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Math Help - elements in extension

  1. #1
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    elements in extension

    g = X^3 + X + 1

    Let E be the extension of F2 with a root alpha of g. Write down a complete list of elements of E without repetitions.

    I've tried by best at this question, but I still don't feel that I fully understand what I'm doing, (which will probably be proved with it being wrong...)

    We know is a route of g, so it follows that is also a root of g. We can apply this repeatedly to find roots and so on.


    Hence
    Hence

    Hence the roots are

    Anyone know if this is OK?

    Thanks in advance
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  2. #2
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    Hello,

    Hm, I don't think it is what you've been asked.

    If there is a root alpha of g, then any \alpha^3 can be transformed in \alpha+1.

    This means that any element of E won't contain any \alpha^3 (or highest powers).

    Actually, the elements of E will be all polynomials of degree < 3, which is the degree of g(x).
    So it's 0, 1, x, x+1, x, x+x, x+1, x+x+1.

    The transition x <-> \alpha can be done because \alpha is an equivalent class of x (sort of).

    I don't know if it is clear

    As I told you in another thread, this is like working in R[X]/(g(x)), which is actually the set of all possible remainders of the division of any polynomial by g(x)...

    Perhaps Isomorphism will be able to explain it better
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  3. #3
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    Thank you Moo Sorry to make you repeat yourself I'm just really really bad at understanding this stuff...I'm much more of a calculus person

    Ok the the next question on this exam paper is:

    Find the other roots of g and the roots of h in E.

    This is the question to what my answer before was to right?! so with my above answer for g and answer below for h is this ok?

    The roots of h are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + 1 and \alpha^2 + \alpha
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  4. #4
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    Quote Originally Posted by hunkydory19 View Post
    g = X^3 + X + 1
    Let E be the extension of F2 with a root alpha of g. Write down a complete list of elements of E without repetitions.
    What about:

    \alpha^4 = \alpha \cdot \alpha^3 = \alpha^2 + \alpha

    \alpha^5 = \alpha \cdot \alpha^4 = \alpha^3 + \alpha^2 = (1+\alpha)+\alpha^2

    \alpha^6 = \alpha \cdot \alpha^5 = \alpha+\alpha^2+\alpha^3 = \alpha+\alpha^2+\alpha + 1 = 1 + \alpha^2

    And so on....
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  5. #5
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    Quote Originally Posted by hunkydory19 View Post
    I've tried by best at this question, but I still don't feel that I fully understand what I'm doing, (which will probably be proved with it being wrong...)

    We know is a route of g, so it follows that is also a root of g.
    You are right up to here.

    Now if \alpha is a root, then \alpha^2,\alpha^4,\alpha^8.... are all roots of g. Why should \alpha^3 be a root?

    Quote Originally Posted by hunkydory19 View Post
    Thank you Moo Sorry to make you repeat yourself I'm just really really bad at understanding this stuff...I'm much more of a calculus person

    Ok the the next question on this exam paper is:

    Find the other roots of g and the roots of h in E.

    This is the question to what my answer before was to right?! so with my above answer for g and answer below for h is this ok?

    The roots of h are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + 1 and \alpha^2 + \alpha

    What is h?
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  6. #6
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    Thanks Iso, right I've had another go...

    So sorry for missing out h by the way, so stupid!

    h =  X^3 + X^2 + 1

    The roots of g are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha

    The roots of h are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1

    This any better? Please say yes!
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  7. #7
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    Quote Originally Posted by hunkydory19 View Post
    Thanks Iso, right I've had another go...

    So sorry for missing out h by the way, so stupid!

    h =  X^3 + X^2 + 1

    The roots of g are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha

    The roots of h are: \alpha, \alpha^2, \alpha + 1, \alpha^2 + \alpha, \alpha^2 + \alpha + 1

    This any better? Please say yes!
    Unfortunately no...

    The roots of g are: \alpha, \alpha^2, \alpha^2 + \alpha
    Why did you add \alpha^3?

    If the elements are from \mathbb{F}_q with characteristic 'p', then the roots are \{\alpha, \alpha^p, \alpha^{p^2},\alpha^{p^3}....\}.

    Since the field we are considering is \mathbb{F}_{2^3} and it has a characteristic 2, the roots of g are \{\alpha, \alpha^2, \alpha^{2^2},\alpha^{2^3}....\}. But observe that since the field has 8 elements, \alpha^8 = \alpha(Why?). So the roots start repeating and thus we truly have only three roots.
    They are \{\alpha, \alpha^2, \alpha^{4}\}

    For h, find one root in \mathbb{F}_{2^3} using the previous exercise of constructing roots. Lets call it \beta. Then the remaining roots are \beta^2,\beta^4

    Understood?

    P.S: A small hint to see if you are going wrong: A cubic can have only 3 roots and you are getting four roots
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  8. #8
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    So for h, \beta is a root so it follows that \beta^2 is also a root.

    Using  \beta^3 = \beta^2 + 1

     \beta^4 = \beta.\beta^3 = \beta(\beta^2 + 1) = \beta^3 + \beta = \beta^2 +\beta + 1

     \beta^8 = (\beta^2 + \beta + 1)^2 = \beta^4 + 1 = \beta(\beta^2 + 1) + 1 = \beta^3 + \beta + 1 = \beta^2 + \beta

    Why am I getting this fourth root? I actually think I will die doing this
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  9. #9
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    Theorem: If p(x) is irreducible n-degree polynomial over F and \alpha is a zero in an extension field then F(\alpha) = \{ a_0+a_1\alpha+...+a_{n-1}\alpha^{n-1} \}.

    This means given F=\mathbb{F}_2 and \alpha solving the irreducible polynomial x^3+x+1 it means \mathbb{F}_2(\alpha) = \{ a+b\alpha+c\alpha^2+d\alpha^3\} where a,b,c,d\in \mathbb{F}_2.

    So can you now list all the elements?
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  10. #10
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    Quote Originally Posted by hunkydory19 View Post
    So for h, \beta is a root so it follows that \beta^2 is also a root.

    Using  \beta^3 = \beta^2 + 1

     \beta^4 = \beta.\beta^3 = \beta(\beta^2 + 1) = \beta^3 + \beta = \beta^2 +\beta + 1

     \beta^8 = (\beta^2 + \beta + 1)^2 = \beta^4 + 1 = \beta(\beta^2 + 1) + 1 = \beta^3 + \beta + 1 = \beta^2 + \beta

    Why am I getting this fourth root? I actually think I will die doing this
    Ok lets go over this carefully

    g(x) = x^3 + x + 1 and g(\alpha) = 0

    The elements of \mathbb{F}_{2^3} are:
    0, \alpha,\alpha^2,\alpha^3(= \alpha + 1),\alpha^4(= \alpha^2 + \alpha),\alpha^5(= \alpha^2 + \alpha + 1),\alpha^6(= \alpha^2 + 1),\alpha^7(= 1)
    So,
    \mathbb{F}_{2^3} = \{0,1,\alpha,\alpha^2,\alpha + 1,\alpha^2 + \alpha ,\alpha^2 + \alpha + 1,\alpha^2 + 1\}

    Now consider h(x) = x^3 + x^2 + 1.
    Dont try a new \beta, but try one of the above elements. We clearly see that h(\alpha^3) = (\alpha^3)^3 + (\alpha^3)^2 + 1 = \alpha^2 + (\alpha^2 + 1) + 1 = 0

    So \alpha^3 is a root of h(x). But this means (\alpha^3)^2 and (\alpha^3)^4 = \alpha^{12} = \alpha^5 are also roots of h(x).

    Thus the \alpha^3, \alpha^6 ,\alpha^5 are the roots of h(x)

    P.S:There ...I actually solved this problem completely for you
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  11. #11
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    Well ring a ding dong its finally hit home and I get it now! 200 years later!

    I didn't know you could actually stick elements in to see if they gave 0...

    Thank you so much for being SO patient with me, I know I wouldn't have been if it was the other way round!
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  12. #12
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    Quote Originally Posted by hunkydory19 View Post
    Well ring a ding dong its finally hit home and I get it now! 200 years later!

    I didn't know you could actually stick elements in to see if they gave 0...

    Thank you so much for being SO patient with me, I know I wouldn't have been if it was the other way round!
    There is a trick to it

    Since \alpha, \alpha^2, \alpha^4 were already roots of g(x), I figured h should have the others and tried \alpha^3
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