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Irreducible polynomials over ring of integers

  1. Nov 19, 2011 #1
    Is it true that polynomials of the form :

    [itex] f_n= x^n+x^{n-1}+\cdots+x^{k+1}+ax^k+ax^{k-1}+\cdots+a[/itex]

    where [itex]\gcd(n+1,k+1)=1[/itex] , [itex] a\in \mathbb{Z^{+}}[/itex] , [itex]a[/itex] is odd number , [itex]a>1[/itex], and [itex]a_1\neq 1[/itex]

    are irreducible over the ring of integers [itex]\mathbb{Z}[/itex]?

    http://en.wikipedia.org/wiki/Eisenstein%27s_criterion" [Broken] cannot be applied to the polynomials of this form.

    Example :

    The polynomial [itex]x^4+x^3+x^2+3x+3[/itex] is irreducible over the integers but none of the criteria above can be applied on this polynomial.

    Thanks in advance...
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 26, 2011 #2
    I hope this helps.

    The polynomial x4+x3+x2+x +1 is the cyclotomic polynomial [itex]\Phi[/itex]5(x). I believe the general polynomial that you described is also the monic cyclotomic polynomial [itex]\Phi[/itex]n(x) in Z[x] with degree [itex]\varphi[/itex](n). These polynomials are irreducible. The proof is a little tedious and not exactly immediate. Check out this paper


    I think it does a pretty good job. Then again, if that general polynomial turns out to not be cyclotomic, I'll have to start over and come up with something new
  4. Nov 26, 2011 #3
    These polynomials are not cyclotomic polynomials.
    [itex]f_n[/itex] can be rewritten into form :

    [itex]f_n=\displaystyle \sum_{i=0}^n x^{i}+(a-1)\cdot \displaystyle \sum_{i=0}^k x^{i} [/itex] ,or

    [itex]f_n=\frac{x^{n+1}+(a-1)x^{k+1}-a}{x-1} [/itex]
  5. Nov 26, 2011 #4
    Maybe if you give a different example it would be easier to see. I don't really see much of a correlation between the general case and the example you gave. Anyways, you can always fall back on the general technique for finding irreduciblity.

    Let I be a proper ideal in the integral domain R and let p(x) be the monic polynomial in R[x]. If the image of p(x) in (R/I)[x] cannot be factored in (R/I)[x] into two polynomials of smaller dgree, then p(x) is irreducible in R[x].

    Start reducing mod some n and see where that gets you. If it's small degree then it's pretty obvious. A higher degree will probably take a little work. Unfortunately, this technique doesn't always work, but I can't think of any other irreducibility criteria.
  6. Nov 26, 2011 #5
    Wait I just thought of another one: Hilbert's Irreducibility Theorem.
  7. Nov 26, 2011 #6
    Also Kronecker's Method
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