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Abstract Algebra/Field Theory question

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello PF!

    Let E be a splitting field of a separable polynomial over F. Define the Norm N: E-->F by:

    N(a) = the product of all q(a) where q is an element of the group Aut(E). I must show that this is a well defined mapping.

    2. Relevant equations


    3. The attempt at a solution
    So I must show that N(a) is an element of a for an arbitrary a in E. To be honest, It suprises me that N(a) would be an element of F, even if a is not an element of F. I mean, by taking the product of all the q(a) where q is an element of the group Aut(E) over F, it seems like it would be entirely possible to get an element that is not in F. Why is this not true? I mean q(a)*q1(a)*......*qr(a) would be in F even if a is not in F? But wouldn't each of qi(a) not be in F? how could the product be?

    Hoping somebody can shed a little light on this for me.
     
  2. jcsd
  3. May 10, 2016 #2

    fresh_42

    Staff: Mentor

    Well, it's been a while since I learned Galois theory, so please be lenient with me.
    Are there informations about the characteristic, can you use the fact that it is a simple extension? And with ##Aut(E)## do you mean ##Aut(E/F)##, i.e. those automorphisms which keep ##F## invariant? What do you know about the automorphisms?

    My textbook defines the norm as the determinant of a linear transformation, a left multiplication in the vector space ##E## over ##F## and shows that the norm of a zero of the minimal polynomial is (up to sign) the constant coefficient of the minimal polynomial, i.e. the product of all zeros. As determinate it is independent of the choice of the basis and thus well-defined. And since the elements of the matrix are all in ##F##, so is the norm.

    Perhaps this helps you a bit and you can build the bridge between the two definitions.
     
  4. May 12, 2016 #3
    Yes, the automorphisms which keep F invariant. Well, since it is the automorphism group of a separable polynomial, I believe that these automorphisms will take the minimal polynomial of any element of E to another root of the same polynomial (It was important for this to be a separable extension or we might not know this for sure I believe). We haven't talked about anything involving characteristics this semester in this course so I highly doubt that it's relevant to the question.

    Here is what seems weird to me: If this is a well defined mapping, then we can take an element in E that is not in F (let's call it a), and then take the product of all q(a) for every q in Aut(E) and apparently this element is in F (otherwise it wouldn't be a well defined mapping). This map is obviously well defined when a is in F, because q(a) = a for every q.... But yeah how could the map be well defined if a isn't in F?
     
  5. May 12, 2016 #4
    Ah man I appreciate your post, you are definitely using some linear algebra knowledge I never learned, determinants being independent of basis and basically every linear algebra thing you said haha, so I doubt that using that definition is necessary but it's still interesting and it's a route i'd like to further explore actually.

    When you said: "that the norm of a zero of the minimal polynomial is (up to sign) the constant coefficient of the minimal polynomial" this is true because it is a separable extension, so the minimal polynomial of each element must factor distintly and into linear factors, so wouldn't that mean that the minimal polynomial of each element in E has order 1? That can't be though... Oh gosh i am so confused haha
     
  6. May 13, 2016 #5

    fresh_42

    Staff: Mentor

    I think you have everything at hand, resp. said already. If the automorphisms map one root to another and you multiply all their images you will probably get the constant term of the minimal polynomial, which is in F. (Just multiply ##(x-id(a))(x-φ_1(a))(x-φ_2(a))....##).
     
  7. May 13, 2016 #6
    Oh wow, I understand now. Multiplying all the roots of the minimal polynomial together will give you the constant term! Thanks homey.
     
  8. May 13, 2016 #7
    I don't know if I should make a new thread, but I have a follow up question closely related to this so I think i'll just post here. What if instead of taking the product of all of the roots of the minimal polynomial we take the sum of them? Could we argue that if the minimal polynomial of a is of order n, then the coefficient on the n-1 term would be the sum of all the roots?
     
  9. May 13, 2016 #8

    fresh_42

    Staff: Mentor

    Yes, correct. It is called trace of the field extension or regular trace. (Again this comes from the same matrix as mentioned above, for the trace of a matrix is the sum of its diagonal elements. You may have a look on https://en.wikipedia.org/wiki/Vieta's_formulas. Perhaps you remember the 2-dimensional case of quadratic polynomials.)
     
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