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[Abstract Algebra] Field and Polynomial Root problem

  1. Feb 20, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Suppose a field F has n elements and [itex]F=(a_1,a_2,...,a_n)[/itex]. Show that the polynomial [itex]w(x)=(x-a_1)(x-a_2)...(x-a_n)+1_F[/itex] has no roots in F, where [itex]1_f[/itex] denotes the multiplicative identity in F.

    2. Relevant equations


    3. The attempt at a solution

    Strategy: We have this polynomial:
    [itex]w(x)=(x-a_1)(x-a_2)...(x-a_n)+1_F[/itex]
    When we set it to 0 we see that the (x-a) terms need to equal -1 for w(x) to have a root.

    But here's my problem.
    I was thinking of trying to use induction on this proof.
    But if n = 0, then we have x+1 = 0 and x = i^2 which is in a field, the complex field.
    That has a root.

    But ok, let's say n = 1. Then we have x-a+1=0 in which case x = 3, a=4 would be a root.

    Must I assume that there is more then n=1 elements?
     
  2. jcsd
  3. Feb 20, 2016 #2

    fresh_42

    Staff: Mentor

    The point is that you already know all possible values for ##x##. What happens if you try them all one by one?
     
  4. Feb 20, 2016 #3

    RJLiberator

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    I'm not quite sure what you mean by this. We know all possible values of x? Sure, but couldn't that be from -infinity to +infinity? How does that help?
    There must be a deeper understanding to what you are alluding at, but my initial thoughts are that it is confusing. We know all possible values for x?
     
  5. Feb 20, 2016 #4

    fresh_42

    Staff: Mentor

    You are looking for a root of ##w(x)##, i.e. an element ##a ∈ \mathbf{F}## such that ##w(a) = 0##.
    Why don't you try all possible ##a## and see what ##w(a)## will get you? There are only finitely many of them, nothing with infinity.
     
  6. Feb 20, 2016 #5

    RJLiberator

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    Okay, I am starting to see where you are going with this.

    w(x) = (x-a)+1 = 0

    w(x)= (x-a1)(x-a2)+1 = 0
    [itex]x^2-x*a_2-x*a_1+a_1*a_2+1= 0[/itex]

    Couldn't we have a1 = -1 and a2 = 1 and x = 0 ?
     
  7. Feb 20, 2016 #6

    fresh_42

    Staff: Mentor

    I think you don't see the forest for all the trees. (Don't know whether this could be said in English, too.)
    Again: What is ##w(a)##? (Forget the ##x## for a second.)
     
  8. Feb 20, 2016 #7

    RJLiberator

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    w(a) = (a-a)+1 = 1.
    So clearly that cannot be a root. So if the product of the (x-a) terms are equal to 0, the +1 disallows it from being a root.

    This also works for more terms then just one. As (a-a) = 0, we see (a-a)(a-a1)(a-a2)....(a-an)+1 = 1.
     
  9. Feb 20, 2016 #8

    fresh_42

    Staff: Mentor

    Yes. ##w(a) = (a-a_1) ... (a-a_n) +1_F## but ##a \in \mathbf{F} = \{a_1, ... , a_n\}##. Therefore ##w(a) = 1_F## and ##1_F \neq 0_F## because ##\mathbf{F}## is a field.
     
  10. Feb 20, 2016 #9

    RJLiberator

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    Ugh, I can't believe I didn't see this obvious result from the start.

    I'm so out of it lately.

    That is completely obvious :p.

    Indeed.
     
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