Results 1 to 9 of 9

Math Help - finite field

  1. #1
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1

    finite field

    I am going to ask a question this time. I have been trying to study some

    abstract algebra, and I must admit it is some wacky stuff.

    I respect NCA, PH, and others who excel at it.

    Here is the question:

    Show that {0,1,x,x^2} forms a field where 0 is the identity for addition and 1 is the multiplicative identity, and where 1+1=0, x^2=x+1.

    Huh?. 1+1=0?. That is completely counter-intuitive. Should I try to construct a group table showing a+b=? and ab=?.

    What is a good way to tackle something like this?. I want to get a handle on the basics before I try anything more advanced, that's for sure.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Quote Originally Posted by galactus View Post
    I am going to ask a question this time. I have been trying to study some

    abstract algebra, and I must admit it is some wacky stuff.

    I respect NCA, PH, and others who excel at it.

    Here is the question:

    Show that {0,1,x,x^2} forms a field where 0 is the identity for addition and 1 is the multiplicative identity, and where 1+1=0, x^2=x+1.

    Huh?. 1+1=0?. That is completely counter-intuitive. Should I try to construct a group table showing a+b=? and ab=?.

    What is a good way to tackle something like this?. I want to get a handle on the basics before I try anything more advanced, that's for sure.
    You could do the multiplication tables, it's the most intuitive way but it's tedious. I'd do it like this:

    Let A=\{ 0,1,x,x^2 \} and define f:A \rightarrow \mathbb{Z}_2 \times \mathbb{Z}_2 where f(0)=(0,0), f(1)=(1,1), f(x)=(1,0), f(x^2 )=(0,1) and \mathbb{Z}_2 \times \mathbb{Z}_2 is considered as an additive group (with operations coordinate-wise) and g:A- \{0 \} \rightarrow \mathbb{Z}_3 (this last considered as an additive group) and g(1)=0, g(x)=1, g(x^2 )=2 It's clear that f and g define bijections, and there is a theorem that says: If h:B \rightarrow G is a bijection where B is a set and G a group, both finite, then there is a unique function *:B \times B \rightarrow B such that (B,*) is a group isomorphic to G and makes h an isomorphism. Such operation is given by a*b=f^{-1}(f(a)f(b)), a^{-1}=f^{-1}(f(a)^{-1}) and e_B=f^{-1}(e_G). It is straightforward to show that this operation induced by f and g on A coincide with the sum and product you have already defined and as such are groups (abelian groups). Now you only have to check that the product distributes over the sum.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I feel embarrassed to ask this, but how would one build the table?. Even though, it is more tedious.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by galactus View Post
    I feel embarrassed to ask this, but how would one build the table?. Even though, it is more tedious.
    The addition table looks like this:

    \begin{array}{c|cccc}+&0&1&x&x^2\\ \hline0&0&1&x&x^2\\ 1&1&0&x^2&x\\ x&x&x^2&0&1\\x^2&x^2&x&1&0\end{array}

    (Notice that 1+x^2=1+(1+x) = 0+x = x. Also x+x = x(1+1) = x\cdot0 = 0; and x+x^2 = x + (x+1) = 0+1 = 1.)

    Do the multiplication table similarly, noticing that x\cdot x^2 = x(x+1) = x^2+x = 1 and x^2\cdot x^2 = (1+x)^2 = 1+x^2 = x.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Wow, thanks a lot, Opalq. I have always wanted to learn some algebra, but was afraid to put in the effort.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    One way of proving that it's a field is just noticing that it's \mathbb{F}_2[X]/(X^2+X+1). Since X^2+X+1 is an irreducible polynomial, the quotient ring \mathbb{F}_2[X]/(X^2+X+1) is a field.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Coincidentally, I was actually looking that irreducible polynomial up, Bruno.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    if you have a little bit background in field theory, then the easiest way would be to use this result that "the" finite field of order q=p^n, where p is a prime, is exactly the set of the roots of the

    polynomial f(t)=t^q - t \in \mathbb{F}_p[t]. then you'd only need to prove that the elements of your set are the roots of the polynomial f(t)=t^4 - t, viewed as an element of \mathbb{F}_2[t], which is very easy:

    obviously f(0)=f(1)=0. also, since x^2=x+1, we have x^4=x and thus f(x)=x^4-x=0. finally from x^4=x we get x^8=x^2 and thus f(x^2)=x^8 - x^2=0.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Thanks NCA, but I have no background in this stuff. That is why I am trying to learn a little. I figured you all would be the best place to start.

    I don't even know what F_{p}[t] stands for.

    I do not expect you to give me an entire lesson. Thanks for your input.

    I have a book...Abstract Algebra, 2nd edition by Hungerford.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Splitting Field of a Polynomial over a Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 1st 2011, 03:45 PM
  2. Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 23rd 2010, 02:19 AM
  3. Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 15th 2010, 02:25 AM
  4. Finite field
    Posted in the Advanced Algebra Forum
    Replies: 18
    Last Post: August 28th 2009, 07:26 PM
  5. Finite Field
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 9th 2009, 09:08 PM

Search Tags


/mathhelpforum @mathhelpforum