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Polynomial Existence and Irreducibility over Rational #'s

  1. Feb 14, 2016 #1

    RJLiberator

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

    Prove that for all n ≥ 1, there exists a polynomial f(x) ∈ ℚ[x] with deg(f(x)) = n such that f(x) is irreducible in ℚ[x].
    2. Relevant equations
    In mathematics, a rational number is any number that can be expressed as the quotient or fractionp/q of two integers, a numeratorp and a non-zero denominatorq


    f(x) is called irreducible if it is not reduicble.
    f(x) is reducible if there exists g(x)*h(x) such that f(x) = g(x)*h(x) and deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x)).

    http://mathworld.wolfram.com/IrreduciblePolynomial.html


    3. The attempt at a solution

    Is it possible to use the quadratic discriminant here to show when no roots exist?

    b^2 - 4ac ≥ 0
    So we say whenever b^2-4ac < 0 then f(x) is irreducible.

    And it would seem clear that you could always make a=c and a^2 > b^2....
    Ouch. Nvm, this doesn't work for higher degree terms actually.

    In that case, hm.

    Could we work with cases, possibly induction?

    Case n = 1. Then by definition f(x) is irreducible.
    Induction step: n+1 can be irreducible.
    Then no, this is not working.


    Is the division algorithm necessary?

    Theorem 7 Let f be a polynomial with rational coefficients in lowest terms so that the numerators of the coefficients are relatively prime (i.e., have no common prime factor). Let d be the least common multiple (lcm) of the denominators of the coefficients of f. Let g = df. Then g has integer coef- ficients. Furthermore, f is irreducible over the rationals if and only if g is irreducible over the integers

    Am I going in the right direction by observing the division algorithm?
     
  2. jcsd
  3. Feb 14, 2016 #2

    fresh_42

    Staff: Mentor

    Have you ever thought about unity roots or just powers of 2, e.g.?
     
    Last edited: Feb 14, 2016
  4. Feb 14, 2016 #3

    RJLiberator

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    Well, with powers of two we use the quadratic discriminant and see that b^2 - 4ac >= 0.

    So we have that when b^2-4ac < 0 we have an irreducible polynomial.
     
  5. Feb 14, 2016 #4

    fresh_42

    Staff: Mentor

    This special discriminate is only for 2nd degree polynomials.
    You only have to point out one irreducible polynomial for each degree.
    So taking algebraic roots that are not in ℚ will do the job, e.g. roots of ##2^n##, e.g.
    Even the roots of ##1^n## will be probably all but at most two outside of ℚ, which has to be proved.
    (Att.: the all but 1 or 2 fact reduces the degree of the searched polynomials!)
     
  6. Feb 15, 2016 #5

    RJLiberator

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    Hm. I understand your point that the discriminant is only for second degree polynomials.

    So I only have to point out one irreducible polynomial for each degree, I agree.

    Algebraic roots that are not in Q will do the job, roots of 2^n. I'm not really following what you mean here.

    I'm struggling to understand this hint, I can understand it is important.
     
  7. Feb 15, 2016 #6

    fresh_42

    Staff: Mentor

    Sorry, I have been too sloppy on that. I meant ##x^n -2 = 0## or ##x^{n-1}-1=0##.
    This delivers ##n## roots, and with Euler's Formula it is easy to show that these roots are not in ℚ. However, ##±1## might be roots of the latter polynomial which are in ℚ! So you have to divide the polynomial by ##x±1## to get the irreducible part. But this irreducible part will then of course be only of degree ##n-1## or ##n-2##.
     
  8. Feb 15, 2016 #7

    RJLiberator

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    Okay, so what we are saying is:

    Take an example polynomial x^n - 2 = 0
    and show that it delivers n roots, say with Euler's Formula.
    Then show that these roots are NOT in Q.
     
  9. Feb 15, 2016 #8

    fresh_42

    Staff: Mentor

    That was the plan, yes.
     
  10. Feb 15, 2016 #9

    RJLiberator

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    So let's try this:

    Observe: x^n-2 = 0.
    x^n = 2
    x = nthroot(2)

    Therefore, for every n this is not possibly in Q.

    We observe now x^1 is irreducible by definition.
    So now, I've shown that there exists a reducible polynomial for every degree n in Q[x].
     
  11. Feb 15, 2016 #10

    Dick

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    You do know that there are polynomials with no roots in Q which are NOT irreducible, yes? Give me an example.
     
  12. Feb 15, 2016 #11

    RJLiberator

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    x^2+1 = 0

    I think I solved this problem in class today. So help is no longer needed, it was the more exhausting proof of my latest homework assignment.
     
  13. Feb 15, 2016 #12

    Dick

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    Ok, but how did you prove it? x^2+1 IS irreducible over Q and has no roots. (x^2+1)(x^2+1) also has no roots. Is it irreducible?
     
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