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Math Help - Field extension to find roots of polynomial

  1. #1
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    Field extension to find roots of polynomial

    Hi, it's me goroner again, now i got some problems in finding field extension, i got to say that this task is not neither hw or anything else near that, it's just practice task..

    I will give my explanation too, and my idea how this could be solved but i can't solve it completely and i don't know if this is the right approach to do it, anyway in my classes the professor spent little time with field extension, and we have solved only 1 task as example and nothing more..

    The task: given the polynomial p(x) = b*x^2+b*x+b in G[x], is it reducible over G, if negative answer find the extension of the field so that this polynomial is reducible to linear factors (to have roots).

    G={a,b} following are the tables for op. + and *:

    + | a | b
    a | a | b
    b | b | a

    * | a | b |
    a | a | a |
    b | a | b |


    First G,+,* must be field so we can do the extension, and it is clearly deterministic that the table for + is somewhat the same like the one for Z2, + with the change of the names of 0 with a and 1 with b, the isomorphism is f = {a if 0, b if 1}, f is bijection since there's only 2 elements and they have unique images, f is homomorphism since : f(0+1) = f(0) + f(1), f(1+1) = f(1)+f(1), from above table this is true, and distributivity holds as seen in the table so f(1+0) = f(1)+f(0) also holds.
    So Z2 is abelian group and isomorphic to our group G,+ => G,+ is abelian group, in order for G,+,* to be field we must have G,* comutative with identity element and inverse for all a E G except for the zero in G for *, and last to check distributivity for * over +.Indeed G,* is comutative, there's identity and that is b, and since a is zero as seen in the table we don't require for it to have inverse, but for b there is inverse and b is self-inverse, since b is identity.Lastly distributivity holds it's not hard to check and i done it my self but i wont explain that here not to get too far=> G,*,+ is field, now we must find extension.

    We need extension cause p(x) has no solution in G, we need solution cause p(x) divided with (x-s) where s is the solution to p(x) , we get unique q(x) and r(x) such that the order of these new polynomials is less then that of p(x), and r(x) must be less in terms of order then (x-s), thus r(x) is constant, if you sub. all this=> p(x)=(x-s)*q(x) + r(x), now p(s)=0*q(x)+r(x) => r(x) = 0 , so =>
    p(x) = (x-s)*q(x), substituting a, and b from our field we cant find such solution to this p(x), so we must extend it and find the minimal extension that will include all the solutions to p(x) in this case there's two, and the new field call it S must have G as proper subfield and also all the elements of G under the operation in G with the new added elements from the extension again must be included in S thus S is field itself..

    First i thought of it this way:

    If i take c to be such element that is solution to the p(x), c must be included in S (the new field we are making) but also a*c , c*a, b*c, c*b, c*c must be defined in S and S must be closed under all this op. but there's problem this only can give me one solution, and i can't determine all the entries in the table for both the ops.

    if i use the formula for finding roots of quadratic polynomial there's again undeterminism, mainly: x1/2 = (-b+-Sqrt(b^2 - 4*b^2))/2*b, (x1/2 = x1 or x2, according to this the signs are used in the quition where says +-), now there's problem in the equation, first what is 4, it's not defined anywhere in my tables, i can however express it as 2*2, but that's in Z2, and here the formula is all mixed up from both my group and Z2, cause there's b in it as well, so it's impossible to use this formula to determine the roots, thus we must use the first approach.

    Let c be root for p(x) => b*c^2 +b*c + b = a, since a is identity in G,+, now because b is identity in favour of * then it's must be true that the above formula is same as c^2+c+b=a, now if i express c^2 = a - c - b => c^2 = a +(-c)+(-b), since a is zero in favour of + => c^2 = -c+(-b), but we have defined inverse for b in G,+, thus => c^2 = -c+b, or c^2 = b+(-c), the inverse of c is unknow yet.. nor the c*c product so in the field extension i must have {a,b, c,...} now i don't know how do go on from here to finish it all i don't have idea how to find the second root and make sure that my new field S the extension of G is closed, actually how to make it minimal and yet closed including the roots of p(x) under S[x]..

    Thanks for any replays!!
    Last edited by goroner; June 10th 2011 at 10:53 AM. Reason: operation tables were not readable
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  2. #2
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    Quote Originally Posted by goroner View Post
    Hi, it's me goroner again, now i got some problems in finding field extension, i got to say that this task is not neither hw or anything else near that, it's just practice task..

    I will give my explanation too, and my idea how this could be solved but i can't solve it completely and i don't know if this is the right approach to do it, anyway in my classes the professor spent little time with field extension, and we have solved only 1 task as example and nothing more..

    The task: given the polynomial p(x) = b*x^2+b*x+b in G[x], is it reducible over G, if negative answer find the extension of the field so that this polynomial is reducible to linear factors (to have roots).

    G={a,b} following are the tables for op. + and *:

    + | a | b
    a | a | b
    b | b | a

    * | a | b |
    a | a | a |
    b | a | b |


    First G,+,* must be field so we can do the extension, and it is clearly deterministic that the table for + is somewhat the same like the one for Z2, + with the change of the names of 0 with a and 1 with b, the isomorphism is f = {a if 0, b if 1}, f is bijection since there's only 2 elements and they have unique images, f is homomorphism since : f(0+1) = f(0) + f(1), f(1+1) = f(1)+f(1), from above table this is true, and distributivity holds as seen in the table so f(1+0) = f(1)+f(0) also holds.
    So Z2 is abelian group and isomorphic to our group G,+ => G,+ is abelian group, in order for G,+,* to be field we must have G,* comutative with identity element and inverse for all a E G except for the zero in G for *, and last to check distributivity for * over +.Indeed G,* is comutative, there's identity and that is b, and since a is zero as seen in the table we don't require for it to have inverse, but for b there is inverse and b is self-inverse, since b is identity.Lastly distributivity holds it's not hard to check and i done it my self but i wont explain that here not to get too far=> G,*,+ is field, now we must find extension.

    We need extension cause p(x) has no solution in G, we need solution cause p(x) divided with (x-s) where s is the solution to p(x) , we get unique q(x) and r(x) such that the order of these new polynomials is less then that of p(x), and r(x) must be less in terms of order then (x-s), thus r(x) is constant, if you sub. all this=> p(x)=(x-s)*q(x) + r(x), now p(s)=0*q(x)+r(x) => r(x) = 0 , so =>
    p(x) = (x-s)*q(x), substituting a, and b from our field we cant find such solution to this p(x), so we must extend it and find the minimal extension that will include all the solutions to p(x) in this case there's two, and the new field call it S must have G as proper subfield and also all the elements of G under the operation in G with the new added elements from the extension again must be included in S thus S is field itself..

    First i thought of it this way:

    If i take c to be such element that is solution to the p(x), c must be included in S (the new field we are making) but also a*c , c*a, b*c, c*b, c*c must be defined in S and S must be closed under all this op. but there's problem this only can give me one solution, and i can't determine all the entries in the table for both the ops.

    if i use the formula for finding roots of quadratic polynomial there's again undeterminism, mainly: x1/2 = (-b+-Sqrt(b^2 - 4*b^2))/2*b, (x1/2 = x1 or x2, according to this the signs are used in the quition where says +-), now there's problem in the equation, first what is 4, it's not defined anywhere in my tables, i can however express it as 2*2, but that's in Z2, and here the formula is all mixed up from both my group and Z2, cause there's b in it as well, so it's impossible to use this formula to determine the roots, thus we must use the first approach.

    Let c be root for p(x) => b*c^2 +b*c + b = a, since a is identity in G,+, now because b is identity in favour of * then it's must be true that the above formula is same as c^2+c+b=a, now if i express c^2 = a - c - b => c^2 = a +(-c)+(-b), since a is zero in favour of + => c^2 = -c+(-b), but we have defined inverse for b in G,+, thus => c^2 = -c+b, or c^2 = b+(-c), the inverse of c is unknow yet.. nor the c*c product so in the field extension i must have {a,b, c,...} now i don't know how do go on from here to finish it all i don't have idea how to find the second root and make sure that my new field S the extension of G is closed, actually how to make it minimal and yet closed including the roots of p(x) under S[x]..

    Thanks for any replays!!


    A piece of advice: write your questions in some few lines and focus it. It is highly unlikely that many people would read so much info (I know I didn't)

    Tonio
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  3. #3
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    I will try to write in few lines, i just like to explain my work and where i stuck my self =) , now assume G is field and read the last paragraph it's about the problem it self..
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  4. #4
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    since your field is isomorphic to Z2, you may as well replace "a" with "0" and "b" with "1".

    so you have a polynomial in Z2[x], x^2 + x + 1, and we want to find an extension of Z2 that contains the roots of this polynomial.

    well, note first of all, that x^2 + x + 1 does not factor in Z2[x], neither 0 nor 1 is a root: 0^2 + 0 + 1 = 1, and 1^2 + 1 + 1 = 1.

    this means that x^2 + x + 1 is irreducible over Z2, consequently, K = Z2[x]/<x^2 + x + 1> is a field.

    let α be the coset of x in this field: α = x + <x^2 + x + 1>. it should be clear that since α^2 = -α - 1 = α + 1,

    our new field has exactly 4 elements: 0, 1, α, α+1 (alternatively, you can view K as a vector space over Z2 with basis {1,α}).

    just for grins, let's calculate (x - α)(x - (α+1)) in K[x]:

    (x - α)(x - (α+1)) = x^2 - αx - (α+1)x + (α)(α+1) = x^2 - (α+α+1)x + (α^2 + α)

    now α+α+1 = α(1+1) + 1 = α(0) + 1 = 1, so we have:

    = x^2 - x + (α^2 + α) = x^2 + x + (α+1+α) = x^2 + x + 1 (since -x = (-1)x = 1x =x, because -1 = 1 in GF2).

    so x^2 + x + 1 factors completely in K[x], and thus has the two roots: α and α+1.
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  5. #5
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    Re: Field extension to find roots of polynomial

    Many thanks Deveno, i got it, i had similar example in the book and i just couldn't find out why a+1 is root as it was just stated that it is and it was shown that the polynomial factors with those two roots, but i didn't know how it got there, problem solved thanks to Deveno!
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