quote]Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting field of the polynomial x^8 - x over Z_2.
It seems like this field has a basis consisting of 3 elements; {1, alpha, beta} satisfying alpha^2=-1. If this is correct, how can I find the remaining beta? If this is wrong, what are the elements of the splitting field of the polynomial x^8 - x over Z_2
Thanks.
[/quote]
in fact if p is prime
for all r
there exist a finite field GP__k with k=p^r elements which are the
roots of X^r-X ; (GP like Galois Field)
it is Z/pZ[X]/Q(X)
with Q(X) in Z/pZ[X], degree r, irreducible
(a such poly Q exist) |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 7:06 am |
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On Nov 1, 11:20 pm, Al2009 <conundr... at (no spam) naver.com> wrote:
[quote]Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting field of the polynomial x^8 - x over Z_2.
It seems like this field has a basis consisting of 3 elements; {1, alpha, beta} satisfying alpha^2=-1.
[/quote]
That hardly describes it. And, no. It cannot possibly be that, since
over Z_2, -1=1, so you would have alpha=1; the polynomial x^2+1 is not
irreducible over Z_2, it factors as (x+1)^2.
[quote]If this is correct,
[/quote]
It's not.
[quote]how can I find the remaining beta? If this is wrong, what are the elements of the splitting field of the polynomial x^8 - x over Z_2
[/quote]
Since there is a unique field of order 8, you can start by finding an
irreducible polynomial of degree 2 over Z_2, and adjoin a root alpha.
Then find an irreducible polynomial of degree 2 over Z_2(alpha), and
adjoint a root beta; that will give you the field you want.
There aren't that many polynomials of degree 2 over Z_2: you just have
x^2, x^2+x, x^2+1, and x^2+x+1. The first three are not irreducible
(you missed the fact that (x+1)^2 = x^2+1 in Z_2[x]). So you really
have but one choice as far as an irreducible polynomial of degree 2
over Z_2, namely x^2+x+1. So your alpha should satisfy alpha^2 = alpha
+1. This gives you Z_2(alpha), a field with four elements.
Now, the possible monic polynomials of degree 2 over Z_2(alpha) are:
x^2,
x^2 + x, x^2+alpha*x, x^2+(alpha+1)x,
x^2+1, x^2+alpha, x^2+(alpha+1),
x^2+x+1, x^2+x+alpha, x^2+x+(alpha+1),
x^2+alpha*x+1, x^2+alpha*x + alpha, x^2+alpha*x+(alpha+1),
x^2+(1+alpha)x+1, x^2+(1+alpha)x+alpha, x^2+(1+alpha)x+(1+alpha).
So, find one that is irreducible, and use that. But be sure it is
irreducible!
--
Arturo Magidin |
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| Achava Nakhash, the Loving Snake... |
| Posted: Mon Nov 02, 2009 8:49 am |
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On Nov 2, 9:06 am, Arturo Magidin <magi... at (no spam) member.ams.org> wrote:
[quote]On Nov 1, 11:20 pm, Al2009 <conundr... at (no spam) naver.com> wrote:
Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting field of the polynomial x^8 - x over Z_2.
It seems like this field has a basis consisting of 3 elements; {1, alpha, beta} satisfying alpha^2=-1.
That hardly describes it. And, no. It cannot possibly be that, since
over Z_2, -1=1, so you would have alpha=1; the polynomial x^2+1 is not
irreducible over Z_2, it factors as (x+1)^2.
If this is correct,
It's not.
how can I find the remaining beta? If this is wrong, what are the elements of the splitting field of the polynomial x^8 - x over Z_2
Since there is a unique field of order 8, you can start by finding an
irreducible polynomial of degree 2 over Z_2, and adjoin a root alpha.
Then find an irreducible polynomial of degree 2 over Z_2(alpha), and
adjoint a root beta; that will give you the field you want.
There aren't that many polynomials of degree 2 over Z_2: you just have
x^2, x^2+x, x^2+1, and x^2+x+1. The first three are not irreducible
(you missed the fact that (x+1)^2 = x^2+1 in Z_2[x]). So you really
have but one choice as far as an irreducible polynomial of degree 2
over Z_2, namely x^2+x+1. So your alpha should satisfy alpha^2 = alpha
+1. This gives you Z_2(alpha), a field with four elements.
Now, the possible monic polynomials of degree 2 over Z_2(alpha) are:
x^2,
x^2 + x, x^2+alpha*x, x^2+(alpha+1)x,
x^2+1, x^2+alpha, x^2+(alpha+1),
x^2+x+1, x^2+x+alpha, x^2+x+(alpha+1),
x^2+alpha*x+1, x^2+alpha*x + alpha, x^2+alpha*x+(alpha+1),
x^2+(1+alpha)x+1, x^2+(1+alpha)x+alpha, x^2+(1+alpha)x+(1+alpha).
So, find one that is irreducible, and use that. But be sure it is
irreducible!
--
Arturo Magidin
[/quote]
Arturo, wouldn't that give you a field of order 16? Degree 2 over
degree 2 is degree 4. You need an irreducible polynomial of degree 3
over Z_2 to get a field of order 8, and I gave 2 such in my post last
night.
Regards,
Achava |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 9:27 am |
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On Nov 2, 12:49 pm, "Achava Nakhash, the Loving Snake"
<ach... at (no spam) hotmail.com> wrote:
[quote]Arturo, wouldn't that give you a field of order 16?
[/quote]
Sigh; yes, it would. You need something of degree 3, as you rightly
point out. So he should look at the polynomials of degree 3 over Z_2
(there's 8 of them) and pick an irreducible one.
Thanks for pointing out the rather dumb mistake...
--
Arturo Magidin |
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| Al2009... |
| Posted: Mon Nov 02, 2009 11:48 am |
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Thanks for all replies.
If I construct the field of order 8 without using a splitting field, how do I construct it?
First of all, Z_8 is not the candidate.
What about this one?
Z_2[alpha], where alpha^3 = 2.
Many thanks. |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 11:53 am |
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On Nov 2, 3:48 pm, Al2009 <conundr... at (no spam) naver.com> wrote:
[quote]Thanks for all replies.
If I construct the field of order 8 without using a splitting field, how do I construct it?
[/quote]
Find an irreducible cubic over Z_2 and adjoin a root.
[quote]First of all, Z_8 is not the candidate.
[/quote]
Of course not; it's not a field.
[quote]What about this one?
Z_2[alpha], where alpha^3 = 2.
[/quote]
What is the difference between 2 and 0, in Z_2?
In a field, if alpha^3 = 0, what is alpha?
It seems to me that you are pretty lost. Perhaps it is time to go
stake out your professor's office hour and get some one-on-one
direction...
--
Arturo Magidin |
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| Timothy Murphy... |
| Posted: Mon Nov 02, 2009 2:30 pm |
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Al2009 wrote:
[quote]Hi, I am trying to find a finite field of order 8.
Since it is a finite field of order 2^3, I need to find the splitting
field of the polynomial x^8 - x over Z_2.
[/quote]
I find that remark slightly odd.
If you are looking for a field of order 8
you only need to find an irreducible polynomial of degree 3,
which is not very difficult.
I don't see why you would want to get into the question of splitting fields.
--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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| Al2009... |
| Posted: Mon Nov 02, 2009 3:45 pm |
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[quote]On Nov 2, 3:48Â pm, Al2009 <conundr... at (no spam) naver.com
wrote:
Thanks for all replies.
If I construct the field of order 8 without using a
splitting field, how do I construct it?
Find an irreducible cubic over Z_2 and adjoin a root.
First of all, Z_8 is not the candidate.
Of course not; it's not a field.
What about this one?
Z_2[alpha], where alpha^3 = 2.
What is the difference between 2 and 0, in Z_2?
In a field, if alpha^3 = 0, what is alpha?
It seems to me that you are pretty lost. Perhaps it
is time to go
stake out your professor's office hour and get some
one-on-one
direction...
--
Arturo Magidin
[/quote]
Oops !
That was my mistake. I tried to find a degree 3 irreducible polynomials but I forgot I had a base field Z_2.
Anyhow, thanks for all your help. |
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| Al2009... |
| Posted: Mon Nov 02, 2009 4:09 pm |
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This is what I understand.
To construct a finite field of order p^n, where p is a prime number, we can find a splitting field of x^q -x over Z_p, where q=p^n.
Or,
we can find a degree n irreducible polynomial over Z_p. So our updated basis for this field has the following form,
{1, alpha, alpha^2, ...., alpha^(n-1)},
where "alpha" is the root of irreducible degree-n polynomial over Z_p.
Furthermore, those two constructions produce isomorphic fields.
Umm, what do you think?
Thanks a lot. |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 5:49 pm |
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On Nov 2, 8:09 pm, Al2009 <conundr... at (no spam) naver.com> wrote:
[quote]This is what I understand.
To construct a finite field of order p^n, where p is a prime number, we can find a splitting field of x^q -x over Z_p, where q=p^n.
Or,
we can find a degree n irreducible polynomial over Z_p. So our updated basis for this field has the following form,
{1, alpha, alpha^2, ...., alpha^(n-1)},
where "alpha" is the root of irreducible degree-n polynomial over Z_p.
[/quote]
Alpha is *a* root of *that* irreducible degree n polynomial.
[quote]Furthermore, those two constructions produce isomorphic fields.
Umm, what do you think?
[/quote]
About what?
--
Arturo Magidin |
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| Timothy Murphy... |
| Posted: Tue Nov 03, 2009 8:39 am |
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Al2009 wrote:
[quote]If I construct the field of order 8 without using a splitting field, how
do I construct it?
[/quote]
Find an irreducible polynomial f(x) in F_2[x], eg f(x) = x^3 + x + 1.
Then the field is F_2[x]/f(x), ie its elements are
the polynomials ax^2 + bx + c,
and these are added or multiplied modulo f(x).
--
Timothy Murphy
e-mail: gayleard /at/ eircom.net
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland |
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