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Homework Help: Char(F)=p; poly NOT irreduc. => must have a root?

  1. May 12, 2010 #1

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    1. The problem statement, all variables and given/known data

    Let F have prime characteristic p and let a be in F. Show that the polynomial f(x)=x^p-a either splits or is irreducible in F[x].

    I was given a hit: "what can you say about all of the roots of f in a splitting field?"


    2. Relevant equations



    3. The attempt at a solution

    Suppose f has a root b. That is, b^p=a. Then using Freshman's dream, we have (x-b)^p=x^p+(-b)^p=x^p-a, and so it splits. So, it suffices to show that if that polynomial is NOT irreducible, then it must have a root. That's where I got stuck. The statement seems to be false for field of zero characteristic, so the fact that Char(F)=p should be somehow used.
     
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  3. May 12, 2010 #2

    Dick

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    Why don't you use the correct binomial expansion instead of the incorrect Freshman's dream?
     
  4. May 12, 2010 #3

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    Because it is indeed correct? In a field F of characteristic p for any two x and y in F we have (x+y)^p=x^p+y^p. This is to see using the mentioned binomial expansion and convincing yourself that p divides every binomial coefficient except the frist and the last. Thus, each term in the expansion will look like p*k*x^r*y^s, where k, r and s are some integers whose values are not important. Since for every x in the field we have px=0, all the terms in the expansion (except the first and the last) dissapear. The freshman's dream is a theorem, not a stupid mistake.
     
  5. May 12, 2010 #4

    Dick

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    Ok, so then what's your question? Doesn't that show x^p-a splits if it has a root, since your argument then implies that it has p roots? The Freshman's dream isn't true in characteristic 0.
     
  6. May 12, 2010 #5

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    If it has a root everything is clear. What is not clear is why the fact that x^p-a is NOT irreducible implies that it has a root.
     
  7. May 12, 2010 #6
    Assume for contradiction that x^p - a = p(x)q(x), where p and q are non-trivial and irreducible. From what you've said, you know what p(x) and q(x) must look like in the splitting field...can you arrive at a contradiction? It will help to assume p is odd (you can take care of char F = 2 separately).
     
  8. May 12, 2010 #7

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    First of all, it is not true that x^p-2 can be represented as a product of only two irreducible polynomials.
     
  9. May 12, 2010 #8

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    How about these arguments. All the roots of x^p-a are p-th roots of a.

    Suppose that x^p-2 is not irreducible and factorize in into irreducibles. All of the will be of degree >=2. Consider any of them. Look at the constant term of that factor.

    It MUST (BUT WHY?!) look like the p-th root of a to some power, say, k. This tells us that a^{k/p} is in the field. Since p is prime, there is a multiplicative inverse of k, say l, and raising to that power (since field is closed under products) we find that a^{1/p} is in the field. Done.
     
  10. May 12, 2010 #9
    Sorry, I misspoke... but I think my argument still works: write x^p - a as a product of irreducibles. In the splitting field each irreducible is of the form (x-b)^k for some k (where b is the root of x^p - a). Look at the irreducible of lowest degree. This will divide the other factors in the splitting field and therefore in F[x] (this is by uniqueness of Euclidean algorithm since both polys are in F[x]), contradicting that they were irreducible in F[x].
     
  11. May 12, 2010 #10

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    Why are you referring to a root of x^p-a as THE root? How do you know there is only one root? What if they behave like roots in the complex field?
     
  12. May 13, 2010 #11
    In your first post you observed that if b is a root of x^p - a (i.e. b^p = a), then (x-b)^p = x^p - b^p = x^p - a. So in the splitting field x^p - a breaks down as (x-b)^p which has only one root (with multiplicity p).
     
  13. May 13, 2010 #12
    Unfortunately I missed the last few weeks of my algebra class during which we discussed fields, so I can't help you directly. You mentioned that the constant term of each irreducible factor must be of the form a^{k/p}. I too do not see why this must be so. Did you read that elsewhere?
     
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