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Finding roots in an extension field

  1. May 16, 2016 #1
    1. The problem statement, all variables and given/known data
    Let q be a root of p(x) = x^3 + x^2 + 1 in an extention field of Z2 (integers modulus 2). Show that Z2(q) is a splitting field of p(x by finding the other roots of p(x)

    hint: this question can be greatly simplified by using the frobenius automorphism to find these zero's

    2. Relevant equations
    frobenius automorphism : f(a) = a^p

    3. The attempt at a solution
    So we are given one root and need to prove that this is enough to when adjoined to the original field to get the other two roots...

    so p(q) = 0
    and p(x) = (x-q)m(x) where m(x) is of degree 2

    I'm a little confused on how to use the frobenius automorphism here or even why it would help at all...
    anyone have any insight for me?
     
  2. jcsd
  3. May 17, 2016 #2

    fresh_42

    Staff: Mentor

    Have you tried to compute ##f(q)##? Is it a root again? You know ##q≠1##.
     
  4. May 17, 2016 #3
    warning: i use p(x) to express the polynomial i'm trying to find the roots of and also f(q) = q^p, where p is the prime that the frobenius automorphism maps to. Sorry for unnecessary confusion >.<...

    wouldn't q^p = q in Z mod 2? I mean, if q is odd then q = q^p = 1, if q is even then q = q^p = 0... perhaps these roots are not integers, i probably should have figured that >.< haha.

    Does the solution have to do with the fact that if we apply the frobenius autmorphism to p(x), it will fix the coefficients of each term?

    f(q) = q^p. I am not sure if this is a root or not. I suppose it must be for the reason I just stated, if i apply the frobenius morphism as so: f(0) = f(p(q)) = p(q^p), so yes, it appears that q^p is a root of p(x)... but how do we know that q^p doesn't equal q? Also, it's still a bit foggy to me how this frobenius morphism is going to help me prove the main question, i don't understand the connection very clearly. Thanks for being so patient with me!
     
  5. May 17, 2016 #4

    fresh_42

    Staff: Mentor

    Well, you have everything at hand, resp. said already. (again ...:smile:)

    Your ##p##'s are really confusing. Let us denote the prime by ##2## and the minimal polynomial ##p(x) = x^3 +x^2 + 1 \in ℤ_2[x]##.
    The good news is we don't have to bother signs and ##a+a=0## for any ##a##.

    You've told me last time that automorphisms map roots to roots.
    Since ##f(q) = q^2## is a root, we obstinately could divide ##p(x)## by ##(x+q)## and next by ##(x+q^2)## and find the third root. With ##p(q)=0## we can get rid of the higher potentials of ##q##. If there will be a remainder somewhere, then we must have done something wrong.

    Edit: the "again" above has been meant as: You know more than you think you know.
     
    Last edited: May 17, 2016
  6. May 17, 2016 #5
    Automorphisms map roots to roots, how interesting. No matter what polynomial in what subfield, the roots will be mapped to roots. Should this be obvious or are automorphisms amazing? I realize that automorphisms are all about preserving basically everything, but still, why is it obvious that a counter example to this would be impossible? Like i understand that automorphisms should always map roots to roots if it fixes every coefficient in the polynomial, but even if these coefficients arn't fixed it still maps roots to roots? o_O cool.
     
  7. May 17, 2016 #6

    fresh_42

    Staff: Mentor

    There's another hint: If ##f## maps roots to roots, what does ##f^2##? (I admit I stubborn calculated the third root and found out afterwards that it is ##f^2(q)##. Silly me.)
     
  8. May 17, 2016 #7

    fresh_42

    Staff: Mentor

    No, I don't think so. Please forget this at once. The Frobenius automorphism you mentioned isn't any automorphism! And you can easily check whether ##q^2## is a root of ##p(x)## by dividing ##p## by ##x+q^2##
    Edit: ... or easier whether ##(q^2)^3 + (q^2)^2 + 1 = 0##.
     
    Last edited: May 17, 2016
  9. May 19, 2016 #8

    fresh_42

    Staff: Mentor

    Let us be more careful here.

    If we have a field extension ##\mathbb{F} \subset \mathbb{E}##, a polynomial ## p(x) \in \mathbb{F}[x]## and an automorphism ##σ \in Aut(\mathbb{E} / \mathbb{F})##, i.e. ##σ(a) = a## for all ##a \in \mathbb{F}##, then for every zero ##z \in \mathbb{E}## of ##p(x)## we get: $$0 = p(z) ⇒ 0 = σ(0) = σ(p(z)) = p(σ(z))$$ So the crucial point here is the invariance of ##\mathbb{F}## under ##σ## that allows us the "root hopping" with automorphisms.
     
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