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Irreducibility of Polynomial

  1. Dec 5, 2012 #1
    Hi. There is a polynomial f = (x^3) + 2(x^2) + 1, f belongs to Q[x]. It will be shown that the polynomial is irreducible by contradiction. If it is reducible, (degree here is three) it must have a root in Q, of the form r/s where (r,s) = 1. Plugging in r/s for variable x will resolve to
    r^3 + 2(r^2)s + (s^3) = 0


    I don't understand the next part of the solution. Why introduce this prime number that divides s.
    --
    Suppose a prime number p divides s.
    This implies p divides 2(r^2)s + (s^2)
    Or, this implies also the above equals 2(r^2)pa + (p^3)(a^3) = p(2(r^2)a + (p^2)(a^3)) which implies p divides (r^3) which also means p divides r which is contradiction because (r,s) = 1
    ---
    So I follow the above steps, but what does prime number p dividing s or r have anything to do with this?


    Next part of solution.
    --> If s = 1, (r^3) + 2(r^2) + 1 = 0 means r((r^2) + 2r) = -1 implies r = 1 or -1, but 1 and -1 are not roots, seen by evaluating.
    Same argument for s = -1.
    This means that r/s is not a root, so not irreducible in Q[x]
    ---
    So I'm lost about this step too. Only thing that I think is that if you let s = 1 or -1, it's as if you are trying to find something about a root in Z...1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Dec 5, 2012 #2

    Dick

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    You can easily show it's irreducible by using the rational roots theorem. http://en.wikipedia.org/wiki/Rational_root_theorem What they doing here is not using the theorem but working through the proof of the theorem for this specific polynomial. Either s has a prime factor which contradicts (r,s)=1 or s=1 or -1. That doesn't work either.
     
  4. Dec 5, 2012 #3
    This is basically a proof by cases for a reduced rational number r/s

    Case 1, s is not a unit in Z (i.e. s is not 1 or -1): Then by prime factorization in Z, there is a prime p which divides s. Yada yada. This contradicts r and s coprime. It sounds like you understand the yada yada part. So Case 1 doesn't work.

    Case 2, s is a unit in Z (i.e. s is either 1 or -1): Then r must also be a unit in Z (do you understand why this must be true?). However neither unit works. So we can't be in Case 2 either.

    Case 1 and Case 2 exhaust all possible cases for rational roots. Neither works. So the root(s) must be irrational.
     
  5. Dec 6, 2012 #4
    I should have mentioned, the example was given as a warm up before learning the rational root test, so I can't make use of it.

    When I have r^3 + 2(r^2)s + (s^3) = 0, if I just show that it also equals
    s(2(r^2) + (s^2)) = -(r^3), isn't this enough to show that s divides r so (s,r) != 1 . So now I eliminated rational numbers with GCD = 1 , ie all of them?
     
  6. Dec 6, 2012 #5

    Dick

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    No, you can't say s*n=r^3 implies that s divides r. 4*2=2^3 but 4 doesn't divide 2. That's exactly why they picked the prime. If p*n=r^3 and p is prime then it's correct to say p divides r.
     
  7. Dec 6, 2012 #6
    omg'z ....

    I see. thank you thank you thank you.
     
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